I've got K feature vectors that all share dimension n but have a variable dimension m (n x m). They all live in a list together.
to_be_padded = []
to_be_padded.append(np.reshape(np.arange(9),(3,3)))
array([[0, 1, 2],
[3, 4, 5],
[6, 7, 8]])
to_be_padded.append(np.reshape(np.arange(18),(3,6)))
array([[ 0, 1, 2, 3, 4, 5],
[ 6, 7, 8, 9, 10, 11],
[12, 13, 14, 15, 16, 17]])
to_be_padded.append(np.reshape(np.arange(15),(3,5)))
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]])
What I am looking for is a smart way to zero pad the rows of these np.arrays such that they all share the same dimension m. I've tried solving it with np.pad but I have not been able to come up with a pretty solution. Any help or nudges in the right direction would be greatly appreciated!
The result should leave the arrays looking like this:
array([[0, 1, 2, 0, 0, 0],
[3, 4, 5, 0, 0, 0],
[6, 7, 8, 0, 0, 0]])
array([[ 0, 1, 2, 3, 4, 5],
[ 6, 7, 8, 9, 10, 11],
[12, 13, 14, 15, 16, 17]])
array([[ 0, 1, 2, 3, 4, 0],
[ 5, 6, 7, 8, 9, 0],
[10, 11, 12, 13, 14, 0]])
You could use np.pad for that, which can also pad 2-D arrays using a tuple of values specifying the padding width, ((top, bottom), (left, right)). For that you could define:
def pad_to_length(x, m):
return np.pad(x,((0, 0), (0, m - x.shape[1])), mode = 'constant')
Usage
You could start by finding the ndarray with the highest amount of columns. Say you have two of them, a and b:
a = np.array([[0, 1, 2],
[3, 4, 5],
[6, 7, 8]])
b = np.array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]])
m = max(i.shape[1] for i in [a,b])
# 5
And then use this parameter to pad the ndarrays:
pad_to_length(a, m)
array([[0, 1, 2, 0, 0],
[3, 4, 5, 0, 0],
[6, 7, 8, 0, 0]])