notebooks/01_python_basics_exercises.ipynb

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    "# Exercises module 1 - python basics\n",
    "-------------------------------------------------------\n",
    "\n",
    "<br>\n",
    "\n",
    "## Exercise 1.1 \n",
    "\n",
    "What does `my_string[0:10:2]` do in?\n",
    "\n",
    "```python\n",
    "my_string = \"0123456789abcdef\"\n",
    "my_string[0:10:2]\n",
    "```\n",
    "\n",
    "What could the `2` mean in that context?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "\n",
    "### Solution:\n",
    "Uncomment and run the cell below to show the solution."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# %load solutions/solution_11.py"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "<br>\n",
    "\n",
    "## Exercise 1.2 \n",
    "\n",
    "I am trying to make a division. Could you make my code work?\n",
    "```python\n",
    "a = 12\n",
    "b = \"36\"\n",
    "division = b/a\n",
    "print(division)\n",
    "```"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "\n",
    "### Solution:\n",
    "Uncomment and run the cell below to show the solution."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# %load solutions/solution_12.py"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "<br>\n",
    "\n",
    "## Exercise 1.3\n",
    "\n",
    "Use the `find()` method of strings to determine the position(s) of the motif \"CTCGA\":\n",
    "    \n",
    "GTGCCCCTCGAGAGGAGGGCGCGCGCCGCGCGCTCGACGCGATCGGCGCTCAGCGAGCGAGCTCCTCGAAGCGATCCGCGCGCGCT\n",
    "\n",
    "What is a limitation of `find()` in this application?\n",
    "\n",
    "> Reminder, a **method** is a function associated to a particular object.\n",
    "> It is called using the syntax **`object.method()`**.\n",
    ">\n",
    "> For instance if your string variable is called `my_str` you would write `my_str.find(...)`."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "\n",
    "### Solution:\n",
    "Uncomment and run the cell below to show the solution."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# %load solutions/solution_13.py"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "<br>\n",
    "\n",
    "## Exercise 1.4 \n",
    "\n",
    "1. Put your favorite quote in a string variable.\n",
    "2. Find a way to replace all spaces by '@' in the quote.\n",
    "\n",
    "> browse the help either in the notebook (`help(str)`), or somewhere online."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "\n",
    "### Solution:\n",
    "Uncomment and run the cell below to show the solution."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# %load solutions/solution_14.py"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "<br>\n",
    "\n",
    "## Exercise 1.5\n",
    "\n",
    "1. Create a dictionary whose key/value pairs are the scientific classification of the [Python genus](https://en.wikipedia.org/wiki/Python_(genus)):\n",
    "\n",
    "\n",
    "| key    | value      |\n",
    "|--------|------------|\n",
    "| Class  | Reptilia   |\n",
    "| Order  | Squamata   |\n",
    "| Family | Pythonidae |\n",
    "| Genus  | Python     |\n",
    "\n",
    "\n",
    "2. **After** having created the dictionary, add a new `Species` entry to it that contains **2 or more**\n",
    "   python species of your choice (you can lookup the link above for inspiration).\n",
    "\n",
    "3. Print the lists of keys and values as two separate lists.  \n",
    "   **Hint:** use the `.keys()` and `.values()` methods of dictionary objects.  \n",
    "   \n",
    "   Your output should look like this:\n",
    "   ```\n",
    "   List of keys: ['Class', 'Order', 'Family', 'Genus', 'Species']\n",
    "   List of values: ['Reptilia', 'Squamata', 'Pythonidae', 'Python', < your species here > ]\n",
    "   ```"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "\n",
    "### Solution:\n",
    "Uncomment and run the cell below to show the solution."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# %load solutions/solution_15.py"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "<br>\n",
    "<br>\n",
    "\n",
    "# Additional exercises\n",
    "--------------------------------\n",
    "\n",
    "## Additional exercise 1.6\n",
    "\n",
    "Consider the two dictionaries below, describing the number of PhD degrees delivered in 1999 in the US (adapted from [nsf data](http://www.nsf.gov/statistics/infbrief/nsf11305/)):"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "science_doc = {\"Agricultural sciences\": 1065,\n",
    "                \"Biochemistry\": 759,\n",
    "                \"Molecular biology\": 716,\n",
    "                \"Neurosciences\": 431,\n",
    "                \"Other biological sciences\": 3675,\n",
    "                \"Computer sciences\": 856,\n",
    "                \"Earth, atmospheric, and ocean sciences\": 706,\n",
    "                \"Mathematics\": 1083,\n",
    "                \"Chemistry\": 2132,\n",
    "                \"Physics and astronomy\": 715,\n",
    "                \"Astrology\": 2012, \n",
    "                \"Psychology\": 3668,\n",
    "                \"Social sciences\": 4063}\n",
    "\n",
    "engineering_doc = {\"Aerospace/aeronautical engineering\": 206,\n",
    "                    \"Chemical engineering\": 576,\n",
    "                    \"Civil engineering\": 506,\n",
    "                    \"Electrical engineering\": 1236,\n",
    "                    \"Industrial/manufacturing engineering\": 211,\n",
    "                    \"Materials science engineering\": 393,\n",
    "                    \"Mechanical engineering\": 786,\n",
    "                    \"Other engineering\": 1416 }\n",
    "\n",
    "humanities_doc = {\"Foreign languages and literature\": 626,\n",
    "                    \"History\": 960,\n",
    "                    \"Letters\": 1516,\n",
    "                    \"Other humanities\": 1934}\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1. Merge these three dictionaries in a `all_doc` dictionary\n",
    "2. What is the length of the `all_doc` dictionary\n",
    "3. Add the \"Health\" doctorates of the `all_doc` dictionary, with 1407 degrees awarded.\n",
    "4. Multiply by 2 the number of Physics and astronomy degrees in `all_doc`.\n",
    "5. Remove the Astrology key from `all_doc`."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "\n",
    "### Solution:\n",
    "Uncomment and run the cell below to show the solution."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# %load solutions/solution_16.py"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "<br>\n",
    "\n",
    "## Additional exercise 1.7\n",
    "\n",
    "Set a variable called `is_divisible` to `True` if 7453102041884640481897 is evenly divisible by 135138672131"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "\n",
    "### Solution:\n",
    "Uncomment and run the cell below to show the solution."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# %load solutions/solution_17.py"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br>\n",
    "<br>\n",
    "\n",
    "## Additional exercise 1.8\n",
    "\n",
    "Here is a list containing the name of all swiss cantons. Your task is to:\n",
    "\n",
    "1. Sort it alphabetically.\n",
    "2. Reverse it.\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "canton_list = [\n",
    "    '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": [
    "<br>\n",
    "\n",
    "### Solution:\n",
    "Uncomment and run the cell below to show the solution."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# %load solutions/solution_18.py"
   ]
  }
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