Is there a simple pad in numpy?

Question

Is there a numpy function that pads an array this way?

import numpy as np

def pad(x, length):
    tmp = np.zeros((length,))
    tmp[:x.shape[0]] = x
    return tmp

x = np.array([1,2,3])
print pad(x, 5)

Output:

[ 1.  2.  3.  0.  0.]

I couldn't find a way to do it with numpy.pad()

score 5 · Accepted Answer · answered Sep 02 '15 at 14:03

5

You can use ndarray.resize():

>>> x = np.array([1,2,3])
>>> x.resize(5)
>>> x
array([1, 2, 3, 0, 0])

Note that this functions behaves differently from numpy.resize(), which pads with repeated copies of the array itself. (Consistency is for people who can't remember everything.)

answered Sep 02 '15 at 14:03

Sven Marnach

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one caveat: it crops the array if `length` is shorter than its shape. – Yariv Sep 03 '15 at 08:24

score 3 · Answer 2 · answered Sep 02 '15 at 14:20

Sven Marnach's suggestion to use ndarray.resize() is probably the simplest way to do it, but for completeness, here's how it can be done with numpy.pad:

In [13]: x
Out[13]: array([1, 2, 3])

In [14]: np.pad(x, [0, 5-x.size], mode='constant')
Out[14]: array([1, 2, 3, 0, 0])

Is there a simple pad in numpy?

2 Answers2