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Fred Baptiste
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Notebooks for Part 3
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‎Part 3/Section 03 - Dictionaries/01 - Creating Python Dictionaries.ipynb‎

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‎Part 3/Section 03 - Dictionaries/02 - Common Operations.ipynb‎

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‎Part 3/Section 03 - Dictionaries/03 - Dictionary Views.ipynb‎

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‎Part 3/Section 03 - Dictionaries/04 - Updating, Merging and Copying.ipynb‎

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‎Part 3/Section 03 - Dictionaries/05 - Custom Classes and Hashing.ipynb‎

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‎Part 3/Section 04 - Coding Exercises/01 - Coding Exercises.ipynb‎

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‎Part 3/Section 04 - Coding Exercises/02 - Coding Exercises - Solution 1.ipynb‎

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{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"### Coding Exercises - Solution 1"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"#### Exercise 1"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Write a Python function that will create and return a dictionary from another dictionary, but sorted by value. You can assume the values are all comparable and have a natural sort order."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"For example, given the following dictionary:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {},
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"outputs": [],
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"source": [
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"composers = {'Johann': 65, 'Ludwig': 56, 'Frederic': 39, 'Wolfgang': 35}"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Your function should return a dictionary that looks like the following:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {},
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"outputs": [],
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"source": [
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"sorted_composers = {'Wolfgang': 35,\n",
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" 'Frederic': 39, \n",
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" 'Ludwig': 56,\n",
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" 'Johann': 65}"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Remember if you are using Jupyter notebook to use `print()` to view your dictionary in it's natural ordering (Jupyter will display your dictionary sorted by key).\n",
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"\n",
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"Also try to keep your code Pythonic - i.e. don't start with an empty dictionary and build it up one key at a time - look for a different, more Pythonic, way of doing it. \n",
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"\n",
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"Hint: you'll likely want to use Python's `sorted` function."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"##### Solution"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"My approach here is to sort the `items()` view using Python's `sorted` function and a custom `key` that uses the dictionary values (or second element of each tuple in the `items` view):"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {},
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"outputs": [],
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"source": [
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"composers = {'Johann': 65, 'Ludwig': 56, 'Frederic': 39, 'Wolfgang': 35}\n",
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"\n",
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"def sort_dict_by_value(d):\n",
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" d = {k: v\n",
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" for k, v in sorted(d.items(), key=lambda el: el[1])}\n",
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" return d"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 4,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"{'Wolfgang': 35, 'Frederic': 39, 'Ludwig': 56, 'Johann': 65}\n"
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]
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}
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],
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"source": [
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"print(sort_dict_by_value(composers))"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Here's a better approach - instead of using a dictionary comprehension, we can simply use the `dict()` function to create a dictionary from the sorted tuples!"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 5,
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"metadata": {},
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"outputs": [],
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"source": [
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"def sort_dict_by_value(d):\n",
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" return dict(sorted(d.items(), key=lambda el: el[1]))"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"And we end up with the same end result:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 6,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"{'Wolfgang': 35, 'Frederic': 39, 'Ludwig': 56, 'Johann': 65}"
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]
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},
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"execution_count": 6,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"sort_dict_by_value(composers)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": []
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.6.7"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}

‎Part 3/Section 04 - Coding Exercises/03 - Coding Exercises - Solution 2.ipynb‎

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{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"### Coding Exercises - Solution 2"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"#### Exercise 2"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Given two dictionaries, `d1` and `d2`, write a function that creates a dictionary that contains only the keys common to both dictionaries, with values being a tuple containg the values from `d1` and `d2`. (Order of keys is not important)."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"For example, given two dictionaries as follows:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {},
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"outputs": [],
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"source": [
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"d1 = {'a': 1, 'b': 2, 'c': 3, 'd': 4}\n",
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"d2 = {'b': 20, 'c': 30, 'y': 40, 'z': 50}"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Your function should return a dictionary that looks like this:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {},
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"outputs": [],
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"source": [
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"d = {'b': (2, 20), 'c': (3, 30)}"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Hint: Remember that `s1 & s2` will return the intersection of two sets."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Again, try to keep your code Pythonic - don't just start with an empty dictionary and build it up one by one - think of a cleaner approach."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"##### Solution"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"My approach here is to use set intersections to find the keys common to both dictionaries.\n",
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"Then I use a dictionary comprehension to build up my new dictionary, making each value in the new dictionary a tuple containing the values from the original dictionaries:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {},
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"outputs": [],
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"source": [
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"d1 = {'a': 1, 'b': 2, 'c': 3, 'd': 4}\n",
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"d2 = {'b': 20, 'c': 30, 'y': 40, 'z': 50}\n",
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"\n",
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"def intersect(d1, d2):\n",
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" d1_keys = d1.keys()\n",
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" d2_keys = d2.keys()\n",
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" keys = d1_keys & d2_keys\n",
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" d = {k: (d1[k], d2[k]) for k in keys}\n",
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" return d"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 4,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"{'b': (2, 20), 'c': (3, 30)}"
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]
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},
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"execution_count": 4,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"intersect(d1, d2)"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.6.7"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}

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