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<?xml version="1.0" encoding="utf-8"?>
<search>
<entry>
<title>修改wsl下绿色背景的文件夹颜色</title>
<link href="/2021/12/12/ckx3e5iv30007nce2im6u169q/"/>
<url>/2021/12/12/ckx3e5iv30007nce2im6u169q/</url>
<content type="html"><![CDATA[<h5 id="修改使用ls时文件夹绿色底色的问题："><a href="#修改使用ls时文件夹绿色底色的问题：" class="headerlink" title="修改使用ls时文件夹绿色底色的问题："></a>修改使用ls时文件夹绿色底色的问题：</h5><p>个人使用的是zsh终端<br>使用<code>ls</code>命令查看时有某种权限的文件夹会显示非常别眼的绿色，像下面的blog这样<br><img src="https://lwh-graph-bed.oss-cn-hangzhou.aliyuncs.com/pic/492495522211253.png" alt="例子"></p><p><strong>修改：</strong><br>使用命令：<code>dircolors -p > ~/.dircolors</code>生成配置文件，用vim打开并修改<code>OTHER_WRITABLE</code>部分。<br>修改为：<code>OTHER_WRITABLE 30;46</code>，保存退出。（当然你也可以选择自己喜欢的颜色）<br>在终端运行以下命令生效：<code>eval "$(dircolors ~/.dircolors)";</code>。</p><p>参考于StackOverflow的这个问题：<a href="https://unix.stackexchange.com/questions/94498/what-causes-this-green-background-in-ls-output" target="_blank" rel="noopener">https://unix.stackexchange.com/questions/94498/what-causes-this-green-background-in-ls-output</a></p><blockquote><p>ps：需要注意的是，如果改变了配色主题：<code>~/.oh-my-zsh/themes</code>下的主题在<code>~/.zshrc</code>中改变，<br>要删除<code>.dircolors</code>文件并重复上面的步骤。</p></blockquote><p>最后的效果图是这样（我选择的主题是<code>fletcherm</code>，另外<code>dpoggi</code>也不错）<br><img src="https://lwh-graph-bed.oss-cn-hangzhou.aliyuncs.com/pic/467961423229679.png" alt="例子"></p>]]></content>
</entry>
<entry>
<title>win+wsl2+terminal+nvim(NvChad)教程</title>
<link href="/2021/12/08/ckx3e5iv50009nce2v1y82cc3/"/>
<url>/2021/12/08/ckx3e5iv50009nce2v1y82cc3/</url>
<content type="html"><![CDATA[<h5 id="一-win-terminal-wsl升级wsl2"><a href="#一-win-terminal-wsl升级wsl2" class="headerlink" title="一. win+terminal+wsl升级wsl2"></a>一. win+terminal+wsl升级wsl2</h5><ol><li><p>前往系统设置-应用-程序和功能-启用或关闭Windows功能，勾选上述目标项，确认后重启电脑<br><img src="https://lwh-graph-bed.oss-cn-hangzhou.aliyuncs.com/pic/265244611211249.png" alt="Windows功能"></p></li><li><p>Microsoft story下载Windows terminal和Ubuntu18.04 LTS</p></li><li><p>打开terminal在Powershell中，输入如下命令：<code>wsl --set-default-version 2</code>，查看版本：<code>wsl -l -v</code>；输出的<code>VERSION</code>下面是2就成功了。</p></li><li><p>如果没有生效，需要安装wsl2：<a href="https://wslstorestorage.blob.core.windows.net/wslblob/wsl_update_x64.msi" target="_blank" rel="noopener">wsl_update_x64</a><br>安装后在terminal输入：<code>wsl --set-version Ubuntu-18.04 2</code>，然后继续输入命令将WSL 2设置为默认版本<code>wsl --set-default-version 2</code>，重新查看版本。</p></li></ol><a id="more"></a><h5 id="二-美化终端界面（zsh）"><a href="#二-美化终端界面（zsh）" class="headerlink" title="二. 美化终端界面（zsh）"></a>二. 美化终端界面（zsh）</h5><p>First：安装Ubuntu后先换源（阿里源）再更新，我用的这个目前没一点问题<br>把前面的全部注释掉后面接这个<br><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">#========================================================================</span></span><br><span class="line">deb http://mirrors.aliyun.com/ubuntu/ bionic main restricted universe multiverse</span><br><span class="line">deb-src http://mirrors.aliyun.com/ubuntu/ bionic main restricted universe multiverse</span><br><span class="line"></span><br><span class="line">deb http://mirrors.aliyun.com/ubuntu/ bionic-security main restricted universe multiverse</span><br><span class="line">deb-src http://mirrors.aliyun.com/ubuntu/ bionic-security main restricted universe multiverse</span><br><span class="line"></span><br><span class="line">deb http://mirrors.aliyun.com/ubuntu/ bionic-updates main restricted universe multiverse</span><br><span class="line">deb-src http://mirrors.aliyun.com/ubuntu/ bionic-updates main restricted universe multiverse</span><br><span class="line"></span><br><span class="line">deb http://mirrors.aliyun.com/ubuntu/ bionic-proposed main restricted universe multiverse</span><br><span class="line">deb-src http://mirrors.aliyun.com/ubuntu/ bionic-proposed main restricted universe multiverse</span><br><span class="line"></span><br><span class="line">deb http://mirrors.aliyun.com/ubuntu/ bionic-backports main restricted universe multiverse</span><br><span class="line">deb-src http://mirrors.aliyun.com/ubuntu/ bionic-backports main restricted universe multiverse</span><br></pre></td></tr></table></figure></p><p>Next：<br><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">sudo apt update</span><br><span class="line">sudo apt-get upgrade</span><br></pre></td></tr></table></figure></p><p>要知道现在有两个系统：powershell和Ubuntu，所以要两个都要安装zsh。</p><blockquote><p>有很多教程，详细的可以自行搜索，像这个：<a href="https://zhuanlan.zhihu.com/p/273237897" target="_blank" rel="noopener">https://zhuanlan.zhihu.com/p/273237897</a><br>或者试着搜索：<code>Windows10上开启WSL2(Windows Subsystem for Linux 2)</code>这类的就行</p></blockquote><p>刚打开Ubuntu后，路径是在<code>mnt/c/Users/<你的账号名></code>，我不知道怎么让它一打开就在<code>~</code>目录，所以每次都要自己cd过去。<br>不过文章里修改机器名称那里，即命令行前缀部分，适用于Ubuntu的方法不适合wsl，直接在设置部分修改自己的电脑名称即可，见：<a href="https://blog.csdn.net/lianghecai52171314/article/details/113064742" target="_blank" rel="noopener">https://blog.csdn.net/lianghecai52171314/article/details/113064742</a><br>另外美化后使用zsh会使pip3失效，所以需要：<code>sudo apt-get install --reinstall python3-pip</code>；再升级pip：<code>pip3 install --upgrade pip</code>；用<code>pip3 -V</code>查看版本。</p><p>ok，美化结束。</p><h5 id="三-nvim前置环境"><a href="#三-nvim前置环境" class="headerlink" title="三. nvim前置环境"></a>三. nvim前置环境</h5><ol><li><strong>安装管理nvim插件的插件：</strong></li></ol><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">git <span class="built_in">clone</span> --depth 1 https://github.com/wbthomason/packer.nvim\</span><br><span class="line"> ~/.<span class="built_in">local</span>/share/nvim/site/pack/packer/start/packer.nvim</span><br></pre></td></tr></table></figure><ol><li><strong>安装node.js</strong></li></ol><p>打开<a href="http://nodejs.cn/download/" target="_blank" rel="noopener">nodejs中文网</a>里的下载 找到64位的文件（下面例子里的版本比较老，不要直接复制）<br>下载：<code>wget https://npm.taobao.org/mirrors/node/v12.14.1/node-v12.14.1-linux-x64.tar.xz</code>；<br>解压：<code>tar -xvf node-v12.14.1-linux-x64.tar.xz</code>；<br>移动：<code>sudo mv ./node-v16.13.1-linux-x64 /usr/local/node</code>；<br>接下来为node和npm创建快捷方式(软链接)<br><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">sudo ln -s /usr/<span class="built_in">local</span>/node/bin/node /usr/bin/node</span><br><span class="line">sudo ln -s /usr/<span class="built_in">local</span>/node/bin/npm /usr/bin/npm</span><br></pre></td></tr></table></figure></p><p><code>node -v</code> 查看node版本号<br>安装cnmp：<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">npm install -g cnpm --registry=https://registry.npm.taobao.org</span><br><span class="line">sudo ln -s /usr/local/node/bin/cnpm /usr/bin/cnpm</span><br></pre></td></tr></table></figure></p><ol><li><strong>C编译环境</strong>（没这个nvim的插件会出各种编译错误）</li></ol><p>运行以下命令以检查是否安装了 GCC，<code>rpm -qa | grep gcc</code>,然后：<br><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">sudo apt-get install yum-utils</span><br><span class="line">sudo apt-get install gcc</span><br></pre></td></tr></table></figure></p><p>编译c环境：<code>sudo apt-get install build-essential</code>，如果失败尝试将 <code>/var/lib/apt/lists/</code> 中的内容全部移除：<code>sudo rm -r /var/lib/apt/lists/*</code>，再执行一次 <code>apt update</code>，再尝试安装所需软件包。如果还失败，就说明前面提供的阿里源已经过气了，你需要自己在阿里社区更换最新的源后再重复编译的操作。</p><ol><li><strong>关于ripgrep的安装</strong> from：<a href="https://github.com/sharkdp/fd#on-ubuntu" target="_blank" rel="noopener">https://github.com/sharkdp/fd#on-ubuntu</a></li></ol><p>If you use an older version of Ubuntu, you can download the latest .deb package from the <a href="https://github.com/sharkdp/fd/releases" target="_blank" rel="noopener">release page</a> and install it via:<br><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">sudo dpkg -i fd_8.3.0_amd64.deb <span class="comment"># adapt version number and architecture</span></span><br></pre></td></tr></table></figure></p><h5 id="四-NvChad"><a href="#四-NvChad" class="headerlink" title="四. NvChad"></a>四. <a href="https://nvchad.github.io/getting-started/setup" target="_blank" rel="noopener">NvChad</a></h5><blockquote><p>我这里还有一个关于插件安装的前置条件，那就是需要一点点魔法，要不然插件根本下载不下来。</p></blockquote><p>这个配置的使用直接看官方文档就可以了，讲的非常详细，我不再赘述。</p><p>讲讲遇到的关于lsp的问题：</p><p>我按照他们的这个<a href="https://nvchad.github.io/config/Lsp%20stuff" target="_blank" rel="noopener">文档</a>调配lsp，第一个方法不成功（然而我真的很喜欢这种，简练优雅且干脆），<br>但是当我查看lsp的info的时候，一直都是<br><img src="https://lwh-graph-bed.oss-cn-hangzhou.aliyuncs.com/pic/305032701211248.png" alt="lsp服务问题"><br>他们给出了解决<code>cmd not executable</code>问题的办法，然而对我没用，后来各种尝试都没用，迫于无奈用了第二种。</p><hr><p>根据教程：<code>hook.add</code>里加入安装lsp的插件：<br><figure class="highlight lua"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">use {</span><br><span class="line"> <span class="string">"williamboman/nvim-lsp-installer"</span>,</span><br><span class="line"> }</span><br></pre></td></tr></table></figure></p><p>或者更详细的设置见文档。</p><p>我的<code>chadrc.lua</code>文件的plugins部分<br><figure class="highlight lua"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line">M.plugins = {</span><br><span class="line"> <span class="built_in">status</span> = {</span><br><span class="line"> dashboard = <span class="literal">true</span>,</span><br><span class="line"> colorizer = <span class="literal">true</span>,</span><br><span class="line"> }</span><br><span class="line"> options = {</span><br><span class="line"> nvimtree = {</span><br><span class="line"> enable_git = <span class="number">0</span>,</span><br><span class="line"> },</span><br><span class="line"> lspconfig = {</span><br><span class="line"> setup_lspconf = <span class="string">"custom.plugins.lspconfig"</span>,</span><br><span class="line"> },</span><br><span class="line"> },</span><br><span class="line"> default_plugin_config_replace = {}</span><br><span class="line">}</span><br></pre></td></tr></table></figure></p><p>接下来<code>mkdir plugins</code>，然后<code>nvim lspconfig.lua</code>写入：<br><figure class="highlight lua"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">local</span> M = {}</span><br><span class="line"></span><br><span class="line">M.setup_lsp = <span class="function"><span class="keyword">function</span><span class="params">(attach, capabilities)</span></span></span><br><span class="line"> <span class="keyword">local</span> lsp_installer = <span class="built_in">require</span> <span class="string">"nvim-lsp-installer"</span></span><br><span class="line"></span><br><span class="line"> lsp_installer.on_server_ready(<span class="function"><span class="keyword">function</span><span class="params">(server)</span></span></span><br><span class="line"> <span class="keyword">local</span> opts = {</span><br><span class="line"> on_attach = attach,</span><br><span class="line"> capabilities = capabilities,</span><br><span class="line"> flags = {</span><br><span class="line"> debounce_text_changes = <span class="number">150</span>,</span><br><span class="line"> },</span><br><span class="line"> settings = {},</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> <span class="keyword">if</span> server.name == <span class="string">"pyright-langserver"</span> <span class="keyword">then</span></span><br><span class="line"> opts.settings = {</span><br><span class="line"> [<span class="string">"pyright-langserver"</span>] = {</span><br><span class="line"> experimental = {</span><br><span class="line"> procAttrMacros = <span class="literal">true</span>,</span><br><span class="line"> },</span><br><span class="line"> },</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> opts.on_attach = <span class="function"><span class="keyword">function</span><span class="params">(client, bufnr)</span></span></span><br><span class="line"> <span class="keyword">local</span> <span class="function"><span class="keyword">function</span> <span class="title">buf_set_keymap</span><span class="params">(...)</span></span></span><br><span class="line"> vim.api.nvim_buf_set_keymap(bufnr, ...)</span><br><span class="line"> <span class="keyword">end</span></span><br><span class="line"></span><br><span class="line"> <span class="comment">-- Run nvchad's attach</span></span><br><span class="line"> attach(client, bufnr)</span><br><span class="line"></span><br><span class="line"> <span class="comment">-- Use nvim-code-action-menu for code actions for rust</span></span><br><span class="line"> buf_set_keymap(bufnr, <span class="string">"n"</span>, <span class="string">"<leader>ca"</span>, <span class="string">"lua vim.lsp.buf.range_code_action()<CR>"</span>, { noremap = <span class="literal">true</span>, silent = <span class="literal">true</span> })</span><br><span class="line"> buf_set_keymap(bufnr, <span class="string">"v"</span>, <span class="string">"<leader>ca"</span>, <span class="string">"lua vim.lsp.buf.range_code_action()<CR>"</span>, { noremap = <span class="literal">true</span>, silent = <span class="literal">true</span> })</span><br><span class="line"> <span class="keyword">end</span></span><br><span class="line"> <span class="keyword">end</span></span><br><span class="line"></span><br><span class="line"> server:setup(opts)</span><br><span class="line"> vim.cmd <span class="string">[[ do User LspAttachBuffers ]]</span></span><br><span class="line"> <span class="keyword">end</span>)</span><br><span class="line"><span class="keyword">end</span></span><br><span class="line"></span><br><span class="line"><span class="keyword">return</span> M</span><br></pre></td></tr></table></figure></p><p>最后重进nvim，安装插件：<code>:PackerInstall</code>,接着安装lsp服务：<code>:LspInstall pyright</code>.</p><p>完结！</p>]]></content>
</entry>
<entry>
<title>连接远程服务器使用git报错：Permission denied (publickey)</title>
<link href="/2021/04/09/ckx3e5ivj000knce2ry5b2pg6/"/>
<url>/2021/04/09/ckx3e5ivj000knce2ry5b2pg6/</url>
<content type="html"><![CDATA[<p>用xshell连的远程服务器，最近几天刚配置好vim，然后开始试着用git，然鹅配置好密钥后权限还是被拒绝了，网上搜索一番试了各种办法还是没有解决，包括什么从修改etc/ssh/sshd_config等等的，这里给出我的解决方法。</p><a id="more"></a><p>前期已经完成的工作说明一下：直接通过xshell在远程服务器上生成密钥</p><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">$ ssh-keygen -t rsa -C <span class="string">'任何随意的名称标记(英文数字等的最好)'</span> -f ~/.ssh/gitee_id_rsa</span><br><span class="line">//像这样</span><br><span class="line">$ ssh-keygen -t rsa -C <span class="string">'[email protected]'</span> -f ~/.ssh/gitee_id_rsa</span><br></pre></td></tr></table></figure><p>其中<code>-f</code>指示生成密钥的路径和名称，因为我这个密钥是用在gitee上的，所以加了gitee前缀。</p><p>随后在 <code>~/.ssh</code> 目录下新建一个config文件，添加如下内容</p><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment"># gitee</span></span><br><span class="line">Host gitee.com</span><br><span class="line">HostName gitee.com</span><br><span class="line">PreferredAuthentications publickey</span><br><span class="line">IdentityFile ~/.ssh/gitee_id_rsa</span><br></pre></td></tr></table></figure><p>最后用这个命令测试：<code>$ ssh -T [email protected]</code>，看到<code>Hi XXX! You've successfully authenticated, but GITEE.COM does not provide shell access.</code>即代表成功。</p><p>最后的最后，由于初次在服务器上使用git，你需要告诉它你是谁，直接运行下面的，会生成.gitconfig：</p><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">$ git config --global user.email <span class="string">"[email protected]"</span></span><br><span class="line">$ git config --global user.name <span class="string">"lwhluvdemo"</span></span><br></pre></td></tr></table></figure><p>PS：这是我学习Linux的第5天，所以在远程服务器上像我这么使用密钥是否涉及安全问题暂不清楚，我使用的服务器仅用来学习和练手，不涉及运营！</p>]]></content>
<tags>
<tag> xshell </tag>
<tag> Linux </tag>
</tags>
</entry>
<entry>
<title>YouComplete安装及报错解决</title>
<link href="/2021/04/08/ckx3e5iuu0002nce26gtem8t9/"/>
<url>/2021/04/08/ckx3e5iuu0002nce26gtem8t9/</url>
<content type="html"><![CDATA[<p>租的华为云的服务器，倒腾了两天才成功。经过这个过程才发现，在国内按照官方教程根本不可能成功，翻了一圈相关blog，也都没起作用。</p><a id="more"></a><p>最初是用的Ubuntu18的版本，vim是8.0，安装ycm后会提示你不能启用：<code>YouCompleteMe unavailable: requires Vim 8.1.2269+</code>，网上唯一的解决方法是GitHub的讨论：<a href="https://github.com/ycm-core/YouCompleteMe/issues/3764" target="_blank" rel="noopener">YouCompleteMe unavailable: requires Vim 8.1.2269+</a>。</p><p>非常可惜，多次尝试后依然没成功。</p><p>不得已重装了 Ubuntu20，此后按照官网的前几步安装配置都没问题。</p><p>最后出问题是在安装子模块上，当你运行<code>python3 ./install.py</code>时会提示这个命令：<code>git submodule update --init --recursive</code>，可惜在国内的网络环境下，成功的可能性基本为0。</p><p>除了一堆GitHub的url，还有类似<code>go.googlesource.com</code>这样的地址，就？？？</p><p>我把报错的网址大概记录了一下：</p><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line">https://github.com/PythonCharmers/python-future</span><br><span class="line">https://github.com/ross/requests-futures</span><br><span class="line">https://github.com/certifi/python-certifi</span><br><span class="line">https://github.com/chardet/chardet</span><br><span class="line">https://github.com/kjd/idna</span><br><span class="line">https://github.com/requests/requests</span><br><span class="line">https://github.com/urllib3/urllib3</span><br><span class="line">https://github.com/Valloric/ycmd</span><br><span class="line">https://go.googlesource.com/tools/ ---> https://github.com/golang/tools</span><br><span class="line">https://github.com/davidhalter/jedi/</span><br></pre></td></tr></table></figure><p>个人觉得唯一靠谱点的方法就是修改网址，类似这个博客：<a href="https://blog.csdn.net/weixin_44740860/article/details/105248387?utm_medium=distribute.pc_relevant.none-task-blog-baidujs_title-1&spm=1001.2101.3001.4242" target="_blank" rel="noopener">从gitee中转安装</a>；或者这个：<a href="https://blog.csdn.net/m0_37604813/article/details/107130881" target="_blank" rel="noopener">修改GitHub后缀</a>，不过个人觉得需要修改的隐藏文件不止<code>.gitmodules</code>，还有<code>./.git/config</code>这个文件，可能还有其它需要修改的地方，比如下载tools这个模块所在的文件我就没找到。如果这样能成功的话最好不过。</p><p>我就是那个又没成功的倒霉催的。</p><p>给出我最后的解决方法：在vim交流群里找了个能成功用ycm的朋友，把他的YouComplete文件夹搬到我的<code>~/.vim/plugged</code>里(我用的是vim-plug管理插件)，好的，当然。。还没有万事大吉呀！别人的文件里插件的相关配置基本都已经改过了，所以会有很多不匹配的命令，这时候我们只需要把<code>plugin</code>和<code>autoload</code>文件夹里的vim文件替换成原始<code>master</code>里的文件，再运行<code>sudo python3 ./install.py --clang-completer</code>编译C就可以了（因为我搬的那个朋友有编译过，有clang文件，所以只能补上这个步骤，否则会报错）。</p><p>这里把原始master的两个文件提取出来：第一个附件是<code>autoload</code>里的vim，第二个附件是<code>plugin</code>里的，替换原来的就好了。</p><p><a href="https://www.yuque.com/attachments/yuque/0/2021/vim/12979166/1617860559406-7237471b-0339-4235-85e0-17f06b150922.vim" target="_blank" rel="noopener">📎youcompleteme.vim</a>，<a href="https://www.yuque.com/attachments/yuque/0/2021/vim/12979166/1617860509389-0f14f5d1-8772-4fa6-a309-15a32ac00a72.vim" target="_blank" rel="noopener">📎youcompleteme.vim</a></p><p>over！</p><p>补充：<code>.vimrc</code>配置文件里的插件管理那一条需要换个地址；例如我的运行：<code>PlugInstall</code>时会在ycm那里出现网址报错，所以我的配置改为报错给出的网址(可能是需要我匹配搬运的人家的ycm文件里的地址吧，反正找半天没找到)</p><figure class="highlight plain"><figcaption><span>basic</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">Plug 'https://git::@hub.fastgit.org/Valloric/YouCompleteMe.git',{'do':'./install.py --tern-completer'}</span><br></pre></td></tr></table></figure><p>如果有不懂的地方，欢迎随时交流！</p>]]></content>
<tags>
<tag> vim </tag>
<tag> YouCompleteMe </tag>
</tags>
</entry>
<entry>
<title>闲聊</title>
<link href="/2019/10/02/ckx3e5ivc000ence27hvfpl2c/"/>
<url>/2019/10/02/ckx3e5ivc000ence27hvfpl2c/</url>
<content type="html"><![CDATA[<p>From：全部来自CVer群里“Leo林”大佬的一些闲聊</p><p>这里好像不少毕业找工作的，说一些建议吧:</p><ol><li>充分评估自己的能力，不要总想着 BAT 这样的大公司。<br>大公司人才济济，说实话，除非天才类型，否则各个位置上都占好了坑，要上位不容易的。<br>大多数毕业生都是大公司用来筛选的基数，给你个 2年高工资无所谓，但一般半年后就开始见分晓，该转岗去填劳力的就强制转，不转的话看你是想去哪？拿着高薪不做事？<br>一般降薪比较难操作，直接开掉比较容易。开老员工容易伤感情，赔偿还多，开工作一两年就简单很多。<br>总之呢，大公司高薪找毕业生其实是一种大撒网策略，不要真以为就是自己值这个钱。</li></ol><a id="more"></a><ol><li>目前人工智能在风头上，但今年产业发展已经开始回归理性，大多数公司在讨论的是怎么落地。这个时候大公司给高工资招的算法，一到两年内冷下来之后就很危险。大公司也不是做慈善的，不会养一批人做纯研究的。如果在公司做的算法和产品脱离了，自己就真要当心想一想了。不讲业绩的部门，也就不受重视，自己前途如何，要心里有数。做研究类的算法，真正能转化为产品的应该不会高于30%。<br>等风口冷下来了，也是大公司就高薪酬年轻人动刀的 时候了</li></ol><p>大公司每年招那么多人，其实就和鲸鱼差不多，先把鱼虾海水啥的都吞进来，然后再过滤掉，那一两年的工资差负担的起。否则为啥每年招那么多毕业生公司还在不停的招，真的是业务发展的需求？看看公司每年的增长率是多少就知道了。拿高薪人人都想，但是真的要搞清楚自己是不是准备被过滤掉的那批。2年后再出来，风口已过，到处都是被大公司筛出来的，还要和应届生再竞争了</p><p>真的是不是白菜价只有公司内部知道，如果凭着风口和老员工薪酬倒挂，那就更危险。要调老员工是大规模的调，公司受不了的，那如果短时间内没有做出大的贡献或者证明不了自己，看公司是情愿把所有的老员工薪酬调的和你一样，还是直接把你开了简单？</p><p>2年跳一次，到底是能力问题还是态度问题呀？到底是被开的还是自己辞职的呀？HR 冒不起这个风险呀<br>跳了以后，除非是行业内大牛，在新公司又要花个几年去铺垫才能上位，上不了位就只能一直待在某一个层级，10年工作还没有进展，以后基本就这样层次了</p><p>公司招人进来，HR 都是会评估的。<br>这个人未来是否可以长期培养，是否资源和态度方面可以向他倾斜，这些都是你背后一只看不见的手。<br>如果定位是招进来干2年活，做做事情，那也不需要花太多精力去照顾，反正迟早要走的</p><p>不是劝退，只是想说，不要只看着大公司的高薪酬。<br>要评估一下自己的能力，是否技术和能力真的好，能在大公司和全国的人才拼一拼。<br>如果自己是 20% 以下的，有比较靠谱的中型公司，有明确的技术前景，也可以考虑。潜心做几年，中小公司反而比较愿意花精力培养新人，当然前提是真的研发为核心的技术公司，而不是贸易公司。</p><p>每个人有每个人的造化，中小型公司也有风险。总得来说运气比实力占的比重大。工作以后也不单看技术，还要看沟通能力。能否把自己做的事情讲出来。只会技术讲不出来的，安安静静做个技术型专家也未尝不可。会技术又能讲得出来的，可以往技术领头人方向走。嘴皮子比较厉害，无心做技术的，做两年最好就考虑转方向</p><p>进了大公司的，头一年要保持些敏感度，和周边的人对比一下，看看自己是否真的有竞争力，一年内主动走的，下家企业基本是能按毕业生的角度来看的。2年后再走的，一般就按社招来判断了，会比较关心你跳槽的原因是什么了</p><p>转哪里就看自己的优势所在了，技术精英总归是少数，做个非精英部队的头部也不错的。如果工作了两年还没搞清楚自己的优势所在，我觉得最好就找个咨询公司帮忙做一下自己的职业规划做个参考。<br>现在好像也蛮多这样的服务的，花点钱值得</p><p>另外，有一个错觉是大公司学技术会比较好，其实大公司大多流程化，长时间可能就只能做一小块事情的。要转岗跳项目组多少有些忌讳。<br>中小公司反而因为缺人，很快就要负责很多事情，累肯定累，但学得东西也多。长远来说，这方面倒是比薪酬其实带来的价值会更大一些</p>]]></content>
<tags>
<tag> 工作 </tag>
</tags>
</entry>
<entry>
<title>双指针</title>
<link href="/2019/09/19/ckx3e5iwg001dnce2okbnecdu/"/>
<url>/2019/09/19/ckx3e5iwg001dnce2okbnecdu/</url>
<content type="html"><![CDATA[<h2 id="26-删除排序数组中的重复项"><a href="#26-删除排序数组中的重复项" class="headerlink" title="26. 删除排序数组中的重复项"></a>26. <a href="https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array/" target="_blank" rel="noopener">删除排序数组中的重复项</a></h2><p>给定一个排序数组，你需要在原地删除重复出现的元素，使得每个元素只出现一次，返回移除后数组的新长度。</p><p>不要使用额外的数组空间，你必须在原地修改输入数组并在使用 $O(1)$ 额外空间的条件下完成。</p><p>示例 1:</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line">给定数组 nums = [<span class="number">1</span>,<span class="number">1</span>,<span class="number">2</span>], </span><br><span class="line"></span><br><span class="line">函数应该返回新的长度 <span class="number">2</span>, 并且原数组 nums 的前两个元素被修改为 <span class="number">1</span>, <span class="number">2</span>。 </span><br><span class="line"></span><br><span class="line">你不需要考虑数组中超出新长度后面的元素。</span><br></pre></td></tr></table></figure><a id="more"></a><p>题目要求两件事：</p><ul><li>统计数组中不同数字数量 $k$ ；</li><li>修改数组前 $k$ 个元素为这些不同数字。</li></ul><p>算法流程：</p><ul><li>第一个指针 $i$ ： 由于数组已经完成排序，因此遍历数组，每遇到 $nums[i] \not= nums[i - 1]$，就说明遇到了新的不同数字，记录之；</li><li>第二个指针 $k$ ： 每遇到新的不同数字时，执行$k+=1$ ，$k$ 指针有两个作用：<ul><li>记录数组中不同数字的数量；</li><li>作为修改数组元素的索引<code>index</code>。</li></ul></li><li>最终，返回 $k$ 即可。</li></ul><h4 id="解答："><a href="#解答：" class="headerlink" title="解答："></a><strong>解答：</strong></h4><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">removeDuplicates</span><span class="params">(self, nums: List[int])</span> -> int:</span></span><br><span class="line"> <span class="keyword">if</span> <span class="keyword">not</span> nums:<span class="keyword">return</span> <span class="number">0</span></span><br><span class="line"> k = <span class="number">1</span></span><br><span class="line"> <span class="keyword">for</span> i <span class="keyword">in</span> range(<span class="number">1</span>,len(nums)):</span><br><span class="line"> <span class="keyword">if</span> nums[i] != nums[i<span class="number">-1</span>]:</span><br><span class="line"> nums[k] = nums[i]</span><br><span class="line"> k += <span class="number">1</span></span><br><span class="line"> <span class="keyword">return</span> k</span><br></pre></td></tr></table></figure><p><strong>复杂度分析</strong></p><ul><li>时间复杂度：$O(n)$，假设数组的长度是 $n$，那么 $i$ 和 $j$ 分别最多遍历 $n$ 步。</li><li>空间复杂度：$O(1)$。</li></ul><h2 id="88-合并两个有序数组"><a href="#88-合并两个有序数组" class="headerlink" title="88. 合并两个有序数组"></a>88. <a href="https://leetcode-cn.com/problems/merge-sorted-array/" target="_blank" rel="noopener">合并两个有序数组</a></h2><p>给定两个有序整数数组 nums1 和 nums2，将 nums2 合并到 nums1 中，使得 num1 成为一个有序数组。</p><p><strong>说明:</strong></p><ul><li>初始化 nums1 和 nums2 的元素数量分别为 m 和 n。</li><li>你可以假设 nums1 有足够的空间（空间大小大于或等于 m + n）来保存 nums2 中的元素</li></ul><p><strong>示例：</strong></p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line">输入:</span><br><span class="line">nums1 = [<span class="number">1</span>,<span class="number">2</span>,<span class="number">3</span>,<span class="number">0</span>,<span class="number">0</span>,<span class="number">0</span>], m = <span class="number">3</span></span><br><span class="line">nums2 = [<span class="number">2</span>,<span class="number">5</span>,<span class="number">6</span>], n = <span class="number">3</span></span><br><span class="line"></span><br><span class="line">输出: [<span class="number">1</span>,<span class="number">2</span>,<span class="number">2</span>,<span class="number">3</span>,<span class="number">5</span>,<span class="number">6</span>]</span><br></pre></td></tr></table></figure><h4 id="解法一：双指针-从前往后"><a href="#解法一：双指针-从前往后" class="headerlink" title="解法一：双指针 / 从前往后"></a><strong>解法一：</strong>双指针 / 从前往后</h4><p>一般而言，对于有序数组可以通过 双指针法 达到$O(n + m)$的时间复杂度。</p><p>最直接的算法实现是将指针<code>p1</code> 置为 <code>nums1</code>的开头， <code>p2</code>为 <code>nums2</code>的开头，在每一步将最小值放入输出数组中。</p><p>由于 <code>nums1</code> 是用于输出的数组，需要将<code>nums1</code>中的前<code>m</code>个元素放在其他地方，也就需要 $O(m)$ 的空间复杂度。</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span><span class="params">(object)</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">merge</span><span class="params">(self, nums1, m, nums2, n)</span>:</span></span><br><span class="line"> <span class="string">"""</span></span><br><span class="line"><span class="string"> :type nums1: List[int]</span></span><br><span class="line"><span class="string"> :type m: int</span></span><br><span class="line"><span class="string"> :type nums2: List[int]</span></span><br><span class="line"><span class="string"> :type n: int</span></span><br><span class="line"><span class="string"> :rtype: void Do not return anything, modify nums1 in-place instead.</span></span><br><span class="line"><span class="string"> """</span></span><br><span class="line"> <span class="comment"># Make a copy of nums1.</span></span><br><span class="line"> nums1_copy = nums1[:m] </span><br><span class="line"> nums1[:] = []</span><br><span class="line"></span><br><span class="line"> <span class="comment"># Two get pointers for nums1_copy and nums2.</span></span><br><span class="line"> p1 = <span class="number">0</span> </span><br><span class="line"> p2 = <span class="number">0</span></span><br><span class="line"> </span><br><span class="line"> <span class="comment"># Compare elements from nums1_copy and nums2</span></span><br><span class="line"> <span class="comment"># and add the smallest one into nums1.</span></span><br><span class="line"> <span class="keyword">while</span> p1 < m <span class="keyword">and</span> p2 < n: </span><br><span class="line"> <span class="keyword">if</span> nums1_copy[p1] < nums2[p2]: </span><br><span class="line"> nums1.append(nums1_copy[p1])</span><br><span class="line"> p1 += <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> nums1.append(nums2[p2])</span><br><span class="line"> p2 += <span class="number">1</span></span><br><span class="line"></span><br><span class="line"> <span class="comment"># if there are still elements to add</span></span><br><span class="line"> <span class="keyword">if</span> p1 < m: </span><br><span class="line"> nums1[p1 + p2:] = nums1_copy[p1:]</span><br><span class="line"> <span class="keyword">if</span> p2 < n:</span><br><span class="line"> nums1[p1 + p2:] = nums2[p2:]</span><br></pre></td></tr></table></figure><p><strong>复杂度分析</strong></p><ul><li>时间复杂度 : $O(n + m)$。</li><li>空间复杂度 : $O(m)$。</li></ul><h4 id="解法二-双指针-从后往前"><a href="#解法二-双指针-从后往前" class="headerlink" title="解法二 : 双指针 / 从后往前"></a>解法二 : 双指针 / 从后往前</h4><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span><span class="params">(object)</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">merge</span><span class="params">(self, nums1, m, nums2, n)</span>:</span></span><br><span class="line"> <span class="string">"""</span></span><br><span class="line"><span class="string"> :type nums1: List[int]</span></span><br><span class="line"><span class="string"> :type m: int</span></span><br><span class="line"><span class="string"> :type nums2: List[int]</span></span><br><span class="line"><span class="string"> :type n: int</span></span><br><span class="line"><span class="string"> :rtype: void Do not return anything, modify nums1 in-place instead.</span></span><br><span class="line"><span class="string"> """</span></span><br><span class="line"> <span class="comment"># two get pointers for nums1 and nums2</span></span><br><span class="line"> p1 = m - <span class="number">1</span></span><br><span class="line"> p2 = n - <span class="number">1</span></span><br><span class="line"> <span class="comment"># set pointer for nums1</span></span><br><span class="line"> p = m + n - <span class="number">1</span></span><br><span class="line"> </span><br><span class="line"> <span class="comment"># while there are still elements to compare</span></span><br><span class="line"> <span class="keyword">while</span> p1 >= <span class="number">0</span> <span class="keyword">and</span> p2 >= <span class="number">0</span>:</span><br><span class="line"> <span class="keyword">if</span> nums1[p1] < nums2[p2]:</span><br><span class="line"> nums1[p] = nums2[p2]</span><br><span class="line"> p2 -= <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> nums1[p] = nums1[p1]</span><br><span class="line"> p1 -= <span class="number">1</span></span><br><span class="line"> p -= <span class="number">1</span></span><br><span class="line"> </span><br><span class="line"> <span class="comment"># add missing elements from nums2</span></span><br><span class="line"> nums1[:p2 + <span class="number">1</span>] = nums2[:p2 + <span class="number">1</span>]</span><br></pre></td></tr></table></figure><p><strong>复杂度分析</strong></p><ul><li>时间复杂度 : $O(n + m)$。</li><li>空间复杂度 : $O(1)$。</li></ul><h2 id="125-验证回文串"><a href="#125-验证回文串" class="headerlink" title="125. 验证回文串"></a>125. <a href="https://leetcode-cn.com/problems/valid-palindrome/" target="_blank" rel="noopener">验证回文串</a></h2><p>给定一个字符串，验证它是否是回文串，只考虑字母和数字字符，可以忽略字母的大小写。</p><p><strong>说明：</strong>本题中，我们将空字符串定义为有效的回文串。</p><p><strong>示例 :</strong></p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入: <span class="string">"A man, a plan, a canal: Panama"</span></span><br><span class="line">输出: true</span><br></pre></td></tr></table></figure><h4 id="解法："><a href="#解法：" class="headerlink" title="解法："></a>解法：</h4><p><strong>思路：</strong></p><ul><li>设置一个头指针和一个尾指针</li><li>头指针从头开始找，找到第一个数字或字符</li><li>尾指针从尾开始找，找到最后一个数字或字符</li><li>比较两者是否相同，不同就return False</li><li>相同则两指针往中间靠</li><li>当 i > j 的时候，return True</li></ul><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">isPalindrome</span><span class="params">(self, s: str)</span> -> bool:</span></span><br><span class="line"> i, j = <span class="number">0</span>, len(s) - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> i < j:</span><br><span class="line"> <span class="comment"># isalnum() 方法检测字符串是否由字母和数字组成</span></span><br><span class="line"> <span class="keyword">while</span> i < len(s) <span class="keyword">and</span> <span class="keyword">not</span> s[i].isalnum():</span><br><span class="line"> i += <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> j > <span class="number">-1</span> <span class="keyword">and</span> <span class="keyword">not</span> s[j].isalnum():</span><br><span class="line"> j -= <span class="number">1</span></span><br><span class="line"> <span class="keyword">if</span> i > j:</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">True</span></span><br><span class="line"> <span class="keyword">if</span> s[i].upper() != s[j].upper():</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">False</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> i += <span class="number">1</span></span><br><span class="line"> j -= <span class="number">1</span></span><br><span class="line"> <span class="keyword">return</span> <span class="literal">True</span></span><br></pre></td></tr></table></figure><p><strong>复杂度分析</strong></p><ul><li>时间复杂度 <strong>O(N)</strong>, 遍历一遍数组。</li><li>空间复杂读 <strong>O(1)</strong>，只使用了常量的空间。</li></ul><p>同思想另一种写法：</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">isPalindrome</span><span class="params">(self, s: str)</span> -> bool:</span></span><br><span class="line"> l = <span class="number">0</span></span><br><span class="line"> r = len(s)<span class="number">-1</span></span><br><span class="line"> <span class="keyword">while</span> l < r:</span><br><span class="line"> <span class="keyword">if</span> s[l].isalnum() <span class="keyword">and</span> s[r].isalnum():</span><br><span class="line"> <span class="keyword">if</span> s[l].upper() == s[r].upper():</span><br><span class="line"> l += <span class="number">1</span></span><br><span class="line"> r -= <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">False</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> <span class="keyword">if</span> <span class="keyword">not</span> s[l].isalnum():</span><br><span class="line"> l += <span class="number">1</span></span><br><span class="line"> <span class="keyword">if</span> <span class="keyword">not</span> s[r].isalnum():</span><br><span class="line"> r -= <span class="number">1</span></span><br><span class="line"> <span class="keyword">return</span> <span class="literal">True</span></span><br></pre></td></tr></table></figure><p>其他解法：</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">isPalindrome</span><span class="params">(self, s: str)</span> -> bool:</span></span><br><span class="line"> <span class="keyword">return</span> (<span class="string">""</span>.join(filter(str.isalnum, s)).lower()[::<span class="number">-1</span>]) == (<span class="string">""</span>.join(filter(str.isalnum, s)).lower())</span><br><span class="line"><span class="comment">########分开写</span></span><br><span class="line">s = <span class="string">""</span>.join(filter(str.isalnum, s)).lower()</span><br><span class="line"><span class="keyword">return</span> s[::<span class="number">-1</span>] == s</span><br></pre></td></tr></table></figure><h2 id="141-环形链表"><a href="#141-环形链表" class="headerlink" title="141. 环形链表"></a>141. <a href="https://leetcode-cn.com/problems/linked-list-cycle/" target="_blank" rel="noopener">环形链表</a></h2><p>给定一个链表，判断链表中是否有环。</p><p>为了表示给定链表中的环，我们使用整数 pos 来表示链表尾连接到链表中的位置（索引从 0 开始）。 如果 pos 是 -1，则在该链表中没有环。</p><p><strong>示例 1：</strong></p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入：head = [<span class="number">3</span>,<span class="number">2</span>,<span class="number">0</span>,<span class="number">-4</span>], pos = <span class="number">1</span></span><br><span class="line">输出：true</span><br><span class="line">解释：链表中有一个环，其尾部连接到第二个节点。</span><br></pre></td></tr></table></figure><h4 id="解法一："><a href="#解法一：" class="headerlink" title="解法一："></a>解法一：</h4><ul><li>设两指针fast slow指向链表头部head，迭代：<ul><li>fast每轮走两步，slow每轮走一步，这样两指针每轮后距离+1;</li><li>若链表中存在环，fast和slow一定会在将来相遇（距离连续+1，没有跳跃）；</li></ul></li><li>若fast走到了链表尾部，则说明链表无环。</li></ul><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span><span class="params">(object)</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">hasCycle</span><span class="params">(self, head)</span>:</span></span><br><span class="line"> fast, slow = head, head</span><br><span class="line"> <span class="keyword">while</span> fast <span class="keyword">and</span> fast.next:</span><br><span class="line"> fast = fast.next.next</span><br><span class="line"> slow = slow.next</span><br><span class="line"> <span class="keyword">if</span> fast == slow: <span class="keyword">return</span> <span class="literal">True</span></span><br><span class="line"> <span class="keyword">return</span> <span class="literal">False</span></span><br></pre></td></tr></table></figure><h4 id="解法二："><a href="#解法二：" class="headerlink" title="解法二："></a>解法二：</h4><p>通过检查一个结点此前是否被访问过来判断链表是否为环形链表。常用的方法是使用哈希表</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span><span class="params">(object)</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">hasCycle</span><span class="params">(self, head)</span>:</span></span><br><span class="line"> <span class="string">"""</span></span><br><span class="line"><span class="string"> :type head: ListNode</span></span><br><span class="line"><span class="string"> :rtype: bool</span></span><br><span class="line"><span class="string"> """</span></span><br><span class="line"> save = set()</span><br><span class="line"> cur = head </span><br><span class="line"> <span class="keyword">while</span> cur <span class="keyword">is</span> <span class="keyword">not</span> <span class="literal">None</span>:</span><br><span class="line"> <span class="keyword">if</span> cur <span class="keyword">in</span> save:</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">True</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> save.add(cur)</span><br><span class="line"> cur = cur.next</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">False</span></span><br></pre></td></tr></table></figure>]]></content>
<tags>
<tag> 数据结构与算法 </tag>
</tags>
</entry>
<entry>
<title></title>
<link href="/2019/09/12/ckx3e5ivh000ince2mbz6rdmb/"/>
<url>/2019/09/12/ckx3e5ivh000ince2mbz6rdmb/</url>
<content type="html"><![CDATA[<h4 id="比赛群学习笔记"><a href="#比赛群学习笔记" class="headerlink" title="比赛群学习笔记"></a>比赛群学习笔记</h4><p>如果某个特征很强很强，那么原样拷贝一下，构造一个新的特征。会怎样？</p><p>分裂次数，首先，每个特征复杂度不同。。自身能分裂的次数就是不定的。</p><p>道理上，这样普遍情况下肯定没用。<br>但是两个特征肯定会有原特征一半左右的特征重要性，那么这个表面看，还是强特呢</p><p>我看的<a href="https://github.com/WillKoehrsen/feature-selector" target="_blank" rel="noopener">这个链接</a>，按照林佬的讲法，这个feature-selector有些思路也是有问题的，我觉得我也要再想想想，模型只有学到，在测试集中依旧有用的信息，才有价值，构建好一致的线下验证集，是成功的一半，其实影响得分最大的因素是验证数据</p><p>线下5折的时候。每折的得分的方差大，还是同一个折，换不同的5个种子的方差大？其实很明显是前者。</p><p>绝大部分误差是数据真实标签带来的。。一般说的线下线上一致，都是指增减一致。而不是分数接近。</p>]]></content>
</entry>
<entry>
<title>pandas技巧</title>
<link href="/2019/09/05/ckx3e5iv00005nce28wdgk6mp/"/>
<url>/2019/09/05/ckx3e5iv00005nce28wdgk6mp/</url>
<content type="html"><![CDATA[<h4 id="用一列的非空值填充另一列对应行的空值"><a href="#用一列的非空值填充另一列对应行的空值" class="headerlink" title="用一列的非空值填充另一列对应行的空值"></a>用一列的非空值填充另一列对应行的空值</h4><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment"># 利用数据框df的name列中的非空值，去填充df的features_1列中对应的NaN。</span></span><br><span class="line">df.loc[df[<span class="string">'features_1'</span>].isnull(),<span class="string">'features_1'</span>]=df[df[<span class="string">'features_1'</span>].isnull()][‘name’]</span><br></pre></td></tr></table></figure><a id="more"></a><h4 id="填充缺失值"><a href="#填充缺失值" class="headerlink" title="填充缺失值"></a>填充缺失值</h4><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">#填充缺失值</span></span><br><span class="line"><span class="keyword">for</span> i <span class="keyword">in</span> string_feature+other_feature:</span><br><span class="line"> mode_num = data[i].mode()[<span class="number">0</span>]</span><br><span class="line"> <span class="keyword">if</span> (mode_num != <span class="number">-1</span>):</span><br><span class="line"> print(i)</span><br><span class="line"> data.loc[data[i] == <span class="number">-1</span>, i] = mode_num</span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> print(<span class="number">-1</span>)</span><br><span class="line"><span class="keyword">for</span> i <span class="keyword">in</span> numeric_feature:</span><br><span class="line"> mean_num = data[i].mean()</span><br><span class="line"> <span class="keyword">if</span> (mean_num != <span class="number">-1</span>):</span><br><span class="line"> print(i)</span><br><span class="line"> data.loc[data[i] == <span class="number">-1</span>, i] = mean_num</span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> print(<span class="number">-1</span>)</span><br></pre></td></tr></table></figure><p>判断两个list是否是父子集关系</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line">a = [<span class="number">1</span>,<span class="number">2</span>]</span><br><span class="line">b = [<span class="number">1</span>,<span class="number">2</span>,<span class="number">3</span>]</span><br><span class="line">c = [<span class="number">0</span>, <span class="number">1</span>]</span><br><span class="line">set(b) > set(a)</span><br><span class="line">set(b) > set(c)</span><br></pre></td></tr></table></figure><p>求列表差值：</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">list(set(a).difference(set(b)))</span><br></pre></td></tr></table></figure><p>按列时间值筛选 Pandas DataFrame</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">import</span> pandas <span class="keyword">as</span> pd</span><br><span class="line"><span class="keyword">import</span> numpy <span class="keyword">as</span> np</span><br><span class="line"><span class="keyword">import</span> io</span><br></pre></td></tr></table></figure><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line">t=<span class="string">"""2015-01-01 04:00:00</span></span><br><span class="line"><span class="string">2015-01-01 05:00:00</span></span><br><span class="line"><span class="string">2015-01-01 06:00:00</span></span><br><span class="line"><span class="string">2015-01-01 07:00:00</span></span><br><span class="line"><span class="string">2015-01-02 04:00:00</span></span><br><span class="line"><span class="string">2015-01-02 05:00:00</span></span><br><span class="line"><span class="string">2015-01-02 06:00:00</span></span><br><span class="line"><span class="string">2015-01-02 07:00:00"""</span></span><br><span class="line">s = pd.read_csv(io.StringIO(t), parse_dates=[<span class="number">0</span>], header=<span class="literal">None</span>, names=[<span class="string">'date'</span>])</span><br><span class="line">s[(s[<span class="string">'date'</span>].dt.hour <= <span class="number">6</span>)&(s[<span class="string">'date'</span>].dt.hour >= <span class="number">2</span>)]</span><br></pre></td></tr></table></figure><p>计算真值</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">np.count_nonzero(df[col].value_counts().reset_index()[col].values>=<span class="number">600</span>)</span><br></pre></td></tr></table></figure><p>每列的最大值：</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">train_df.ix[train_df.idxmax()].iloc[:,<span class="number">60</span>:<span class="number">120</span>].max()</span><br></pre></td></tr></table></figure><p> 同类特征正则匹配：</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment"># 同类特征正则匹配测试</span></span><br><span class="line">df = pd.DataFrame(np.array(([<span class="number">1</span>, <span class="number">2</span>, <span class="number">3</span>,<span class="number">4</span>,<span class="number">4</span>,<span class="number">4</span>,<span class="number">4</span>,<span class="number">8</span>,<span class="number">8</span>,<span class="number">8</span>], [<span class="number">4</span>, <span class="number">5</span>, <span class="number">6</span>,<span class="number">5</span>,<span class="number">5</span>,<span class="number">5</span>,<span class="number">5</span>,<span class="number">7</span>,<span class="number">7</span>,<span class="number">7</span>])),</span><br><span class="line"> index=[<span class="string">'mouse'</span>, <span class="string">'rabbit'</span>],</span><br><span class="line"> columns=[<span class="string">'WebInfo_1'</span>, <span class="string">'WebInfo_2'</span>, <span class="string">'WebInfo_3'</span>,<span class="string">'ProductInfo_1'</span>,</span><br><span class="line"> <span class="string">'ProductInfo_2'</span>,<span class="string">'ProductInfo_3'</span>,<span class="string">'ProductInfo_4'</span>,<span class="string">'UserInfo_1'</span>,</span><br><span class="line"> <span class="string">'UserInfo_2'</span>,<span class="string">'UserInfo_3'</span>])</span><br><span class="line">df_WebInfo = df.filter(regex=<span class="string">"WebInfo_\d{1,3}"</span>)</span><br><span class="line">df_ProductInfo = df.filter(regex=<span class="string">"ProductInfo_\d{1,3}"</span>)</span><br><span class="line">df_UserInfo = df.filter(regex=<span class="string">"UserInfo_\d{1,3}"</span>)</span><br></pre></td></tr></table></figure><p>将一列的值填充到另一列同行的缺失值中</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">train.loc[train[<span class="string">'member_id'</span>].isnull(),<span class="string">'member_id'</span>] = train[train[<span class="string">'member_id'</span>]</span><br><span class="line"> .isnull()][<span class="string">'customer_id'</span>]</span><br></pre></td></tr></table></figure>]]></content>
<tags>
<tag> 机器学习 </tag>
<tag> pandas </tag>
</tags>
</entry>
<entry>
<title>keras_bert</title>
<link href="/2019/08/29/ckx3e5iv20006nce26zco643k/"/>
<url>/2019/08/29/ckx3e5iv20006nce26zco643k/</url>
<content type="html"><![CDATA[<h3 id="keras-bert"><a href="#keras-bert" class="headerlink" title="keras_bert"></a>keras_bert</h3><p><a href="https://colab.research.google.com/github/CyberZHG/keras-bert/blob/master/demo/tune/keras_bert_classification_tpu.ipynb" target="_blank" rel="noopener">分类示例</a>中在IMDB数据集上对模型进行了微调以适应新的分类任务。</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment"># 会下载Bert模型</span></span><br><span class="line">!pip install -q keras-bert keras-rectified-adam</span><br><span class="line">!wget -q https://storage.googleapis.com/bert_models/<span class="number">2018</span>_10_18/uncased_L<span class="number">-12</span>_H<span class="number">-768</span>_A<span class="number">-12.</span>zip</span><br><span class="line"><span class="comment"># !wget -q https://storage.googleapis.com/bert_models/2018_11_03/chinese_L-12_H-768_A-12.zip # 中文</span></span><br><span class="line">!unzip -o uncased_L<span class="number">-12</span>_H<span class="number">-768</span>_A<span class="number">-12.</span>zip</span><br></pre></td></tr></table></figure><a id="more"></a><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment"># 加载路径</span></span><br><span class="line"><span class="keyword">import</span> os</span><br><span class="line"></span><br><span class="line">pretrained_path = <span class="string">'chinese_L-12_H-768_A-12'</span></span><br><span class="line">config_path = os.path.join(pretrained_path, <span class="string">'bert_config.json'</span>)</span><br><span class="line">checkpoint_path = os.path.join(pretrained_path, <span class="string">'bert_model.ckpt'</span>)</span><br><span class="line">vocab_path = os.path.join(pretrained_path, <span class="string">'vocab.txt'</span>)</span><br></pre></td></tr></table></figure><h4 id="第二种方式"><a href="#第二种方式" class="headerlink" title="第二种方式"></a>第二种方式</h4><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">from</span> keras_bert.datasets <span class="keyword">import</span> get_pretrained, PretrainedList</span><br><span class="line">model_path = get_pretrained(PretrainedList.chinese_base)</span><br><span class="line">paths = get_checkpoint_paths(model_path)</span><br><span class="line">model = load_trained_model_from_checkpoint(paths.config, paths.checkpoint, seq_len=<span class="number">10</span>)</span><br><span class="line">model.summary(line_length=<span class="number">120</span>)</span><br><span class="line">token_dict = load_vocabulary(paths.vocab)</span><br></pre></td></tr></table></figure><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line">tokenizer = Tokenizer(token_dict)</span><br><span class="line">text = <span class="string">'语言模型'</span></span><br><span class="line">tokens = tokenizer.tokenize(text)</span><br><span class="line">print(<span class="string">'Tokens:'</span>, tokens)</span><br><span class="line">indices, segments = tokenizer.encode(first=text, max_len=<span class="number">10</span>)</span><br><span class="line"></span><br><span class="line">predicts = model.predict([np.array([indices]), np.array([segments])])[<span class="number">0</span>]</span><br><span class="line"><span class="keyword">for</span> i, token <span class="keyword">in</span> enumerate(tokens):</span><br><span class="line"> print(token, predicts[i].tolist()[:<span class="number">5</span>])</span><br></pre></td></tr></table></figure><h3 id="tensorboard-可视化训练过程"><a href="#tensorboard-可视化训练过程" class="headerlink" title="tensorboard 可视化训练过程"></a>tensorboard 可视化训练过程</h3><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">from</span> tensorflow.python <span class="keyword">import</span> keras</span><br><span class="line"><span class="keyword">from</span> kashgari.tasks.classification <span class="keyword">import</span> BiGRU_Model</span><br><span class="line"><span class="keyword">from</span> kashgari.callbacks <span class="keyword">import</span> EvalCallBack</span><br><span class="line"></span><br><span class="line"><span class="keyword">import</span> logging</span><br><span class="line">logging.basicConfig(level=<span class="string">'DEBUG'</span>)</span><br><span class="line"></span><br><span class="line">model = BiGRU_Model()</span><br><span class="line"></span><br><span class="line">tf_board_callback = keras.callbacks.TensorBoard(log_dir=<span class="string">'./logs'</span>, update_freq=<span class="number">1000</span>)</span><br><span class="line"></span><br><span class="line"><span class="comment"># 这是 Kashgari 内置回调函数，会在训练过程计算精确度，召回率和 F1</span></span><br><span class="line">eval_callback = EvalCallBack(kash_model=model,</span><br><span class="line"> valid_x=valid_x,</span><br><span class="line"> valid_y=valid_y,</span><br><span class="line"> step=<span class="number">5</span>)</span><br><span class="line"><span class="comment"># model = CNNModel()</span></span><br><span class="line">model.fit(train_x,</span><br><span class="line"> train_y,</span><br><span class="line"> val_x,</span><br><span class="line"> val_y,</span><br><span class="line"> batch_size=<span class="number">100</span>,</span><br><span class="line"> callbacks=[eval_callback, tf_board_callback])</span><br></pre></td></tr></table></figure><p>在项目目录运行下面代码即可启动 tensorboard 查看可视化效果</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ tensorboard --log-dir logs</span><br></pre></td></tr></table></figure>]]></content>
<tags>
<tag> keras </tag>
<tag> bert </tag>
</tags>
</entry>
<entry>
<title>数据结构与算法</title>
<link href="/2019/08/23/ckx3e5ivf000hnce2h3cuwrtg/"/>
<url>/2019/08/23/ckx3e5ivf000hnce2h3cuwrtg/</url>
<content type="html"><![CDATA[<h4 id="大O"><a href="#大O" class="headerlink" title="大O"></a>大O</h4><p>大O表示法中的常量有时候事关重大，这就是快速排序比合并排序快的原因所在。 </p><p>比较简单查找和二分查找时，常量几乎无关紧要，因为列表很长时， O(log n)的速度比O(n)<br>快得多。 </p><h4 id="散列函数"><a href="#散列函数" class="headerlink" title="散列函数"></a>散列函数</h4><p>“将输入映射到数字”。 </p><ul><li>模拟映射关系；</li><li>防止重复；</li><li>缓存/记住数据，以免服务器再通过处理来生成它们。 </li></ul>]]></content>
<tags>
<tag> 数据结构 </tag>
</tags>
</entry>
<entry>
<title>pyspark</title>
<link href="/2019/08/22/ckx3e5iv6000ance2l5v94qhb/"/>
<url>/2019/08/22/ckx3e5iv6000ance2l5v94qhb/</url>
<content type="html"><![CDATA[<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">from</span> pyspark <span class="keyword">import</span> SparkConf, SparkContext</span><br><span class="line"><span class="keyword">from</span> pyspark.sql <span class="keyword">import</span> SparkSession,SQLContext,DataFrame,Row</span><br><span class="line">spark= SparkSession.builder.master(<span class="string">'local'</span>).appName(<span class="string">'test'</span>).getOrCreate()</span><br><span class="line">sc = spark.sparkContext</span><br><span class="line"></span><br><span class="line">lines = sc.textFile(<span class="string">"data.txt"</span>)</span><br></pre></td></tr></table></figure><p>相关操作：<a href="https://www.jianshu.com/p/4cd22eda363f" target="_blank" rel="noopener">PySpark之RDD入门</a>.</p><p>测试：</p><script type="math/tex; mode=display">\text { If } f\left(\mathbf{u}, \mathbf{x}^{*}\right) \leq 0 \forall \mathbf{u} \in \mathcal{U}_{\epsilon}, \quad \text { then } \mathbb{P}^{*}\left(f\left(\tilde{\mathbf{u} }, \mathbf{x}^{*}\right) \leq 0\right) \geq 1-\epsilon \tag{2}</script><p>测试2：<script type="math/tex">\text { If } f\left(\mathbf{u}, \mathbf{x}^{*}\right) \leq 0 \forall \mathbf{u} \in \mathcal{U}_{\epsilon}, \quad \text { then } \mathbb{P}^{*}\left(f\left(\tilde{\mathbf{u} }, \mathbf{x}^{*}\right) \leq 0\right) \geq 1-\epsilon \tag{2}</script>.</p><p>测试3：这是$x_y^t$的展示。</p><p>version-02</p><figure class="highlight js"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">formatText</span>(<span class="params">text</span>) </span>{</span><br><span class="line"> <span class="comment">// Fit kramed's rule: $$ + \1 + $$</span></span><br><span class="line"> <span class="keyword">return</span> text.replace(<span class="regexp">/`\$(.*?)\$`/g</span>, <span class="string">'$$$$$1$$$$'</span>);</span><br><span class="line">}</span><br></pre></td></tr></table></figure><script src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML"></script>]]></content>
<tags>
<tag> pyspark </tag>
</tags>
</entry>
<entry>
<title>算法竞赛一些问题总结</title>
<link href="/2019/08/20/ckx3e5ivl000mnce2hb6ki3jy/"/>
<url>/2019/08/20/ckx3e5ivl000mnce2hb6ki3jy/</url>
<content type="html"><![CDATA[<h3 id="L1和L2范数都是用来避免过拟合问题，那么请问在使用的时候二者有何区别呢？"><a href="#L1和L2范数都是用来避免过拟合问题，那么请问在使用的时候二者有何区别呢？" class="headerlink" title="L1和L2范数都是用来避免过拟合问题，那么请问在使用的时候二者有何区别呢？"></a>L1和L2范数都是用来避免过拟合问题，那么请问在使用的时候二者有何区别呢？</h3><ul><li>L1范数是对系数<code>绝对值之和的约束</code>。梯度会比较陡，通过求导可以知道，当某些特征作用不大的时候，会使得对应的系数为0，主要用于稀疏特征的约束，比如one-hot编码。</li><li>L2范数是对系数<code>平方和的约束</code>，从而简化模型。梯度会比较缓，通过求导可以知道，系数很难达到0，因此L2范数不仅能够简化模型，也不会像L1范数一样，迫使模型忽略某些特征，一般来说，如果特征不是很稀疏，可以使用L2。</li><li>L1正则假设参数符合拉普拉斯分布，L2正则假设参数符合正态分布。从梯度来看，L2正则对数值大的参数惩罚大，对数值小的参数惩罚小，而L1对参数的惩罚恒定。于是L1容易把不太重要的特征的参数“惩罚”到0，从而产生稀疏的解。</li></ul><a id="more"></a><h3 id="比赛做特征工程是一个一个特征加进去试试比较好，还是先做完一堆特征然后再挑比较好？"><a href="#比赛做特征工程是一个一个特征加进去试试比较好，还是先做完一堆特征然后再挑比较好？" class="headerlink" title="比赛做特征工程是一个一个特征加进去试试比较好，还是先做完一堆特征然后再挑比较好？"></a>比赛做特征工程是一个一个特征加进去试试比较好，还是先做完一堆特征然后再挑比较好？</h3><p> 首先要大致清楚模型能学到什么样的特征。</p><p> 然后当你做一个特征的时候，你要想清楚为什么这样尝试？为什么你认为这个特征有用，比如，以前的类似比赛用过类似的特征；通过对业务的理解进行推理出来的；通过数据观察(EDA)，发现该特征下，label的表现差异比较大（如果特征是随机数的时候，label理论上差异是最小的），一个具有良好依据的特征解释，是特征有效的根本。</p><p> 最直接有效的特征验证的方法就是代入模型验证，所以一个合理的验证策略变得极为重要。这里一定要留意的一个问题就是，要正确理解什么是“抖动”，抖动存在很多地方，一个随机种子，不同的列的顺序，行的顺序，线上线下数据集等。你所期望的结果，一般需要考虑以下数据集中自身存在的抖动收益。<br>最后就是比较重要的代码效率问题了。</p><p> 一个高效率的运行框架和调试方案。会大大加快你的想法的测试。【大约1min 验证一个特征】</p><h3 id="数据探索-林有夕的分享"><a href="#数据探索-林有夕的分享" class="headerlink" title="数据探索-林有夕的分享"></a>数据探索-林有夕的分享</h3><p>数据探索主要通过一些统计指标来分析数据的分布。观察数据和结果的影响。</p><p>例如观察数值特征的分布，均值，最值等，观察类别特征的频率分布等。以及一些交叉分布，比如平均每个用户看过的广告数之类的。</p><p>一般数据探索有如下几个好处：</p><ol><li>让你更加了解数据的情况。验证是否符合自己所想的逻辑。</li><li>通过观察特征和标签的分布关系。初步验证该特征是否和标签存在明显的关系。（表现在不同标签下，特征分布差异较大），在一些数据量比较大，模型迭代效率较低的场景下，可以通过该方法初步验证特征是否有意义。</li><li>对于比赛后期，大家关于数据现实意义中有道理的特征和方案都尝试差不多的情况下，数据探索就尤其重要了。因为往往最后的提分点都是一些数据固有的特点，而非业务特点。（比如，通过观察你有可能会发现一些采样错误。一些异常数据点（比如地铁停运、极端天气等影响的label）等等。这些单纯靠猜业务是很难想到的。</li></ol>]]></content>
<tags>
<tag> 机器学习 </tag>
<tag> 算法竞赛 </tag>
</tags>
</entry>
<entry>
<title>EDA笔记</title>
<link href="/2019/08/15/ckx3e5iul0000nce2w2qhtphh/"/>
<url>/2019/08/15/ckx3e5iul0000nce2w2qhtphh/</url>
<content type="html"><![CDATA[<p>控制大小、改变X轴字段方向</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">plt.figure(figsize=(<span class="number">9</span>,<span class="number">6</span>))</span><br><span class="line">plt.xticks(fontsize=<span class="number">12</span>,rotation=<span class="number">45</span>)</span><br></pre></td></tr></table></figure><p>改变x轴标签：</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">loc, labels = plt.xticks()</span><br><span class="line">loc, labels = loc, [<span class="string">"not know"</span>, <span class="string">"network"</span>, <span class="string">"WiFi"</span>, <span class="string">"cellular"</span>, <span class="string">"2G"</span>, <span class="string">"3G"</span>, <span class="string">"4G"</span>,<span class="string">"no"</span>]</span><br><span class="line">plt.xticks(loc, labels,fontsize=<span class="number">12</span>)</span><br></pre></td></tr></table></figure><a id="more"></a><p>二分类标签（0，1）分布：</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">sns.countplot(train_df[<span class="string">'favorite'</span>].sort_values())</span><br></pre></td></tr></table></figure><p>单个变量（连续或者类别）：</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">sns.distplot(train.revenue)</span><br><span class="line"><span class="comment"># blog</span></span><br></pre></td></tr></table></figure><p>如果连续变量不服从正太分布：</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">train[<span class="string">'logRevenue'</span>] = np.log1p(train[<span class="string">'revenue'</span>])</span><br><span class="line">sns.distplot(train[<span class="string">'logRevenue'</span>] )</span><br><span class="line"><span class="comment">#------EDA, Feature Engineering, LGB+XGB+CAT：电影预测比赛</span></span><br></pre></td></tr></table></figure><p>（最后需要变换回来）</p><p>对两个特征的连续变量：</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">sns.jointplot(x=<span class="string">"runtime"</span>, y=<span class="string">"revenue"</span>, data=train, height=<span class="number">11</span>, ratio=<span class="number">4</span>, color=<span class="string">"g"</span>)</span><br><span class="line">plt.show()</span><br><span class="line"><span class="comment">#------EDA, Feature Engineering, LGB+XGB+CAT：电影预测比赛</span></span><br></pre></td></tr></table></figure><p>对类别属性：</p><p>单个类别值（按顺序）计数：例如时间类特征</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br></pre></td><td class="code"><pre><span class="line">plt.figure(figsize=(<span class="number">20</span>,<span class="number">12</span>))</span><br><span class="line">sns.countplot(train[<span class="string">'release_month'</span>].sort_values())</span><br><span class="line">plt.title(<span class="string">"Release Month Count"</span>,fontsize=<span class="number">20</span>)</span><br><span class="line">loc, labels = plt.xticks()</span><br><span class="line">loc, labels = loc, [<span class="string">"Jan"</span>, <span class="string">"Feb"</span>, <span class="string">"Mar"</span>, <span class="string">"Apr"</span>, <span class="string">"May"</span>, <span class="string">"Jun"</span>, <span class="string">"Jul"</span>, <span class="string">"Aug"</span>, <span class="string">"Sep"</span>, <span class="string">"Oct"</span>, <span class="string">"Nov"</span>, <span class="string">"Dec"</span>]</span><br><span class="line">plt.xticks(loc, labels,fontsize=<span class="number">20</span>)</span><br><span class="line">plt.show()</span><br><span class="line"><span class="comment">#----------kaggle_baseline反欺诈算法挑战赛：IEEE</span></span><br></pre></td></tr></table></figure><p>例如二分类问题：</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment"># 对正负标签值查看关于某个特征的数值分布</span></span><br><span class="line">plt.figure(figsize=(<span class="number">9</span>,<span class="number">6</span>))</span><br><span class="line">sns.countplot(hue=<span class="string">'label'</span>,x=<span class="string">'ntt'</span>,data=All_data_for_train)</span><br><span class="line">loc, labels = plt.xticks()</span><br><span class="line">loc, labels = loc, [<span class="string">"not know"</span>, <span class="string">"network"</span>, <span class="string">"WiFi"</span>, <span class="string">"cellular"</span>, <span class="string">"2G"</span>, <span class="string">"3G"</span>, <span class="string">"4G"</span>,<span class="string">"no"</span>]</span><br><span class="line">plt.xticks(loc, labels,fontsize=<span class="number">12</span>)</span><br><span class="line"><span class="comment"># plt.xticks(fontsize=12,rotation=45)</span></span><br><span class="line">plt.title(<span class="string">"ntt Count"</span>,fontsize=<span class="number">15</span>)</span><br><span class="line"><span class="comment">#----------2019移动广告反欺诈算法挑战赛</span></span><br></pre></td></tr></table></figure><p>或者，以正负标签为X轴，观察其对另一特征的数值特点</p><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">sns.catplot(x=<span class="string">"has_homepage"</span>, y=<span class="string">"revenue"</span>, data=train)</span><br><span class="line">plt.title(<span class="string">'Revenue of movies with and without homepage'</span>)</span><br><span class="line"><span class="comment">#------EDA, Feature Engineering, LGB+XGB+CAT：电影预测比赛</span></span><br></pre></td></tr></table></figure>]]></content>
<tags>
<tag> EDA </tag>
<tag> 机器学习 </tag>
</tags>
</entry>
<entry>
<title>用GitHub Pages 和 Hexo搭建博客</title>
<link href="/2019/07/30/ckx3e5iwe001cnce27nbb4e43/"/>
<url>/2019/07/30/ckx3e5iwe001cnce27nbb4e43/</url>
<content type="html"><](你的链接地址)</span><br></pre></td></tr></table></figure><h4 id="评论"><a href="#评论" class="headerlink" title="评论"></a>评论</h4><p>设置<code>gitalk.id</code>实现多个页面共用一个评论框</p><p>在文章的Front-matter中：</p><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line">---</span><br><span class="line">gitalk:</span><br><span class="line"> id: /wiki/material-x/</span><br><span class="line">---</span><br></pre></td></tr></table></figure><h4 id="页面配置"><a href="#页面配置" class="headerlink" title="页面配置"></a>页面配置</h4><p>见<a href="https://xaoxuu.com/wiki/material-x/front-matter/index.html" target="_blank" rel="noopener">官方文档</a>.</p><h4 id="图床"><a href="#图床" class="headerlink" title="图床"></a>图床</h4><blockquote><p><a href="https://xaoxuu.com/wiki/vim-cn.sh//" target="_blank" rel="noopener">vim-cn.sh</a>.是一个十分干净好用的图床，本站以及博客中的所有图片都是使用的这个图床。所以，这是一个快速批量<strong>上传图片</strong>的脚本。（勿滥用）</p></blockquote><h3 id="关于添加评论插件和设置个性化域名遇到的错误"><a href="#关于添加评论插件和设置个性化域名遇到的错误" class="headerlink" title="关于添加评论插件和设置个性化域名遇到的错误"></a>关于添加评论插件和设置个性化域名遇到的错误</h3><hr><p>要设置个性化域名，为自己的域名做这些设置（为gitalk评论插件做准备）<img src="https://img.vim-cn.com/9c/76a0e975954ba7777552998278f695839ac226.png" alt>，然后从GitHub的博客存储库的设置最下面<code>Custom domain</code>填入自己的域名（我自己的是<code>lwhluvdemo.ml</code>），再在本地博客根目录的<code>source</code>下新建一个无后缀CNAME文件填入自己的个性域名，否则每次<code>hexo g -d</code>都会重置。</p><p>GitHub个人主页Settings->Developer settings->OAuth Apps，具体可以参考这个文档的</p><p><a href="https://xaoxuu.com/wiki/material-x/third-party-services/index.html" target="_blank" rel="noopener">Gitalk</a>，获得建立评论区<code>clientID</code>和<code>clientSecret</code>，之后在<code>Homepage URL</code>填自己博客的地址，例如我的博客：<a href="https://lwhluvdemo.github.io/">https://lwhluvdemo.github.io/</a></p><p>以及<code>Authorization callback URL</code>填写自己的域名地址，例如我的域名：<a href="https://lwhluvdemo.ml/" target="_blank" rel="noopener">https://lwhluvdemo.ml/</a>.</p><p>两个地址后面都要加 <strong>/</strong>。</p><p><strong>如果前面是按照我的设置来的话，一般不会有hexo配置gitalk 评论后无法初始化创建issue的问题</strong></p><h4 id="解决Error-Validation-Failed错误"><a href="#解决Error-Validation-Failed错误" class="headerlink" title="解决Error: Validation Failed错误"></a>解决Error: Validation Failed错误</h4><p>最后是根目录下<code>_config.yml</code>配置文件的改变，主要是将<code>permalink</code>的title变为id来避免超过50个字符的限制</p><blockquote><p>网上流传的很流行的md5方法等，我。。。。一个也不会用，主要是他们说的在哪哪添加文件，我根本就没有那些路径，所以想用也没办法。</p><p><a href="https://github.com/gitalk/gitalk/issues/102" target="_blank" rel="noopener">大部分与MD5相关的答案可以从这里找到</a>。</p></blockquote><figure class="highlight html"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line"># URL</span><br><span class="line">## If your site is put in a subdirectory, set url as 'http://yoursite.com/child' and root as '/child/'</span><br><span class="line">url: https://lwhluvdemo.github.io</span><br><span class="line">root: /</span><br><span class="line">permalink: :year/:month/:day/:id/</span><br><span class="line">permalink_defaults:</span><br></pre></td></tr></table></figure><figure class="highlight html"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br></pre></td><td class="code"><pre><span class="line">gitalk:</span><br><span class="line"> clientID: **************</span><br><span class="line"> clientSecret: ***********************</span><br><span class="line"> repo: yourname.github.io</span><br><span class="line"> owner: yourname</span><br><span class="line"> admin: [yourname]</span><br><span class="line"> id: decodeURI(window.location.pathname)</span><br></pre></td></tr></table></figure><p>一般提交后需要几分钟来重新部署，立刻刷新网页不会变……等等就好。</p><p>补充：</p><ul><li><a href="https://blog.csdn.net/IT__666/article/details/107053069?spm=1001.2101.3001.6650.3&utm_medium=distribute.pc_relevant.none-task-blog-2%7Edefault%7EOPENSEARCH%7Edefault-3.no_search_link&depth_1-utm_source=distribute.pc_relevant.none-task-blog-2%7Edefault%7EOPENSEARCH%7Edefault-3.no_search_link" target="_blank" rel="noopener">数学公式显示</a>。</li><li><a href="https://zhuanlan.zhihu.com/p/89803184" target="_blank" rel="noopener">Hexo版本升级</a>。</li></ul>]]></content>
<tags>
<tag> Hexo </tag>
<tag> GitHub Pages </tag>
</tags>
</entry>
</search>