mirror of
https://github.com/Andreaierardi/Master-DataScience-Notes.git
synced 2024-07-06 21:51:46 +02:00
711 lines
18 KiB
Plaintext
711 lines
18 KiB
Plaintext
{
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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": 19,
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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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"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\n"
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]
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}
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],
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"source": [
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"n = 10\n",
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"for i in range(1,n+1):\n",
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" print(n,\"x\",i*2,\"=\",n*i*2)\n"
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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 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": 86,
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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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"10 x 2 = 40\n",
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"10 x 4 = 80\n",
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"10 x 6 = 120\n",
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"10 x 8 = 160\n",
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"10 x 10 = 200\n",
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"10 x 12 = 240\n",
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"10 x 14 = 280\n",
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"10 x 16 = 320\n",
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"10 x 18 = 360\n",
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"10 x 20 = 400\n"
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]
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}
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],
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"source": [
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"n = 10\n",
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"for i in range(2,n*2+1, 2):\n",
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" print(n,\"x\",i,\"=\",n*i*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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"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": 34,
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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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"10 x 1 = 20\n",
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"10 x 2 = 40\n",
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"10 x 3 = 60\n",
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"10 x 4 = 80\n",
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"10 x 5 = 100\n",
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"10 x 6 = 120\n",
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"10 x 7 = 140\n",
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"10 x 8 = 160\n",
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"10 x 9 = 180\n",
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"10 x 10 = 200\n"
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]
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},
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{
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"data": {
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"text/plain": [
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"['10x1=20',\n",
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" '10x2=40',\n",
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" '10x3=60',\n",
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" '10x4=80',\n",
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" '10x5=100',\n",
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" '10x6=120',\n",
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" '10x7=140',\n",
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" '10x8=160',\n",
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" '10x9=180',\n",
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" '10x10=200']"
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]
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},
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"execution_count": 34,
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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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"n = 10\n",
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"[print(n,\"x\",i,\"=\",n*i*2) for i in range(1,n+1)]\n",
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"\n",
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"# To get a list of strings\n",
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"[str(n)+\"x\"+str(i)+\"=\"+str(n*i*2) for i in range(1,n+1)]\n"
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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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"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": 63,
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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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"['10 x 1 = 20',\n",
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" '10 x 2 = 40',\n",
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" '10 x 3 = 60',\n",
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" '10 x 4 = 80',\n",
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" '10 x 5 = 100',\n",
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" '10 x 6 = 120',\n",
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" '10 x 7 = 140',\n",
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" '10 x 8 = 160',\n",
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" '10 x 9 = 180',\n",
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" '10 x 10 = 200']"
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]
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},
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"execution_count": 63,
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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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"s = \"{0} x {1} = {2}\"\n",
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"[s.format(*[n,i,n*i*2]) for i in range(1,n+1)]\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": 64,
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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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"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\n"
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]
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}
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],
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"source": [
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"#OR \n",
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"s = \"{0} x {1} = {2}\"\n",
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"for i in range(1,n+1):\n",
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" print(s.format(*[n,i*2,n*i*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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"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": 79,
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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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"'10 x 2 = 2010 x 4 = 4010 x 6 = 6010 x 8 = 8010 x 10 = 10010 x 12 = 12010 x 14 = 14010 x 16 = 16010 x 18 = 18010 x 20 = 200'"
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]
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},
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"execution_count": 79,
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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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"l = \"\"\n",
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"s = \"{0} x {1} = {2}\"\n",
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"#[s.format(*[n,i,n*i*2]) for i in range(1,n+1)]\n",
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"l.join([s.format(*[n,i*2,n*i*2]) for i in range(1,n+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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"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": 101,
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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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"'10 x 1 = 20 \\n10 x 2 = 40 \\n10 x 3 = 60 \\n10 x 4 = 80 \\n10 x 5 = 100 \\n10 x 6 = 120 \\n10 x 7 = 140 \\n10 x 8 = 160 \\n10 x 9 = 180 \\n10 x 10 = 200 \\n'"
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]
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},
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"execution_count": 101,
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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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"l = \"\"\n",
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"s = \"{0} x {1} = {2} {3}\"\n",
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"#[s.format(*[n,i,n*i*2]) for i in range(1,n+1)]\n",
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"l.join([s.format(*[n,i,n*i*2],'\\n') for i in range(1,n+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 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": 102,
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"metadata": {
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"scrolled": false
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},
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"outputs": [
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{
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"data": {
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"text/plain": [
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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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"execution_count": 102,
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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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"l = [(10, 20, 40), (40, 50, 60), (70, 80, 90)]\n",
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"\n",
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"res =[]\n",
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"for tup in l:\n",
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" res+=[(tup[0],tup[1],tup[0]*tup[1])]\n",
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"res"
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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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"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": 15,
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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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"[(10, 20, 100, 200), (40, 50, 2000), (70, 80, 100, 200, 300, 5600)]"
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]
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},
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"execution_count": 15,
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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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"l = [(10, 20, 100, 40), (40, 50, 60), (70, 80, 100, 200, 300, 90)]\n",
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"\n",
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"res =[]\n",
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"for tup in l:\n",
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" tmp = (tup[0:len(tup)-1]) + (tup[0]*tup[1],) \n",
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" res+=[tmp]\n",
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" \n",
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"res"
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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 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": 72,
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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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"Smallest word: A \n",
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"Largest word: quick \n"
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]
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}
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],
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"source": [
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"string = \"A quick red fox\"\n",
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"lis =[]\n",
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"minw=\"\"\n",
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"maxw=\"\"\n",
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"tmp = \"\"\n",
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"for w in string:\n",
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" tmp+=w \n",
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" if(w==\" \" or string.index(w)==len(string)-1):\n",
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" lis+=[tmp]\n",
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" tmp = \"\"\n",
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"\n",
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"if len(lis)>0:\n",
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" minw= lis[0]\n",
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"for i in lis:\n",
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" if(len(i)< len(minw)):\n",
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" minw = i\n",
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" if(len(i)> len(maxw)):\n",
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" maxw = i\n",
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" \n",
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"print(\"Smallest word: \" + minw + \"\\nLargest word: \" + maxw)"
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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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"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": 73,
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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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"Smallest word: A\n",
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"Largest word: quick\n"
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]
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}
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],
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"source": [
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"string = \"A quick red fox\"\n",
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"lis = string.split(\" \")\n",
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"minw= lis[0]\n",
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"maxw=\"\"\n",
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"for i in lis:\n",
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" if(len(i)< len(minw)):\n",
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" minw = i\n",
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" if(len(i)> len(maxw)):\n",
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" maxw = i\n",
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" \n",
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"print(\"Smallest word: \" + minw + \"\\nLargest word: \" + maxw)"
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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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"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": 75,
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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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"Smallest word: A\n",
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"Largest word: quick\n"
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]
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}
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],
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"source": [
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"string = \"A quick red fox\"\n",
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"lis = string.split(\" \") \n",
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"print(\"Smallest word: \" + min(lis,key=len) + \"\\nLargest word: \" + max(lis,key=len))"
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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 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",
|
|
"\n",
|
|
"<img src=\"files/es2.svg\" width = \"90%\">"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"metadata": {},
|
|
"source": [
|
|
"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."
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": 112,
|
|
"metadata": {},
|
|
"outputs": [
|
|
{
|
|
"data": {
|
|
"text/plain": [
|
|
"[[30, 56, 25], [33]]"
|
|
]
|
|
},
|
|
"execution_count": 112,
|
|
"metadata": {},
|
|
"output_type": "execute_result"
|
|
}
|
|
],
|
|
"source": [
|
|
"ls = [[10, 20], [40], [30, 56, 25], [10, 20], [33], [40]]\n",
|
|
"\n",
|
|
"no_dup = ls.copy()\n",
|
|
"\n",
|
|
"for i in range(len(ls)):\n",
|
|
" for j in range(len(ls)):\n",
|
|
" #print(\"compare :\" + str(ls[i]) +\" - \"+ str(ls[j]))\n",
|
|
" if(i!=j and ls[i]==ls[j]):\n",
|
|
" no_dup.remove(ls[i])\n",
|
|
"\n",
|
|
"no_dup"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"metadata": {},
|
|
"source": [
|
|
"We can simplify the above code using in a smarter way the conditional statements:"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"metadata": {},
|
|
"outputs": [],
|
|
"source": []
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"metadata": {},
|
|
"source": [
|
|
"Other common ways people use to tackle duplicates include:\n",
|
|
"- 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",
|
|
"- 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",
|
|
"\n",
|
|
"\n",
|
|
"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``."
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"metadata": {},
|
|
"outputs": [],
|
|
"source": []
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"metadata": {},
|
|
"source": [
|
|
"Compare with the following examples:"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"metadata": {},
|
|
"outputs": [],
|
|
"source": []
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"metadata": {},
|
|
"source": [
|
|
"## Exercise 5\n",
|
|
"\n",
|
|
"Consider the following list of student records:\n",
|
|
"\n",
|
|
"``students = [{'id': 1, 'success': True, 'name': 'Theo'},\n",
|
|
" {'id': 2, 'success': False, 'name': 'Alex'},\n",
|
|
" {'id': 3, 'success': True, 'name': 'Ralph'},\n",
|
|
" {'id': 4, 'success': True, 'name': 'Ralph'}\n",
|
|
" {'id': 5, 'success': False, 'name': 'Theo'}]``\n",
|
|
" \n",
|
|
"We want to write a program to get the different values associated with \"name\" key.\n",
|
|
"\n",
|
|
"With the above list, the program should output ``['Theo', 'Alex', 'Ralph']``."
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": 148,
|
|
"metadata": {},
|
|
"outputs": [
|
|
{
|
|
"data": {
|
|
"text/plain": [
|
|
"['Theo', 'Ralph', 'Ralph']"
|
|
]
|
|
},
|
|
"execution_count": 148,
|
|
"metadata": {},
|
|
"output_type": "execute_result"
|
|
}
|
|
],
|
|
"source": [
|
|
"students = [{'id': 1, 'success': True, 'name': 'Theo'},\n",
|
|
" {'id': 2, 'success': False, 'name': 'Alex'},\n",
|
|
" {'id': 3, 'success': True, 'name': 'Ralph'},\n",
|
|
" {'id': 4, 'success': True, 'name': 'Ralph'},\n",
|
|
" {'id': 5, 'success': False, 'name': 'Theo'}]\n",
|
|
"\n",
|
|
"lis = []\n",
|
|
"for dic in students:\n",
|
|
" if(dic[\"success\"] == True):\n",
|
|
" lis+=[dic[\"name\"]]\n",
|
|
"lis\n",
|
|
"\n"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"metadata": {},
|
|
"source": [
|
|
"We recognize again the pattern that list comprehensions implement, then we can use them:"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"metadata": {},
|
|
"outputs": [],
|
|
"source": [
|
|
"#OR\n",
|
|
"[dic[\"name\"] for dic in students if(dic[\"success\"] == True)]"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"metadata": {},
|
|
"source": [
|
|
"Finally, we can exploit what we learned from Exercise 4, using for instance the ``set()`` function."
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"metadata": {},
|
|
"outputs": [],
|
|
"source": []
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": null,
|
|
"metadata": {},
|
|
"outputs": [],
|
|
"source": []
|
|
}
|
|
],
|
|
"metadata": {
|
|
"kernelspec": {
|
|
"display_name": "Python 3",
|
|
"language": "python",
|
|
"name": "python3"
|
|
},
|
|
"language_info": {
|
|
"codemirror_mode": {
|
|
"name": "ipython",
|
|
"version": 3
|
|
},
|
|
"file_extension": ".py",
|
|
"mimetype": "text/x-python",
|
|
"name": "python",
|
|
"nbconvert_exporter": "python",
|
|
"pygments_lexer": "ipython3",
|
|
"version": "3.7.4"
|
|
}
|
|
},
|
|
"nbformat": 4,
|
|
"nbformat_minor": 2
|
|
}
|