You can use np.pad for this:
In [1]: import numpy as np
In [2]: x = np.ones((3, 3))
In [3]: np.pad(x, [(0, 0), (0, 1)], mode='constant')
Out[3]:
array([[ 1., 1., 1., 0.],
[ 1., 1., 1., 0.],
[ 1., 1., 1., 0.]])
Each item of the pad_width (second) argument is a tuple of bef
Hi all,
This should be an easy one but I can not come up with a good solution.
Given an ndarray with a shape of (..., X) I wish to zero-pad it to have
a shape of (..., X + K), presumably obtaining a new array in the process.
My best solution this far is to use
np.zeros(curr.shape[:-1] + (curr
