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 before, after for each axis. I've only padded the end of the last axis, but if you wanted to pad both "sides" of it:

In [4]: np.pad(x, [(0, 0), (1, 1)], mode='constant') Out[4]: array([[ 0., 1., 1., 1., 0.], [ 0., 1., 1., 1., 0.], [ 0., 1., 1., 1., 0.]])

Hope that helps, -Joe

On Fri, Jan 3, 2014 at 6:58 AM, Freddie Witherden freddie@witherden.orgwrote:

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.shape[-1] + K,))

followed by an assignment. However, this seems needlessly cumbersome. I looked at np.pad but it does not seem to provide a means of just padding a single axis easily.

Regards, Freddie.

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