11: Numpy - Slicing and Indexing

import numpy as np

Indexing and Slicing

The operation is the same as for lists

A = np.array([[1, 2, 3], [4, 5, 6]])
print(A)

[[1 2 3]
 [4 5 6]]

# To access row 0, column 1
A[0, 1] 

2

# To select blocks in row (0-1) column (0-1)
A[0:2, 0:2]

array([[1, 2],
       [4, 5]])

A[0:2, 0:2] = 10
print(A)

[[10 10  3]
 [10 10  6]]

Boolean Indexing

A = np.array([[1, 2, 3], [4, 5, 6]])
 
print(A<5) # Boolean mask
 
print(A[A < 5]) # subset filtered by the boolean mask
 
A[A<5] = 4 # convert the selected values.
print(A)

[[ True  True  True]
 [ True False False]]
[1 2 3 4]
[[4 4 4]
 [4 5 6]]

Exercises and Solutions

Exercise 1

Fill the 4 blocks of the middle with “1’s

B = np.zeros((4, 4))
B

array([[0., 0., 0., 0.],
       [0., 0., 0., 0.],
       [0., 0., 0., 0.],
       [0., 0., 0., 0.]])

# SOLUTION
B[1:3 , 1:3] = 1
B

array([[0., 0., 0., 0.],
       [0., 1., 1., 0.],
       [0., 1., 1., 0.],
       [0., 0., 0., 0.]])

Exercise 2

Fill in the table of “1” (every other row, every other column)

C = np.zeros((5, 5))
C

array([[0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0.]])

# SOLUTION
C[::2, ::2] = 1
C

array([[1., 0., 1., 0., 1.],
       [0., 0., 0., 0., 0.],
       [1., 0., 1., 0., 1.],
       [0., 0., 0., 0., 0.],
       [1., 0., 1., 0., 1.]])

Exercise 3

On the image below, perform a slicing to keep only half of the image (in its center) and replace all pixels > 150 by pixels = 255

from scipy import misc
import matplotlib.pyplot as plt
face = misc.face(gray=True)
plt.imshow(face, cmap=plt.cm.gray)
plt.show()
face.shape

(768, 1024)

SOLUTION

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x, y = face.shape
zoom_face = face[x//4 : -x//4, y //4: -y//4] # resize by dividing each dimension by 4 (integer division)
zoom_face[zoom_face>150] = 255 # boolean indexing
plt.imshow(zoom_face, cmap=plt.cm.gray)
plt.show()