NumPy Exercises Solutions
Contents
NumPy Exercises Solutions¶
These are my solutions that I wrote while working on the article NumPy Examples: Forty-Five Practice Questions To Make You an Expert. I’ve tried to list the original questions in the comments.
import numpy as np
Array Creation¶
NumPy supports a variety of methods for array creation.
# How would you create a one dimensional NumPy array of the numbers from 10 to 100, counting by 10?
np.arange(10, 110, 10)
array([ 10, 20, 30, 40, 50, 60, 70, 80, 90, 100])
# How could you do the same thing with a standard Python range and a list?
np.array([i for i in range(10, 110, 10)])
array([ 10, 20, 30, 40, 50, 60, 70, 80, 90, 100])
# What does np.array() do (no arguments passed to constructor). Guess: exception?
# np.array()
# How might you construct a NumPy array of capital letters (A-Z)
arr = np.array(['fee', 'fi', 'fo', 'fumble-dee-dum'])
arr.dtype
dtype('<U14')
arr = np.array([chr(i) for i in range(ord('A'), ord('Z') + 1)])
arr.dtype
arr[0].dtype
dtype('<U1')
# How would you create a ten-element array of all zeros?
print(np.zeros(10))
# Of all ones?
print(np.ones(10))
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[1. 1. 1. 1. 1. 1. 1. 1. 1. 1.]
# What is the default data dype for the np.zeros function?
# Guess is float64
df = np.zeros(10)
df.dtype
dtype('float64')
# How can you create an array of 10 random integers?
np.random.randint(1, 6, 10)
array([1, 2, 2, 2, 3, 3, 2, 1, 4, 1])
# How can you create a normal distribution of 10 numbers, centered on 5?
# Note, the 1 in center represents mu, size of STD DEV, and is arbitrary.
np.random.normal(5, 1, 10)
array([4.73632454, 5.1021039 , 4.55851208, 6.13236385, 5.95943556,
5.59175941, 5.21378803, 4.89446183, 3.94987317, 4.86283442])
# How can you create an array of 10 random numbers in the range 0 - 1?
np.random.rand(10)
array([0.11020477, 0.70654687, 0.96058828, 0.44059613, 0.08900653,
0.70157386, 0.00543054, 0.02419428, 0.66998649, 0.81528517])
Multidimensional arrays¶
np.zeros((3, 5))
array([[0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0.]])
myarray = np.arange(1, 13).reshape(3, 4)
myarray
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])
myarray[1,2]
7
myarray.ndim
2
myarray.reshape(2,6)
array([[ 1, 2, 3, 4, 5, 6],
[ 7, 8, 9, 10, 11, 12]])
arr
array(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M',
'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'],
dtype='<U1')
myarray.shape
(3, 4)
myarray.reshape(2,6)
array([[ 1, 2, 3, 4, 5, 6],
[ 7, 8, 9, 10, 11, 12]])
myarray
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])
x = myarray
x
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])
x[0,0] = 42
myarray
array([[42, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])
# Create a 3x4 array of random integers between 1 and 10
np.random.randint(1, 11, (3,4))
array([[ 3, 8, 2, 2],
[10, 2, 7, 8],
[ 3, 9, 5, 3]])
# ... or:
arr = np.random.randint(1,11,12).reshape(3,4)
arr
array([[4, 5, 9, 1],
[1, 9, 1, 6],
[8, 7, 9, 2]])
arr - arr
array([[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]])
z_list = [z for z in range(0,5)]
y_list = [z_list for y in range(0,4)]
x_list = [y_list for x in range(0,3)]
x_array = np.array(x_list)
# My guess is x_array.shape == 3,4,5. Correct but show tuple brackets would be better
x_array.shape
(3, 4, 5)
# What is the value of x_array.ndim?
x_array.ndim
3
# Given the following:
arr = np.arange(0,6).reshape(2,3)
arr
# How could you convert it to an array that looks like
# array([[0,3], [1,4], [2,5]))
# My answer either arr.transpose() or arr.T
# (arr.transpose() == arr.T).all()
# Note difference is mutablity? NO! Both return a view. Need to copy.
arr.transpose()
array([[0, 3],
[1, 4],
[2, 5]])
four_by_five = np.arange(1, 21).reshape(4,5)
four_by_five
array([[ 1, 2, 3, 4, 5],
[ 6, 7, 8, 9, 10],
[11, 12, 13, 14, 15],
[16, 17, 18, 19, 20]])
# Write an exoression to return the first row
four_by_five[0]
array([1, 2, 3, 4, 5])
# Write an exoression to return the last row
four_by_five[-1]
array([16, 17, 18, 19, 20])
# What does four_by_five[2,3] return? My answer, 14 (scalar)
four_by_five[2,3]
14
# What does four_by_five[3,2] return? My answer, 18 (scalar)
four_by_five[3,2]
18
# How could you return the first column? It will be a (four-element array ending with 16.) My answer four_by_five[:,0]
four_by_five[:,0]
array([ 1, 6, 11, 16])
# What does four_by_five[:, 2:4] return?
# My answer the last two columns -- wrong, third and fourth columns
four_by_five[:, 2:4]
array([[ 3, 4],
[ 8, 9],
[13, 14],
[18, 19]])
# Corrected question
# What does four_by_five[:, 3:5] return?
four_by_five[:, 3:5]
array([[ 4, 5],
[ 9, 10],
[14, 15],
[19, 20]])
# Write an expression to return the last two columns of the middle two rows.
four_by_five[1:3, 3:]
array([[ 9, 10],
[14, 15]])
one_dim = np.arange(1,6)
one_dim
array([1, 2, 3, 4, 5])
# What would be the result of one_dim * 2
# My answer array([2, 4, 6, 8, 10])
one_dim * 2
array([ 2, 4, 6, 8, 10])
# What would be the result of one_dim + np.arange(5, 0, -1)?
# My answer: array([ 6, 6, 6, 6, 6])
one_dim + np.arange(5, 0, -1)
array([6, 6, 6, 6, 6])
# How many zeros are in the array returned by one_dim - one_dim ?
# My answer: 5
one_dim - one_dim
array([0, 0, 0, 0, 0])
# What is the result of one_dim > 2 ?
# My answer: array([F, F, T, T, T]) (abbreviated)
one_dim > 2
array([False, False, True, True, True])
# For NumPy arrays, logical operations are done with the operators "&" and "|",
# rather than the usual Python "and" and "or". Given that, can you evaluate this expression?
# (one_dim > 4) | (one_dim == 1)
# My answer: array([True, False, False, False, True])
(one_dim > 4) | (one_dim == 1)
array([ True, False, False, False, True])
# What is the result of -one_dim?
# My answer array([-1, -2, -3, -4, -5])
-one_dim
array([-1, -2, -3, -4, -5])
# np.absolute take the absolute value of each element. Given that, what would the result be of the following expression:
# My answer 4,5
np.absolute(-(one_dim[3:]))
array([4, 5])
# What is returned by one_dim.sum()?
# My answer: 15
one_dim.sum()
15
# What is the value of one_dim.mean() ?
print(one_dim)
one_dim.mean() # The median!
[1 2 3 4 5]
3.0
# Given the following...
arr = np.array([0., .5, 1.0, 1.5, 2.0]) * np.pi
arr
array([0. , 1.57079633, 3.14159265, 4.71238898, 6.28318531])
# What are the "approximate" values for
# np.around(np.sin(arr), 0)
# My answer: array(0, 1, 0, -1, 0)
np.around(np.sin(arr), 0)
array([ 0., 1., 0., -1., -0.])
# What are the approximate values for
# np.around(np.cos(arr), 0)
# My answer: array(1, 0, -1, 0, 1)
np.around(np.cos(arr), 0)
array([ 1., 0., -1., -0., 1.])
# You're asked to save the following two arrays as is to a file, "data.npz". The arrays should be named as they are here in the file. How could you do it?
people = np.array(["John", "Jenniffer", "Helen", "Miryam"])
languages = np.array([2, 2, 1, 1])
# My answer np.savez("data.npz", people=people, languages=languages)
np.savez("data.npz", people=people, languages=languages)
# How could you load the files again into two new variables, people2 and languages2
arrays = np.load("data.npz")
people2 = arrays["people"]
languages2 = arrays["languages"]
print(people2)
print(languages2)
['John' 'Jenniffer' 'Helen' 'Miryam']
[2 2 1 1]
# Given
# arr = np.arange(1,13).reshape(3,4)
# Save it to a csv file, "myrray.csv".
arr = np.arange(1,13).reshape(3,4)
np.savetxt("myarray.csv", arr, delimiter=",")
# How would you load it back into arr2?
# My guess: arr2 = np.loadtxt("myarray.csv", delimiter=",")
arr2 = np.loadtxt("myarray.csv", delimiter=",")
arr2
array([[ 1., 2., 3., 4.],
[ 5., 6., 7., 8.],
[ 9., 10., 11., 12.]])
# Given
lumberjack = np.array("I'm a lumberjack and I'm OK I sleep all night and I work all day".split(" "))
lumberjack
array(["I'm", 'a', 'lumberjack', 'and', "I'm", 'OK', 'I', 'sleep', 'all',
'night', 'and', 'I', 'work', 'all', 'day'], dtype='<U10')
# How could you capitalize the first character of each string?
# Guessed np.capitalize(lumberjack) wrong
np.char.capitalize(lumberjack)
array(["I'm", 'A', 'Lumberjack', 'And', "I'm", 'Ok', 'I', 'Sleep', 'All',
'Night', 'And', 'I', 'Work', 'All', 'Day'], dtype='<U10')
# What would you expect the value of np.char.capitalize(lunberjack)[2] to be?
# My answer "Lumberjack"
np.char.capitalize(lumberjack)[2]
'Lumberjack'
# How could you surround each string with an initial and final asterisk character (*)?
# Guessed add takes n arguments, wrong, need two arguments at a time
np.char.add(np.char.add("*", lumberjack), "*")
array(["*I'm*", '*a*', '*lumberjack*', '*and*', "*I'm*", '*OK*', '*I*',
'*sleep*', '*all*', '*night*', '*and*', '*I*', '*work*', '*all*',
'*day*'], dtype='<U12')
# np.where can be used to make selections. How can we use this to create a smaller array of those strings that have a length >= 5?
# My guess np.where(lumberjack, np.length(lumberjack) >= 5) Way off.
np.where(np.char.str_len(lumberjack) >=5)
(array([2, 7, 9]),)
lumberjack[np.where(np.char.str_len(lumberjack) >=5)]
array(['lumberjack', 'sleep', 'night'], dtype='<U10')