python how to pad numpy array with zeros

NumPy 1.7.0 (when numpy.pad was added) is pretty old now (it was released in 2013) so even though the question asked for a way without using that function I thought it could be useful to know how that could be achieved using numpy.pad.

It's actually pretty simple:

>>> import numpy as np>>> a = np.array([[ 1.,  1.,  1.,  1.,  1.],...               [ 1.,  1.,  1.,  1.,  1.],...               [ 1.,  1.,  1.,  1.,  1.]])>>> np.pad(a, [(0, 1), (0, 1)], mode='constant')array([[ 1.,  1.,  1.,  1.,  1.,  0.],       [ 1.,  1.,  1.,  1.,  1.,  0.],       [ 1.,  1.,  1.,  1.,  1.,  0.],       [ 0.,  0.,  0.,  0.,  0.,  0.]])

In this case I used that 0 is the default value for mode='constant'. But it could also be specified by passing it in explicitly:

>>> np.pad(a, [(0, 1), (0, 1)], mode='constant', constant_values=0)array([[ 1.,  1.,  1.,  1.,  1.,  0.],       [ 1.,  1.,  1.,  1.,  1.,  0.],       [ 1.,  1.,  1.,  1.,  1.,  0.],       [ 0.,  0.,  0.,  0.,  0.,  0.]])

Just in case the second argument ([(0, 1), (0, 1)]) seems confusing: Each list item (in this case tuple) corresponds to a dimension and item therein represents the padding before (first element) and after (second element). So:

[(0, 1), (0, 1)]         ^^^^^^------ padding for second dimension ^^^^^^-------------- padding for first dimension  ^------------------ no padding at the beginning of the first axis     ^--------------- pad with one "value" at the end of the first axis.

In this case the padding for the first and second axis are identical, so one could also just pass in the 2-tuple:

>>> np.pad(a, (0, 1), mode='constant')array([[ 1.,  1.,  1.,  1.,  1.,  0.],       [ 1.,  1.,  1.,  1.,  1.,  0.],       [ 1.,  1.,  1.,  1.,  1.,  0.],       [ 0.,  0.,  0.,  0.,  0.,  0.]])

In case the padding before and after is identical one could even omit the tuple (not applicable in this case though):

>>> np.pad(a, 1, mode='constant')array([[ 0.,  0.,  0.,  0.,  0.,  0.,  0.],       [ 0.,  1.,  1.,  1.,  1.,  1.,  0.],       [ 0.,  1.,  1.,  1.,  1.,  1.,  0.],       [ 0.,  1.,  1.,  1.,  1.,  1.,  0.],       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.]])

Or if the padding before and after is identical but different for the axis, you could also omit the second argument in the inner tuples:

>>> np.pad(a, [(1, ), (2, )], mode='constant')array([[ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],       [ 0.,  0.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],       [ 0.,  0.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],       [ 0.,  0.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.]])

However I tend to prefer to always use the explicit one, because it's just to easy to make mistakes (when NumPys expectations differ from your intentions):

>>> np.pad(a, [1, 2], mode='constant')array([[ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],       [ 0.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],       [ 0.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],       [ 0.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.]])

Here NumPy thinks you wanted to pad all axis with 1 element before and 2 elements after each axis! Even if you intended it to pad with 1 element in axis 1 and 2 elements for axis 2.

I used lists of tuples for the padding, note that this is just "my convention", you could also use lists of lists or tuples of tuples, or even tuples of arrays. NumPy just checks the length of the argument (or if it doesn't have a length) and the length of each item (or if it has a length)!

Very simple, you create an array containing zeros using the reference shape:

result = np.zeros(b.shape)# actually you can also use result = np.zeros_like(b) # but that also copies the dtype not only the shape

and then insert the array where you need it:

result[:a.shape[0],:a.shape[1]] = a

and voila you have padded it:

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

You can also make it a bit more general if you define where your upper left element should be inserted

result = np.zeros_like(b)x_offset = 1  # 0 would be what you wantedy_offset = 1  # 0 in your caseresult[x_offset:a.shape[0]+x_offset,y_offset:a.shape[1]+y_offset] = aresultarray([[ 0.,  0.,  0.,  0.,  0.,  0.],       [ 0.,  1.,  1.,  1.,  1.,  1.],       [ 0.,  1.,  1.,  1.,  1.,  1.],       [ 0.,  1.,  1.,  1.,  1.,  1.]])

but then be careful that you don't have offsets bigger than allowed. For x_offset = 2 for example this will fail.

If you have an arbitary number of dimensions you can define a list of slices to insert the original array. I've found it interesting to play around a bit and created a padding function that can pad (with offset) an arbitary shaped array as long as the array and reference have the same number of dimensions and the offsets are not too big.

def pad(array, reference, offsets):    """    array: Array to be padded    reference: Reference array with the desired shape    offsets: list of offsets (number of elements must be equal to the dimension of the array)    """    # Create an array of zeros with the reference shape    result = np.zeros(reference.shape)    # Create a list of slices from offset to offset + shape in each dimension    insertHere = [slice(offset[dim], offset[dim] + array.shape[dim]) for dim in range(a.ndim)]    # Insert the array in the result at the specified offsets    result[insertHere] = a    return result

And some test cases:

import numpy as np# 1 Dimensiona = np.ones(2)b = np.ones(5)offset = [3]pad(a, b, offset)# 3 Dimensionsa = np.ones((3,3,3))b = np.ones((5,4,3))offset = [1,0,0]pad(a, b, offset)

I understand that your main problem is that you need to calculate d=b-a but your arrays have different sizes. There is no need for an intermediate padded c

You can solve this without padding:

import numpy as npa = np.array([[ 1.,  1.,  1.,  1.,  1.],              [ 1.,  1.,  1.,  1.,  1.],              [ 1.,  1.,  1.,  1.,  1.]])b = np.array([[ 3.,  3.,  3.,  3.,  3.,  3.],              [ 3.,  3.,  3.,  3.,  3.,  3.],              [ 3.,  3.,  3.,  3.,  3.,  3.],              [ 3.,  3.,  3.,  3.,  3.,  3.]])d = b.copy()d[:a.shape[0],:a.shape[1]] -=  aprint d

Output:

[[ 2.  2.  2.  2.  2.  3.] [ 2.  2.  2.  2.  2.  3.] [ 2.  2.  2.  2.  2.  3.] [ 3.  3.  3.  3.  3.  3.]]