There is a wrong answer in 16

[atomAltera]

Two answers are given to 16, one is using np.pad and the second is using indexing.

Z = np.ones((5,5))
Z = np.pad(Z, pad_width=1, mode='constant', constant_values=0)
print(Z)

# Using fancy indexing
Z[:, [0, -1]] = 0
Z[[0, -1], :] = 0
print(Z)

But the first one adds a new border and the second one is just replaces current border.

[rougier]

You're right, we should use Z[1:-1,1:-1] instead of Z with the pad function. Can you make a PR ?

[atomAltera]

Z[1:-1,1:-1] = 0 would not help here. As I know, using indexing we can only replace values in existing rows/columns/cells.
The questions is How to add a border (filled with 0's) around an existing array?. And I think this either could not be solved using indexing, or it would be too complex.

[rougier]

The question is ambiguous but we can take advantage of it by simply distinguishing internal and external border. What do you think?

[rougier]

The question is ambiguous but we can take advantage of it by simply distinguishing internal and external border. What do you think?

[atomAltera]

I think answer with pad function is more preferred because padding can’t/hard to be done with indexing. To me the question is clear: add a border to existing array. So, all existing data have to be preserved and an extra rows/columns have to be added

[ghost]

import numpy as np
x = np.ones((3,3))
print("Original array:")
print(x)
print("0 on the border and 1 inside in the array")
x = np.pad(x, pad_width=1, mode='constant', constant_values=0)
print(x)
#use this

[rougier]

Again I would distinguish external vs internal border in the answer. Can you make a PR?