Zero pad numpy array

Question

What's the more pythonic way to pad an array with zeros at the end?

def pad(A, length):     ...  A = np.array([1,2,3,4,5]) pad(A, 8)    # expected : [1,2,3,4,5,0,0,0] 

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In my real use case, in fact I want to pad an array to the closest multiple of 1024. Ex: 1342 => 2048, 3000 => 3072

Answer 1

numpy.pad with constant mode does what you need, where we can pass a tuple as second argument to tell how many zeros to pad on each size, a (2, 3) for instance will pad 2 zeros on the left side and 3 zeros on the right side:

Given A as:

A = np.array([1,2,3,4,5])  np.pad(A, (2, 3), 'constant') # array([0, 0, 1, 2, 3, 4, 5, 0, 0, 0]) 

It's also possible to pad a 2D numpy arrays by passing a tuple of tuples as padding width, which takes the format of ((top, bottom), (left, right)):

A = np.array([[1,2],[3,4]])  np.pad(A, ((1,2),(2,1)), 'constant')  #array([[0, 0, 0, 0, 0],           # 1 zero padded to the top #       [0, 0, 1, 2, 0],           # 2 zeros padded to the bottom #       [0, 0, 3, 4, 0],           # 2 zeros padded to the left #       [0, 0, 0, 0, 0],           # 1 zero padded to the right #       [0, 0, 0, 0, 0]]) 

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For your case, you specify the left side to be zero and right side pad calculated from a modular division:

B = np.pad(A, (0, 1024 - len(A)%1024), 'constant') B # array([1, 2, 3, ..., 0, 0, 0]) len(B) # 1024 

For a larger A:

A = np.ones(3000) B = np.pad(A, (0, 1024 - len(A)%1024), 'constant') B # array([ 1.,  1.,  1., ...,  0.,  0.,  0.])  len(B) # 3072