{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Exercises\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exercise 1.1 \n",
    "What does `myString[0:10:2]` do?  \n",
    "What could the `2` mean in that context?\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "02468\n",
      "019876543210\n"
     ]
    }
   ],
   "source": [
    "myString = '012345678910'\n",
    "print(myString[0:10:2])\n",
    "# The 2 is the step/increment between each selected index in the sequence. In this case, every 2nd letter is kept.\n",
    "# This allows the following neat trick to reverse a string: have steps of -1\n",
    "print(myString[::-1])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "\n",
    "## Exercise 1.2 \n",
    "I am trying to make a division. Could you make my code work?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "3.0\n"
     ]
    }
   ],
   "source": [
    "a=12\n",
    "b='36'\n",
    "# The problem here, is that \"a\" in an int(), and \"b\" is a str().\n",
    "# -> we have to convert \"b\" to a number type (int or float) in order for the division to work.\n",
    "division = int(b)/a\n",
    "print(division)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "\n",
    "## Exercise 1.3 \n",
    "1. Put your favorite quote in a string variable."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "myQuote=\"- Regarde-moi, mon cher, et dis quelle espérance Pourrait bien me laisser cette protubérance !\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "2. Find a way to replace all spaces by '@' in the quote.\n",
    "Since `myQuote` is a string, we look at `help(str)` to search for methods available for string objects. We can see that there is a `replace()` method that allows replacing characters within a string."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "- Regarde-moi, mon cher, et dis quelle espérance Pourrait bien me laisser cette protubérance !\n",
      "-@Regarde-moi,@mon@cher,@et@dis@quelle@espérance@Pourrait@bien@me@laisser@cette@protubérance@!\n"
     ]
    }
   ],
   "source": [
    "modifedQuote = myQuote.replace(' ','@')\n",
    "print(myQuote)\n",
    "print(modifedQuote)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "\n",
    "## Exercise 1.4\n",
    "Consider the two dictionnaries below, describing the number of doctorates delivered in 1999 in the US (adapted from [nsf data](http://www.nsf.gov/statistics/infbrief/nsf11305/)):"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [],
   "source": [
    "science_doc = {\"Agricultural sciences\"\t: 1065,\n",
    "                \"Biochemistry\"\t: 759,\n",
    "                \"Molecular biology\"\t: 716,\n",
    "                \"Neurosciences\"\t: 431,\n",
    "                \"Other biological sciences\"\t: 3675,\n",
    "                \"Computer sciences\"\t: 856,\n",
    "                \"Earth, atmospheric, and ocean sciences\"\t: 706,\n",
    "                \"Mathematics\"\t: 1083,\n",
    "                \"Chemistry\"\t: 2132,\n",
    "                \"Physics and astronomy\"\t: 715,\n",
    "                \"Astrology\" : 2012, \n",
    "                \"Psychology\"\t: 3668,\n",
    "                \"Social sciences\"\t: 4063}\n",
    "\n",
    "engineering_doc = {\"Aerospace/aeronautical engineering\"\t: 206,\n",
    "                    \"Chemical engineering\"\t: 576,\n",
    "                    \"Civil engineering\"\t: 506,\n",
    "                    \"Electrical engineering\"\t: 1236,\n",
    "                    \"Industrial/manufacturing engineering\"\t: 211,\n",
    "                    \"Materials science engineering\"\t: 393,\n",
    "                    \"Mechanical engineering\"\t: 786,\n",
    "                    \"Other engineering\"\t: 1416 }\n",
    "\n",
    "humanities_doc = {\"Foreign languages and literature\"\t : 626,\n",
    "                    \"History\"\t : 960,\n",
    "                    \"Letters\"\t : 1516,\n",
    "                    \"Other humanities\"\t : 1934}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1. merge these three dictionnaries in a `all_doc` dictionnary.\n",
    "Looking at `help(dict)`, we find there is an `update` method that \"Update D from dict/iterable E and F.\"\n",
    "This method adds key-value pairs to a dicitonnary from another dictionnary.\n",
    "   "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "{'Agricultural sciences': 1065, 'Biochemistry': 759, 'Molecular biology': 716, 'Neurosciences': 431, 'Other biological sciences': 3675, 'Computer sciences': 856, 'Earth, atmospheric, and ocean sciences': 706, 'Mathematics': 1083, 'Chemistry': 2132, 'Physics and astronomy': 715, 'Astrology': 2012, 'Psychology': 3668, 'Social sciences': 4063, 'Aerospace/aeronautical engineering': 206, 'Chemical engineering': 576, 'Civil engineering': 506, 'Electrical engineering': 1236, 'Industrial/manufacturing engineering': 211, 'Materials science engineering': 393, 'Mechanical engineering': 786, 'Other engineering': 1416, 'Foreign languages and literature': 626, 'History': 960, 'Letters': 1516, 'Other humanities': 1934}\n",
      "Are the two dict identical? True\n"
     ]
    }
   ],
   "source": [
    "all_doc = {}\n",
    "all_doc.update(science_doc)\n",
    "all_doc.update(engineering_doc)\n",
    "all_doc.update(humanities_doc)\n",
    "print(all_doc)\n",
    "\n",
    "# Alternatively, we could also copy one of the dict as a starting point with the \"copy()\" method of dict:\n",
    "all_doc_2 = science_doc.copy()\n",
    "all_doc_2.update(engineering_doc)\n",
    "all_doc_2.update(humanities_doc)\n",
    "print('Are the two dict identical?', all_doc == all_doc_2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "2. what is the length of the `all_doc` dictionnary"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Length of the dicitonnary: 25\n"
     ]
    }
   ],
   "source": [
    "print( \"Length of the dicitonnary:\" , len( all_doc ) )"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "3. add the \"Health\" doctorates of the `all_doc` dictionnary, with 1407 doctorates awarded"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [],
   "source": [
    "all_doc['Health'] = 1407"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "4. multiply by 2 the number of \"Physics and astronomy\" doctorates in `all_doc`"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Physics and astronomy : 715\n",
      "Physics and astronomy : 1430\n"
     ]
    }
   ],
   "source": [
    "key = 'Physics and astronomy'\n",
    "print(key , ':' , all_doc[key] )\n",
    "all_doc[key] = all_doc[key] * 2             # Alternatively: all_doc[key] *= 2\n",
    "print(key , ':' , all_doc[key] )"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "5. remove the Astology key from `all_doc`\n",
    "Look at `help(dict)`: there is a pop method that \"remove specified key\"."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "{'Agricultural sciences': 1065, 'Biochemistry': 759, 'Molecular biology': 716, 'Neurosciences': 431, 'Other biological sciences': 3675, 'Computer sciences': 856, 'Earth, atmospheric, and ocean sciences': 706, 'Mathematics': 1083, 'Chemistry': 2132, 'Physics and astronomy': 1430, 'Psychology': 3668, 'Social sciences': 4063, 'Aerospace/aeronautical engineering': 206, 'Chemical engineering': 576, 'Civil engineering': 506, 'Electrical engineering': 1236, 'Industrial/manufacturing engineering': 211, 'Materials science engineering': 393, 'Mechanical engineering': 786, 'Other engineering': 1416, 'Foreign languages and literature': 626, 'History': 960, 'Letters': 1516, 'Other humanities': 1934, 'Health': 1407}\n"
     ]
    }
   ],
   "source": [
    "all_doc.pop(\"Astrology\")\n",
    "print(all_doc)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "\n",
    "## Exercise 1.5\n",
    "Set a variable called `isDivisible` to `True` if 7453102041884640481897 is evenly divisible by 135138672131"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "True\n"
     ]
    }
   ],
   "source": [
    "# The % operator gives the remainder of the euclidian division of two numbers.\n",
    "# A number x is evenly divisible by y if the remainder of the euclidian division is equal to 0,\n",
    "# or, to put it in semi-code : x % y is equal to 0.\n",
    "# We then use the \"==\" comparison operator to return a booleann value (True or False).\n",
    "isDivisible = (7453102041884640481897 % 135138672131) == 0\n",
    "print(isDivisible)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "\n",
    "## Exercise 1.6\n",
    "Use the `find()` method to determine the position(s) of the motif \"CTCGA\" in the following sequence:\n",
    "    \n",
    "GTGCCCCTCGAGAGGAGGGCGCGCGCCGCGCGCTCGACGCGATCGGCGCTCAGCGAGCGAGCTCCTCGAAGCGATCCGCGCGCGCT\n",
    "\n",
    "Is there a limitation here?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "6\n",
      "32\n",
      "64\n",
      "-1\n",
      "[6, 32, 64]\n"
     ]
    }
   ],
   "source": [
    "seq = \"GTGCCCCTCGAGAGGAGGGCGCGCGCCGCGCGCTCGACGCGATCGGCGCTCAGCGAGCGAGCTCCTCGAAGCGATCCGCGCGCGCT\"\n",
    "print(seq.find(\"CTCGA\"))\n",
    "\n",
    "# Limitation:\n",
    "# find() only returns the 1st occurence of the motif... \n",
    "# To find other occurrences of \"CTCGA\", we can use the optionnal argument \"start\" of the \n",
    "# \"find()\"\" method and pass the index of the last found motif + 1:\n",
    "print(seq.find(\"CTCGA\", 7))\n",
    "print(seq.find(\"CTCGA\", 33))\n",
    "print(seq.find(\"CTCGA\", 65))     # Note that \"find()\" returns -1 when no match is found.\n",
    "\n",
    "# Of course, it would be nicer to automate this:\n",
    "# Here is a sneak peak at how to do this using loops, which we'll see in the next lesson.\n",
    "positions = []              # Instantiate a list were we will store the motif positions\n",
    "pos = seq.find('CTCGA')\n",
    "while pos != -1 :           # While the motif is found...\n",
    "    positions.append(pos)   # ... add the motif to the result list.\n",
    "    pos = seq.find('CTCGA' , pos + 1) # Attempt to find another motif\n",
    "print(positions)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "\n",
    "## Exercise 1.7\n",
    "1. Build a list containing the name of all swiss cantons."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [],
   "source": [
    "cantonList = ['Zürich',\n",
    "'Bern',\n",
    "'Luzern',\n",
    "'Uri',\n",
    "'Schwyz',\n",
    "'Obwalden',\n",
    "'Nidwalden',\n",
    "'Glarus',\n",
    "'Zug',\n",
    "'Fribourg',\n",
    "'Solothurn',\n",
    "'Basel-Stadt',\n",
    "'Basel-Landschaft',\n",
    "'Schaffhausen',\n",
    "'Appenzell Ausserrhoden',\n",
    "'Appenzell Innerrhoden',\n",
    "'St. Gallen',\n",
    "'Graubünden',\n",
    "'Aargau',\n",
    "'Thurgau',\n",
    "'Ticino',\n",
    "'Vaud',\n",
    "'Valais',\n",
    "'Neuchâtel',\n",
    "'Geneva',\n",
    "'Jura']"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "2. sort it alphabetically\n",
    "Looking at `help(list)`, we can see that there is a `sort` method that does just this:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "['Aargau', 'Appenzell Ausserrhoden', 'Appenzell Innerrhoden', 'Basel-Landschaft', 'Basel-Stadt', 'Bern', 'Fribourg', 'Geneva', 'Glarus', 'Graubünden', 'Jura', 'Luzern', 'Neuchâtel', 'Nidwalden', 'Obwalden', 'Schaffhausen', 'Schwyz', 'Solothurn', 'St. Gallen', 'Thurgau', 'Ticino', 'Uri', 'Valais', 'Vaud', 'Zug', 'Zürich']\n"
     ]
    }
   ],
   "source": [
    "cantonList.sort()\n",
    "print(cantonList)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "3. reverse it\n",
    "Again, looking at `help(list)` shows that there is a `reverse` method that does this:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "['Zürich', 'Zug', 'Vaud', 'Valais', 'Uri', 'Ticino', 'Thurgau', 'St. Gallen', 'Solothurn', 'Schwyz', 'Schaffhausen', 'Obwalden', 'Nidwalden', 'Neuchâtel', 'Luzern', 'Jura', 'Graubünden', 'Glarus', 'Geneva', 'Fribourg', 'Bern', 'Basel-Stadt', 'Basel-Landschaft', 'Appenzell Innerrhoden', 'Appenzell Ausserrhoden', 'Aargau']\n"
     ]
    }
   ],
   "source": [
    "cantonList.reverse()\n",
    "print(cantonList)\n",
    "# Alternatively, you can use the slicing trick seen earlier: cantonList[::-1]"
   ]
  }
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