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Andrea Ierardi
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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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"# Exercises in Python\n",
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"\n",
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"In the first three lessons, you have seen:\n",
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"\n",
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"- lists (including list comprehensions);\n",
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"- tuples;\n",
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"- flow control (selection and loop statements);\n",
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"- modules;\n",
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"- dictionaries;\n",
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"\n",
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"We will revise the above topics with five exercises.\n",
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"\n",
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"## Exercise 1\n",
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"\n",
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"Write a program to create a multiplication table, from 2 to 20 with a step of 2, of a number.\n",
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"\n",
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"For instance, given ``n = 10``, the program should output:\n",
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"\n",
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"``10 x 2 = 20\n",
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"10 x 4 = 40\n",
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"10 x 6 = 60\n",
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"10 x 8 = 80\n",
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"10 x 10 = 100\n",
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"10 x 12 = 120\n",
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"10 x 14 = 140\n",
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"10 x 16 = 160\n",
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"10 x 18 = 180\n",
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"10 x 20 = 200``"
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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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"We can simply address the required task using a ``for`` loop and ``range``: it represents an immutable sequence of numbers and is commonly used for looping a specific number of times in for loops."
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"We recall that ``range`` also accept a ``step`` argument. Then, we improve the code avoiding the ``if`` statement and setting ``step = 2``.\n",
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"\n",
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"We also get rid of the variable ``v``: if you don't need to use its value later, just don't create it."
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Finally, we write the same program in one line using list comprehensions: they provide a compact way to filter elements from a sequence and they implement the following for loop\n",
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"\n",
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"``result = []\n",
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"for <variable> in <sequence>:\n",
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" if <condition>:\n",
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" result.append(<expression>)``\n",
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" \n",
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"in the following equivalent form\n",
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"\n",
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"``[<expression> for <variable> in <sequence> if <condition>]``\n",
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"\n",
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"In our case, we avoid the filtering part and we can obtain results as:"
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"To obtain the same pretty printing, we create a formatted string inside the list comprehension."
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Finally, we use the ``join()`` method of strings: it concatenates each element of an iterable (such our list) to a string and returns the concatenated string. The syntax is ``string.join(iterable)``. An example:\n"
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"We concatenate with ``\\n`` to go to the next line, then we print."
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Exercise 2\n",
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"\n",
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"Write a Python program to replace the last value of the tuples in a list with the product of the respective first two elements of the tuple. Suppose that the list is composed only by tuples of three integers.\n",
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"\n",
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"For instance, given in input the list ``l = [(10, 20, 40), (40, 50, 60), (70, 80, 90)]``, the program should output the following list of tuples\n",
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"``[(10, 20, 200), (40, 50, 2000), (70, 80, 5600)]``."
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Now, suppose that we want to address the same task as above, but the input list is now composed of tuples with a variable number of elements.\n",
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"\n",
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"For instance, consider the list ``l = [(10, 20, 100, 40), (40, 50, 60), (70, 80, 100, 200, 300, 90)]``, the program should output the following list of tuples ``[(10, 20, 100, 200), (40, 50, 2000), (70, 80, 100, 200, 300, 5600)]``.\n",
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"\n",
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"We can write a one-line expression using the slice operator, whose syntax is ``[start:stop:step]``."
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Exercise 3\n",
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"\n",
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"Write a program to find the smallest and the largest word in a given string.\n",
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"\n",
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"For instance, consider the string ``string = \"A quick red fox\"``. The program should output\n",
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"\n",
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"``Smallest word: A\n",
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"Largest word: quick``\n",
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"\n",
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"A possible strategy is:\n",
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"\n",
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"- creating a list containing all the words of the sentence;\n",
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"- loop on the list and compute both the smallest and the longest word at the same time.\n",
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"\n",
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"<img src=\"files/es1.svg\" width=\"40%\">"
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Let's refine the above code. Point 1 can be adressed using the Python ``split()`` built-in method of strings.\n",
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"\n",
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"The ``split()`` method splits a string into a list. You can specify the separator, and default separator is any whitespace."
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Also Point 2 can be adressed in a smarter way. We can use the built-in functions ``min()`` and ``max()``, which respectively returns the smallest and largest of the input values.\n",
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"\n",
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"Such functions provide a parameter named ``key``, which allow to set a function to indicate the sort order. We must specify ``key = len``, as the default ordering for strings is the lexicographic one."
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"At the end of the day, the exercise can be solved in the following way:"
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Exercise 4\n",
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"\n",
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"Write a Python program to remove duplicates from a list of lists.\n",
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"\n",
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"For instance, given in input the list ``ls = [[10, 20], [40], [30, 56, 25], [10, 20], [33], [40]]``, the program should output the following list without duplicates: ``[[10, 20], [40], [30, 56, 25], [33]]``.\n",
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"\n",
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"<img src=\"files/es2.svg\" width = \"90%\">"
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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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"We initialize a new empty list named ``ls_no_dup``. We can address the exercise using two ``for`` loops: with the first one we pick an element from the original list, and with the second one we check if there is another equal element in the ``ls_no_dup`` list. If the element we are currently considering is yet present in ``ls_no_dup`` we don't add it again, otherwise we add it."
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"We can simplify the above code using in a smarter way the conditional statements:"
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Other common ways people use to tackle duplicates include:\n",
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"- dictionaries: the ``fromkeys()`` method of ``dict`` returns a dictionary with the specified keys. If we cast the dictionary, we obtain a list with no duplicate values.\n",
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"- sets: the ``set()`` function, return a set whose does not allow duplicates, by its mathematical definition. Again, if we cast the set to a list, we obtain a list with no duplicate values.\n",
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"\n",
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"\n",
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"Here we can't adopt the latter solutions: you can't use a list as the key in a ``dict``, since ``dict`` keys need to be immutable. The same holds for ``set``."
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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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"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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Compare with the following examples:"
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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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"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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Exercise 5\n",
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"\n",
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"Consider the following list of student records:\n",
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"\n",
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"``students = [{'id': 1, 'success': True, 'name': 'Theo'},\n",
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" {'id': 2, 'success': False, 'name': 'Alex'},\n",
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" {'id': 3, 'success': True, 'name': 'Ralph'},\n",
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" {'id': 4, 'success': True, 'name': 'Ralph'}\n",
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" {'id': 5, 'success': False, 'name': 'Theo'}]``\n",
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" \n",
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"We want to write a program to get the different values associated with \"name\" key.\n",
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"\n",
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"With the above list, the program should output ``['Theo', 'Alex', 'Ralph']``."
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"We recognize again the pattern that list comprehensions implement, then we can use them:"
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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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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Finally, we can exploit what we learned from Exercise 4, using for instance the ``set()`` function."
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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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"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": "3.7.6"
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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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