Sets.ipynb

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      "language": "python"
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  "cells": [
    {
      "cell_type": "markdown",
      "source": "<center>\n    <img src=\"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/IDSNlogo.png\" width=\"300\" alt=\"cognitiveclass.ai logo\">\n</center>\n\n# Sets in Python\n\nEstimated time needed: **20** minutes\n\n## Objectives\n\nAfter completing this lab you will be able to:\n\n*   Work with sets in Python, including operations and logic operations.\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "<h2>Table of Contents</h2>\n<div class=\"alert alert-block alert-info\" style=\"margin-thttps://op/?utm_medium=Exinfluencer&utm_source=Exinfluencer&utm_content=000026UJ&utm_term=10006555&utm_id=NA-SkillsNetwork-Channel-SkillsNetworkCoursesIBMDeveloperSkillsNetworkPY0101ENSkillsNetwork19487395-2021-01-01: 20px\">\n    <ul>\n        <li>\n            <a href=\"#Sets\">Sets</a>\n            <ul>\n                <li><a href=\"#Set-Content\">Set Content</a></li>\n                <li><a href=\"#Set-Operations\">Set Operations</a></li>\n                <li><a href=\"#Sets-Logic-Operations\">Sets Logic Operations</a></li>\n            </ul>\n        </li>\n        <li>\n            <a href=\"#Quiz-on-Sets\">Quiz on Sets</a>\n        </li>\n    </ul>\n\n</div>\n\n<hr>\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "## Sets\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "### Set Content\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "A set is a unique collection of objects in Python. You can denote a set with a pair of curly brackets <b>{}</b>. Python will automatically remove duplicate items:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Create a set\n\nset1 = {\"pop\", \"rock\", \"soul\", \"hard rock\", \"rock\", \"R&B\", \"rock\", \"disco\"}\nset1",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 1,
          "output_type": "execute_result",
          "data": {
            "text/plain": "{'R&B', 'disco', 'hard rock', 'pop', 'rock', 'soul'}"
          },
          "metadata": {}
        }
      ],
      "execution_count": 1
    },
    {
      "cell_type": "markdown",
      "source": "The process of mapping is illustrated in the figure:\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "<img src=\"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/labs/Module%202/images/SetsUnique.png\" width=\"1100\">\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "You can also  create a set from a list as follows:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Convert list to set\n\nalbum_list = [ \"Michael Jackson\", \"Thriller\", 1982, \"00:42:19\", \\\n              \"Pop, Rock, R&B\", 46.0, 65, \"30-Nov-82\", None, 10.0]\nalbum_set = set(album_list)             \nalbum_set",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 2,
          "output_type": "execute_result",
          "data": {
            "text/plain": "{'00:42:19',\n 10.0,\n 1982,\n '30-Nov-82',\n 46.0,\n 65,\n 'Michael Jackson',\n None,\n 'Pop, Rock, R&B',\n 'Thriller'}"
          },
          "metadata": {}
        }
      ],
      "execution_count": 2
    },
    {
      "cell_type": "markdown",
      "source": "Now let us create a set of  genres:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Convert list to set\n\nmusic_genres = set([\"pop\", \"pop\", \"rock\", \"folk rock\", \"hard rock\", \"soul\", \\\n                    \"progressive rock\", \"soft rock\", \"R&B\", \"disco\"])\nmusic_genres",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 3,
          "output_type": "execute_result",
          "data": {
            "text/plain": "{'R&B',\n 'disco',\n 'folk rock',\n 'hard rock',\n 'pop',\n 'progressive rock',\n 'rock',\n 'soft rock',\n 'soul'}"
          },
          "metadata": {}
        }
      ],
      "execution_count": 3
    },
    {
      "cell_type": "markdown",
      "source": "### Set Operations \n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "Let us go over set operations, as these can be used to change the set. Consider the set <b>A</b>:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Sample set\n\nA = set([\"Thriller\", \"Back in Black\", \"AC/DC\"])\nA",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 4,
          "output_type": "execute_result",
          "data": {
            "text/plain": "{'AC/DC', 'Back in Black', 'Thriller'}"
          },
          "metadata": {}
        }
      ],
      "execution_count": 4
    },
    {
      "cell_type": "markdown",
      "source": "We can add an element to a set using the <code>add()</code> method:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Add element to set\n\nA.add(\"NSYNC\")\nA",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 5,
          "output_type": "execute_result",
          "data": {
            "text/plain": "{'AC/DC', 'Back in Black', 'NSYNC', 'Thriller'}"
          },
          "metadata": {}
        }
      ],
      "execution_count": 5
    },
    {
      "cell_type": "markdown",
      "source": "If we add the same element twice, nothing will happen as there can be no duplicates in a set:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Try to add duplicate element to the set\n\nA.add(\"NSYNC\")\nA",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 7,
          "output_type": "execute_result",
          "data": {
            "text/plain": "{'AC/DC', 'Back in Black', 'NSYNC', 'Thriller'}"
          },
          "metadata": {}
        }
      ],
      "execution_count": 7
    },
    {
      "cell_type": "markdown",
      "source": "We can remove an item from a set using the <code>remove</code> method:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Remove the element from set\n\nA.remove(\"NSYNC\")\nA",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 8,
          "output_type": "execute_result",
          "data": {
            "text/plain": "{'AC/DC', 'Back in Black', 'Thriller'}"
          },
          "metadata": {}
        }
      ],
      "execution_count": 8
    },
    {
      "cell_type": "markdown",
      "source": "We can verify if an element is in the set using the <code>in</code> command:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Verify if the element is in the set\n\n\"AC/DC\" in A",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 9,
          "output_type": "execute_result",
          "data": {
            "text/plain": "True"
          },
          "metadata": {}
        }
      ],
      "execution_count": 9
    },
    {
      "cell_type": "markdown",
      "source": "### Sets Logic Operations\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "Remember that with sets you can check the difference between sets, as well as the symmetric difference, intersection, and union:\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "Consider the following two sets:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Sample Sets\n\nalbum_set1 = set([\"Thriller\", 'AC/DC', 'Back in Black'])\nalbum_set2 = set([ \"AC/DC\", \"Back in Black\", \"The Dark Side of the Moon\"])",
      "metadata": {
        "trusted": true
      },
      "outputs": [],
      "execution_count": 10
    },
    {
      "cell_type": "markdown",
      "source": "<img src=\"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/labs/Module%202/images/SetsSamples.png\" width=\"650\">\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Print two sets\n\nalbum_set1, album_set2",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 11,
          "output_type": "execute_result",
          "data": {
            "text/plain": "({'AC/DC', 'Back in Black', 'Thriller'},\n {'AC/DC', 'Back in Black', 'The Dark Side of the Moon'})"
          },
          "metadata": {}
        }
      ],
      "execution_count": 11
    },
    {
      "cell_type": "markdown",
      "source": "As both sets contain <b>AC/DC</b> and <b>Back in Black</b> we represent these common elements with the intersection of two circles.\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "<img src=\"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/labs/Module%202/images/SetsLogic.png\" width=\"650\">\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "You can find the intersect of two sets as follow using <code>&</code>:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Find the intersections\n\nintersection = album_set1 & album_set2\nintersection",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 12,
          "output_type": "execute_result",
          "data": {
            "text/plain": "{'AC/DC', 'Back in Black'}"
          },
          "metadata": {}
        }
      ],
      "execution_count": 12
    },
    {
      "cell_type": "markdown",
      "source": "You can find all the elements that are only contained in <code>album_set1</code> using the <code>difference</code> method:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Find the difference in set1 but not set2\n\nalbum_set1.difference(album_set2)  ",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 13,
          "output_type": "execute_result",
          "data": {
            "text/plain": "{'Thriller'}"
          },
          "metadata": {}
        }
      ],
      "execution_count": 13
    },
    {
      "cell_type": "markdown",
      "source": "You only need to consider elements in <code>album_set1</code>; all the elements in <code>album_set2</code>, including the intersection, are not included.\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "<img src=\"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/labs/Module%202/images/SetsLeft.png\" width=\"650\">\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "The elements in <code>album_set2</code> but not in <code>album_set1</code> is given by:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "album_set2.difference(album_set1)  ",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 14,
          "output_type": "execute_result",
          "data": {
            "text/plain": "{'The Dark Side of the Moon'}"
          },
          "metadata": {}
        }
      ],
      "execution_count": 14
    },
    {
      "cell_type": "markdown",
      "source": "<img src=\"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/labs/Module%202/images/SetsRight.png\" width=\"650\">\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "You can also find the intersection of <code>album_list1</code> and <code>album_list2</code>, using the <code>intersection</code> method:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Use intersection method to find the intersection of album_list1 and album_list2\n\nalbum_set1.intersection(album_set2)   ",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 15,
          "output_type": "execute_result",
          "data": {
            "text/plain": "{'AC/DC', 'Back in Black'}"
          },
          "metadata": {}
        }
      ],
      "execution_count": 15
    },
    {
      "cell_type": "markdown",
      "source": "This corresponds to the intersection of the two circles:\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "<img src=\"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/labs/Module%202/images/SetsIntersect.png\" width=\"650\">\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "The union corresponds to all the elements in both sets, which is represented by coloring both circles:\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "<img src=\"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/labs/Module%202/images/SetsUnion.png\" width=\"650\">\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "The union is given by:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Find the union of two sets\n\nalbum_set1.union(album_set2)",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 16,
          "output_type": "execute_result",
          "data": {
            "text/plain": "{'AC/DC', 'Back in Black', 'The Dark Side of the Moon', 'Thriller'}"
          },
          "metadata": {}
        }
      ],
      "execution_count": 16
    },
    {
      "cell_type": "markdown",
      "source": "And you can check if a set is a superset or subset of another set, respectively, like this:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Check if superset\n\nset(album_set1).issuperset(album_set2)   ",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 17,
          "output_type": "execute_result",
          "data": {
            "text/plain": "False"
          },
          "metadata": {}
        }
      ],
      "execution_count": 17
    },
    {
      "cell_type": "code",
      "source": "# Check if subset\n\nset(album_set2).issubset(album_set1)     ",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 18,
          "output_type": "execute_result",
          "data": {
            "text/plain": "False"
          },
          "metadata": {}
        }
      ],
      "execution_count": 18
    },
    {
      "cell_type": "markdown",
      "source": "Here is an example where <code>issubset()</code> and <code>issuperset()</code> return true:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Check if subset\n\nset({\"Back in Black\", \"AC/DC\"}).issubset(album_set1) ",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 19,
          "output_type": "execute_result",
          "data": {
            "text/plain": "True"
          },
          "metadata": {}
        }
      ],
      "execution_count": 19
    },
    {
      "cell_type": "code",
      "source": "# Check if superset\n\nalbum_set1.issuperset({\"Back in Black\", \"AC/DC\"})   ",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 20,
          "output_type": "execute_result",
          "data": {
            "text/plain": "True"
          },
          "metadata": {}
        }
      ],
      "execution_count": 20
    },
    {
      "cell_type": "markdown",
      "source": "<hr>\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "## Quiz on Sets\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "Convert the list <code>\\['rap','house','electronic music', 'rap']</code> to a set:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Write your code below and press Shift+Enter to execute\nset(['rap','house','electronic music','rap'])",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 21,
          "output_type": "execute_result",
          "data": {
            "text/plain": "{'electronic music', 'house', 'rap'}"
          },
          "metadata": {}
        }
      ],
      "execution_count": 21
    },
    {
      "cell_type": "markdown",
      "source": "<details><summary>Click here for the solution</summary>\n\n```python\nset(['rap','house','electronic music','rap'])\n\n```\n\n</details>\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "<hr>\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "Consider the list <code>A = \\[1, 2, 2, 1]</code> and set <code>B = set(\\[1, 2, 2, 1])</code>, does <code>sum(A) == sum(B)</code>?\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Write your code below and press Shift+Enter to execute\nA = [1, 2, 2, 1]  \nB = set([1, 2, 2, 1])\nprint(\"the sum of A is:\", sum(A))\nprint(\"the sum of B is:\", sum(B))",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "name": "stdout",
          "text": "the sum of A is: 6\nthe sum of B is: 3\n",
          "output_type": "stream"
        }
      ],
      "execution_count": 22
    },
    {
      "cell_type": "markdown",
      "source": "<details><summary>Click here for the solution</summary>\n\n```python\nA = [1, 2, 2, 1]  \nB = set([1, 2, 2, 1])\nprint(\"the sum of A is:\", sum(A))\nprint(\"the sum of B is:\", sum(B))\n\n```\n\n</details>\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "<hr>\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "Create a new set <code>album_set3</code> that is the union of <code>album_set1</code> and <code>album_set2</code>:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Write your code below and press Shift+Enter to execute\n\nalbum_set1 = set([\"Thriller\", 'AC/DC', 'Back in Black'])\nalbum_set2 = set([ \"AC/DC\", \"Back in Black\", \"The Dark Side of the Moon\"])\nalbum_set3 = album_set1.union(album_set2)\nalbum_set3",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 23,
          "output_type": "execute_result",
          "data": {
            "text/plain": "{'AC/DC', 'Back in Black', 'The Dark Side of the Moon', 'Thriller'}"
          },
          "metadata": {}
        }
      ],
      "execution_count": 23
    },
    {
      "cell_type": "markdown",
      "source": "<details><summary>Click here for the solution</summary>\n\n```python\nalbum_set3 = album_set1.union(album_set2)\nalbum_set3\n\n```\n\n</details>\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "<hr>\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "Find out if <code>album_set1</code> is a subset of <code>album_set3</code>:\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "# Write your code below and press Shift+Enter to execute\nalbum_set1.issubset(album_set3)",
      "metadata": {
        "trusted": true
      },
      "outputs": [
        {
          "execution_count": 24,
          "output_type": "execute_result",
          "data": {
            "text/plain": "True"
          },
          "metadata": {}
        }
      ],
      "execution_count": 24
    },
    {
      "cell_type": "markdown",
      "source": "<details><summary>Click here for the solution</summary>\n\n```python\nalbum_set1.issubset(album_set3)\n\n```\n\n</details>\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "<hr>\n<h2>The last exercise!</h2>\n<p>Congratulations, you have completed your first lesson and hands-on lab in Python. \n<hr>\n",
      "metadata": {}
    },
    {
      "cell_type": "markdown",
      "source": "## Author\n\n<a href=\"https://www.linkedin.com/in/joseph-s-50398b136/?utm_medium=Exinfluencer&utm_source=Exinfluencer&utm_content=000026UJ&utm_term=10006555&utm_id=NA-SkillsNetwork-Channel-SkillsNetworkCoursesIBMDeveloperSkillsNetworkPY0101ENSkillsNetwork19487395-2021-01-01\" target=\"_blank\">Joseph Santarcangelo</a>\n\n## Other contributors\n\n<a href=\"https://www.linkedin.com/in/jiahui-mavis-zhou-a4537814a?utm_medium=Exinfluencer&utm_source=Exinfluencer&utm_content=000026UJ&utm_term=10006555&utm_id=NA-SkillsNetwork-Channel-SkillsNetworkCoursesIBMDeveloperSkillsNetworkPY0101ENSkillsNetwork19487395-2021-01-01\">Mavis Zhou</a>\n\n<!--## Change Log--!>\n\n<!--| Date (YYYY-MM-DD) | Version | Changed By | Change Description                 |\n--!>\n<!--| ----------------- | ------- | ---------- | ---------------------------------- |\n--!>\n<!--| 2023-11-01        | 2.2     | Abhishek Gagneja | Updated instructions  |\n--!>\n<!--| 2022-01-10        | 2.1     | Malika     | Removed the readme for GitShare    |\n--!>\n<!--| 2020-08-26        | 2.0     | Lavanya    | Moved lab to course repo in GitLab |\n--!>\n\n## <h3 align=\"center\"> © IBM Corporation 2023. All rights reserved. <h3/>\n",
      "metadata": {}
    },
    {
      "cell_type": "code",
      "source": "",
      "metadata": {},
      "outputs": [],
      "execution_count": null
    }
  ]
}