I want to know how I can pad a 2D numpy array with zeros using python 2.6.6 with numpy version 1.5.0. Sorry! But these are my limitations. Therefore I cannot use np.pad. For example, I want to pad a with zeros such that its shape matches b. The reason why I want to do this is so I can do:

b-a

such that

>>> a

array([[ 1., 1., 1., 1., 1.],

[ 1., 1., 1., 1., 1.],

[ 1., 1., 1., 1., 1.]])

>>> b

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.]])

>>> c

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]])

The only way I can think of doing this is appending, however, this seems pretty ugly. is there a cleaner solution possibly using b.shape?

I had to clean it up a bit, and this is what I got:

def pad(array, reference_shape, offsets):

""" array: Array to be padded

reference_shape: tuple of size of ndarray to create

offsets: list of offsets (number of elements must be equal

to the dimension of the array) will throw a ValueError if

offsets is too big and the reference_shape cannot handle

the offsets """

# 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(offsets[dim], offsets[dim] +

array.shape[dim]) for dim in range(array.ndim)]

# Insert the array in the result at the specified offsets

result[insertHere] = array

return result