Lecture 6 bis : Additional exercices

December, 2021 - François HU

Master of Science - EPITA

This lecture is available here: https://curiousml.github.io/

Exercices

These exercices are considered "normal-hard" exercices.

Exercice 1: lists

Create a function insert, that produces the following results:

>>> l = [0, 9, 3, 10]
>>> print(insert(l, -1, 1))
[0, -1, 9, 3, 10]
>>> print(insert(l, -1, 2))
[0, 9, -1, 3, 10]
>>> print(insert(l, "epita", 4))
[0, 9, 3, 10, 'epita']
>>> print(insert(l, "epita", 10))
[0, 9, 3, 10, 'epita']
>>> print(l)
[0, 9, 3, 10]

def insert(l, v, i):
    l = l[:i] + [v] + l[i:]
    return(l)

l = [0, 9, 3, 10]
print(insert(l, -1, 1))
print(insert(l, -1, 2))
print(insert(l, "epita", 4))
print(insert(l, "epita", 10))
print(l)

[0, -1, 9, 3, 10]
[0, 9, -1, 3, 10]
[0, 9, 3, 10, 'epita']
[0, 9, 3, 10, 'epita']
[0, 9, 3, 10]

Exercice 2: dictionaries

we have the following list of dictionaries

paris = [{"District": 3, "Date":2007, "Pop":34576},
         {"District": 5, "Date":2007, "Pop":62664},
         {"District": 6, "Date":2007, "Pop":45332},
         {"District": 7, "Date":2007, "Pop":57410},
         {"District": 8, "Date":2007, "Pop":39165},
         {"District": 9, "Date":2007, "Pop":58632},
         {"District": 10, "Date":2007, "Pop":93373},
         {"District": 11, "Date":2007, "Pop":151421},
         {"District": 12, "Date":2007, "Pop":142425},
         {"District": 13, "Date":2007, "Pop":179213},
         {"District": 14, "Date":2007, "Pop":134382},
         {"District": 15, "Date":2007, "Pop":232247},
         {"District": 16, "Date":2007, "Pop":159706},
         {"District": 17, "Date":2007, "Pop":164673},
         {"District": 18, "Date":2007, "Pop":191523},
         {"District": 19, "Date":2007, "Pop":184038},
         {"District": 20, "Date":2007, "Pop":194018}]

1. write a script that add a field Rounded_pop which discard the last 3 digits. For example we transform the value 34576 to 34

for p in paris:
    p["Rounded_pop"] = p["Pop"]//1000
paris

[{'District': 3, 'Date': 2007, 'Pop': 34576, 'Rounded_pop': 34},
 {'District': 5, 'Date': 2007, 'Pop': 62664, 'Rounded_pop': 62},
 {'District': 6, 'Date': 2007, 'Pop': 45332, 'Rounded_pop': 45},
 {'District': 7, 'Date': 2007, 'Pop': 57410, 'Rounded_pop': 57},
 {'District': 8, 'Date': 2007, 'Pop': 39165, 'Rounded_pop': 39},
 {'District': 9, 'Date': 2007, 'Pop': 58632, 'Rounded_pop': 58},
 {'District': 10, 'Date': 2007, 'Pop': 93373, 'Rounded_pop': 93},
 {'District': 11, 'Date': 2007, 'Pop': 151421, 'Rounded_pop': 151},
 {'District': 12, 'Date': 2007, 'Pop': 142425, 'Rounded_pop': 142},
 {'District': 13, 'Date': 2007, 'Pop': 179213, 'Rounded_pop': 179},
 {'District': 14, 'Date': 2007, 'Pop': 134382, 'Rounded_pop': 134},
 {'District': 15, 'Date': 2007, 'Pop': 232247, 'Rounded_pop': 232},
 {'District': 16, 'Date': 2007, 'Pop': 159706, 'Rounded_pop': 159},
 {'District': 17, 'Date': 2007, 'Pop': 164673, 'Rounded_pop': 164},
 {'District': 18, 'Date': 2007, 'Pop': 191523, 'Rounded_pop': 191},
 {'District': 19, 'Date': 2007, 'Pop': 184038, 'Rounded_pop': 184},
 {'District': 20, 'Date': 2007, 'Pop': 194018, 'Rounded_pop': 194}]

1. write a script that store the values of Rounded_pop into a list named populations

populations = []
for p in paris:
    populations.append(p["Rounded_pop"])
populations

[34, 62, 45, 57, 39, 58, 93, 151, 142, 179, 134, 232, 159, 164, 191, 184, 194]

1. sort by ascending order the list populations

populations.sort()
populations

[34, 39, 45, 57, 58, 62, 93, 134, 142, 151, 159, 164, 179, 184, 191, 194, 232]

Exercice 3: strings

we have the following list of strings

list_of_strings = [
    'EPITA', 'propose', 'trois', 'programmes',
    'spécialisés', 'Master', 'of', 'Science', 
    'dans', 'un', 'environnement', 'international.',
    'Cette', 'formation', 'professionnalisante', 'de',
    '18', 'mois', 'intégralement', 'en', 'anglais',
    'pour','les','étudiants','français', 'et', 'internationaux,', 
    'permet', 'de', 'se', 'spécialiser', 'dans', 'un', 
    'domaine', 'après', '3', 'ou', '4', 'ans', 'd’études.']

Write a script that create a variable named text which concatenante the items of list_of_strings with exactly one space between words.

text = ' '.join(list_of_strings)
text

'EPITA propose trois programmes spécialisés Master of Science dans un environnement international. Cette formation professionnalisante de 18 mois intégralement en anglais pour les étudiants français et internationaux, permet de se spécialiser dans un domaine après 3 ou 4 ans d’études.'

# alternatively

text = list_of_strings[0]
for s in list_of_strings[1:]:
    text += ' '
    text += s
text

'EPITA propose trois programmes spécialisés Master of Science dans un environnement international. Cette formation professionnalisante de 18 mois intégralement en anglais pour les étudiants français et internationaux, permet de se spécialiser dans un domaine après 3 ou 4 ans d’études.'

Exercice 4: arrays

1. Generate the following array without explicit loops and without manually fill in the numbers. We will name it arr1:

[[  0  25 100 225 400]
 [  1  36 121 256 441]
 [  4  49 144 289 484]
 [  9  64 169 324 529]
 [ 16  81 196 361 576]]

import numpy as np

arr1 = np.arange(25).reshape(5,-1).T**2
arr1

array([[  0,  25, 100, 225, 400],
       [  1,  36, 121, 256, 441],
       [  4,  49, 144, 289, 484],
       [  9,  64, 169, 324, 529],
       [ 16,  81, 196, 361, 576]])

1. Extract the diagonal values of arr1 (do not generate it) and call it diag1

[  0,  36, 144, 324, 576]

diag1 = arr1[np.arange(5), np.arange(5)]
diag1

array([  0,  36, 144, 324, 576])

1. From arr1, extract the submatrix:

[[225 100  25]
 [289 144  49]
 [256 121  36]]

and name it subarr1.

subarr1 = arr1[[0, 2, 1], :][:, [3, 2, 1]]
print(subarr1)

[[225 100  25]
 [289 144  49]
 [256 121  36]]

Exercice 5: arrays

1. Generate the following array without explicit loops and without manually fill in the numbers. We will name it arr2:

[[10  9  8  7  6]
 [ 9  8  7  6  5]
 [ 8  7  6  5  4]
 [ 7  6  5  4  3]
 [ 6  5  4  3  2]]

arr = np.arange(5, 0, -1)
arr2 = np.tile(arr, (5, 1)).T + arr
print(arr2)

[[10  9  8  7  6]
 [ 9  8  7  6  5]
 [ 8  7  6  5  4]
 [ 7  6  5  4  3]
 [ 6  5  4  3  2]]

1. Replace all the even values of arr2 by 0 without explicit loops. You should have:

   [[0 9 0 7 0]
   [9 0 7 0 5]
   [0 7 0 5 0]
   [7 0 5 0 3]
   [0 5 0 3 0]]

arr2[arr2%2==0] = 0
print(arr2)

[[0 9 0 7 0]
 [9 0 7 0 5]
 [0 7 0 5 0]
 [7 0 5 0 3]
 [0 5 0 3 0]]

Exercice 6: plotting

1. Generate and (scatter) plot 2000 random points in the space $[−10,0]×[1,2]$ with the color orange. You should have:

import numpy as np
import matplotlib.pyplot as plt

n = 2000
x = -10 * np.random.rand(n)
y = np.random.rand(n) + 1

plt.scatter(x, y, color="orange");

1. color the points inside the space $[−4,-2]×[1.2,1.8]$ or $[−8,-6]×[1.2,1.8]$ by green. You should have:

import numpy as np
import matplotlib.pyplot as plt

ind1 = (x<=-2) & (x>=-4) & (y<=1.8) & (y>=1.2)
ind2 = (x<=-6) & (x>=-8) & (y<=1.8) & (y>=1.2)
x_green = x[ind1 | ind2]
y_green = y[ind1 | ind2]

plt.scatter(x, y, color="orange");
plt.scatter(x_green, y_green, color="green");

Exercice 7: data manipulation

1. Import the Iris and defra_consumptiondatasets as dataframes and name them respectively iris and cons. You should have the following first 5 rows for each dataframe:

import pandas as pd

cons = pd.read_csv('data/defra_consumption.csv', sep=';', index_col=0)
iris = pd.read_csv('data/Iris.csv', sep=',', index_col="Id")

cons.head()

	England	Wales	Scotland	N Ireland
Cheese	105	103	103	66
Carcass meat	245	227	242	267
Other meat	685	803	750	586
Fish	147	160	122	93
Fats and oils	193	235	184	209

iris.head()

	SepalLengthCm	SepalWidthCm	PetalLengthCm	PetalWidthCm	Species
Id
1	5.1	3.5	1.4	0.2	Iris-setosa
2	4.9	3.0	1.4	0.2	Iris-setosa
3	4.7	3.2	1.3	0.2	Iris-setosa
4	4.6	3.1	1.5	0.2	Iris-setosa
5	5.0	3.6	1.4	0.2	Iris-setosa

1. For the dataframe iris (aside from the column species):
   • discard the Cm at the end of each column name
   • convert values to minimeter (instead of centimiter)

iris.columns = ["SepalLength", "SepalWidth", "PetalLength", "PetalWidth", "Species"]
iris.iloc[:, :-1] *= 10
iris.head()

	SepalLength	SepalWidth	PetalLength	PetalWidth	Species
Id
1	51.0	35.0	14.0	2.0	Iris-setosa
2	49.0	30.0	14.0	2.0	Iris-setosa
3	47.0	32.0	13.0	2.0	Iris-setosa
4	46.0	31.0	15.0	2.0	Iris-setosa
5	50.0	36.0	14.0	2.0	Iris-setosa

1. For the dataframe cons, convert columns to percentages of the totals. Therefore the first row should be:

cons = (cons/cons.sum(axis=0))*100
cons.head()

	England	Wales	Scotland	N Ireland
Cheese	1.315130	1.202288	1.316462	0.902996
Carcass meat	3.068637	2.649702	3.093047	3.653031
Other meat	8.579659	9.373176	9.585890	8.017513
Fish	1.841182	1.867632	1.559305	1.272404
Fats and oils	2.417335	2.743084	2.351738	2.859488