notebooks/answers/3 - Control Structures.ipynb

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   "source": [
    "# Control Structures"
   ]
  },
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   "source": [
    "## Simple for loop \n",
    "\n",
    "Write a ``for`` loop which iterates over the list of breakfast items \"sausage\", \"eggs\", \"bacon\" and \"spam\" and prints out the name of item\n"
   ]
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   "source": [
    "breakfast = [\"sausage\", \"eggs\", \"bacon\", \"spam\"]\n",
    "for item in breakfast:\n",
    "    print(item)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Write then a ``for`` which loop determines the squares of the odd\n",
    "integers up to 10. Use the ``range()`` function."
   ]
  },
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   "source": [
    "squares = []\n",
    "for i in range(1, 10, 2):\n",
    "    squares.append(i**2)\n",
    "print(squares)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Looping through a dictionary\n",
    "Write a loop that prints out the names of the fruits in the dictionary containing the fruit prices."
   ]
  },
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   "source": [
    "fruits = {'banana' : 5, 'strawberry' : 7, 'pineapple' : 3}\n",
    "for fruit in fruits:\n",
    "    print(fruit)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Next, write a loop that sums up the prices."
   ]
  },
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   "source": [
    "sum = 0\n",
    "for price in fruits.values():\n",
    "    sum += price\n",
    "print(sum)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## While loop\n",
    "\n",
    "Fibonacci numbers are a sequence of integers defined by the recurrence relation\n",
    "```\n",
    "\t\tF[n] = F[n-1] + F[n-2]\n",
    "```\n",
    "with the initial values F[0]=0, F[1]=1. \n",
    "Create a list of Fibonacci numbers F[n] < 100 using a while loop."
   ]
  },
  {
   "cell_type": "code",
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   "outputs": [],
   "source": [
    "f = [0, 1]\n",
    "while True:\n",
    "    new = f[-1] + f[-2]\n",
    "    if new > 100:\n",
    "        break\n",
    "    f.append(new)\n",
    "print(f)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## If - else\n",
    "\n",
    "Write a control structure which checks whether an integer is\n",
    "negative, zero, or belongs to the prime numbers 3,5,7,11,17\n",
    "and perform e.g. corresponding print statement.\n",
    "\n",
    "Use keyword ``in`` when checking for belonging to prime numbers."
   ]
  },
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   "source": [
    "number = 7\n",
    "if number < 0:\n",
    "    print(\"Negative\")\n",
    "elif number == 0:\n",
    "    print(\"Zero\")\n",
    "elif number in [3, 5, 7, 11, 17]:\n",
    "    print(\"Prime\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Advanced exercises\n",
    "\n",
    "Don't worry if you don't have time to finish all of these. They are not essential."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Looping through multidimensional lists\n",
    "\n",
    "Start from a two dimensional list of (x,y) value pairs, and sort it according to y values. (Hint: you may need to create a temporary list)."
   ]
  },
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   "source": [
    "xys =  [[2, 3], [0, -1], [4, -2], [1, 6]]\n",
    "tmp = []\n",
    "for x, y in xys:\n",
    "    tmp.append([y,x])\n",
    "tmp.sort()\n",
    "for i, (y,x) in enumerate(tmp):\n",
    "    xys[i] = [x,y]\n",
    "print(xys)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Next, create a new list containing only the sorted y values."
   ]
  },
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   "outputs": [],
   "source": [
    "ys = []\n",
    "for x, y in xys:\n",
    "    ys.append(y)\n",
    "print(ys)"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Finally, create a new list consisting of sums the (x,y) pairs where both x and y are positive."
   ]
  },
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   "source": [
    "sums = []\n",
    "for x, y in xys:\n",
    "    if x > 0 and y > 0:\n",
    "        sums.append(x + y)\n",
    "print(sums)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "List comprehension is often convenient in this kind of situations:"
   ]
  },
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   "source": [
    "xys = [[2, 3], [0, -1], [4, -2], [1, 6]]\n",
    "tmp = [[y, x] for x, y in xys]\n",
    "tmp.sort()\n",
    "xys = [[x, y] for y, x in tmp]\n",
    "\n",
    "# One liner is possible but not very readable anymore:\n",
    "xys = [[x, y] for y, x in sorted([[ytmp, xtmp] for xtmp, ytmp in xys])]\n",
    "\n",
    "# Summing positives with one liner is ok:\n",
    "sums = [x+y for x,y in xys if x > 0 and y > 0]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## FizzBuzz\n",
    "\n",
    "This is a classic job interview question. Depending on the interviewer or interviewee it can filter out up to 95% of the interviewees for a position. The task is not difficult but it's easy to make simple mistakes.\n",
    "\n",
    "If a number is divisible by 3, instead of the number print \"Fizz\", if a number is divisible by 5, print \"Buzz\" and if the number is divisible by both 3 and 5, print \"FizzBuzz\". "
   ]
  },
  {
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     "text": [
      "1\n",
      "2\n",
      "Fizz\n",
      "3\n",
      "4\n",
      "Buzz\n",
      "5\n",
      "Fizz\n",
      "6\n",
      "7\n",
      "8\n",
      "Fizz\n",
      "9\n",
      "Buzz\n",
      "10\n",
      "11\n",
      "Fizz\n",
      "12\n",
      "13\n",
      "14\n",
      "FizzBuzz\n",
      "15\n",
      "16\n",
      "17\n",
      "Fizz\n",
      "18\n",
      "19\n",
      "Buzz\n",
      "20\n",
      "Fizz\n",
      "21\n",
      "22\n",
      "23\n",
      "Fizz\n",
      "24\n",
      "Buzz\n",
      "25\n",
      "26\n",
      "Fizz\n",
      "27\n",
      "28\n",
      "29\n",
      "FizzBuzz\n",
      "30\n",
      "31\n",
      "32\n",
      "Fizz\n",
      "33\n",
      "34\n",
      "Buzz\n",
      "35\n",
      "Fizz\n",
      "36\n",
      "37\n",
      "38\n",
      "Fizz\n",
      "39\n",
      "Buzz\n",
      "40\n",
      "41\n",
      "Fizz\n",
      "42\n",
      "43\n",
      "44\n",
      "FizzBuzz\n",
      "45\n",
      "46\n",
      "47\n",
      "Fizz\n",
      "48\n",
      "49\n",
      "Buzz\n",
      "50\n",
      "Fizz\n",
      "51\n",
      "52\n",
      "53\n",
      "Fizz\n",
      "54\n",
      "Buzz\n",
      "55\n",
      "56\n",
      "Fizz\n",
      "57\n",
      "58\n",
      "59\n",
      "FizzBuzz\n",
      "60\n",
      "61\n",
      "62\n",
      "Fizz\n",
      "63\n",
      "64\n",
      "Buzz\n",
      "65\n",
      "Fizz\n",
      "66\n",
      "67\n",
      "68\n",
      "Fizz\n",
      "69\n",
      "Buzz\n",
      "70\n",
      "71\n",
      "Fizz\n",
      "72\n",
      "73\n",
      "74\n",
      "FizzBuzz\n",
      "75\n",
      "76\n",
      "77\n",
      "Fizz\n",
      "78\n",
      "79\n",
      "Buzz\n",
      "80\n",
      "Fizz\n",
      "81\n",
      "82\n",
      "83\n",
      "Fizz\n",
      "84\n",
      "Buzz\n",
      "85\n",
      "86\n",
      "Fizz\n",
      "87\n",
      "88\n",
      "89\n",
      "FizzBuzz\n",
      "90\n",
      "91\n",
      "92\n",
      "Fizz\n",
      "93\n",
      "94\n",
      "Buzz\n",
      "95\n",
      "Fizz\n",
      "96\n",
      "97\n",
      "98\n",
      "Fizz\n",
      "99\n",
      "Buzz\n",
      "100\n"
     ]
    }
   ],
   "source": [
    "for number in range(1, 101):\n",
    "    if number % 3 == 0 and number % 5 == 0:\n",
    "        print(\"FizzBuzz\")\n",
    "    elif number % 3 == 0:\n",
    "        print(\"Fizz\")\n",
    "    elif number % 5 == 0:\n",
    "        print(\"Buzz\")\n",
    "    print(number)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "*Food for thought:* How do people commonly fail this test and why?"
   ]
  },
  {
   "cell_type": "code",
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   "outputs": [],
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Breaking\n",
    "\n",
    "The python `random` module generates pseudorandom numbers.\n",
    "\n",
    "Write a *while* loop that runs until \n",
    "the output of random.random() is below 0.1 and `break` when \n",
    "the value is below 0.1."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "done\n"
     ]
    }
   ],
   "source": [
    "import random\n",
    "\n",
    "while True:\n",
    "    value = random.random() \n",
    "    if value < 0.1:\n",
    "        break\n",
    "print(\"done\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## List comprehension\n",
    "\n",
    "Using a list comprehension create a new list, `temperatures_kelvin` from following Celsius temperatures and convert them by adding the value 273.15 to each."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "temperatures_celsius = [0, -15, 20.15, 13.3, -5.2]\n",
    "\n",
    "temperatures_kelvin = [c+273.15 for c in temperatures_celsius]"
   ]
  }
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