zero padding numpy array

As numpy have to know size of an array just prior to its initialization, best solution would be a numpy based constructor for such case. Sadly, as far as I know, there is none.

Probably not ideal, but slightly faster solution will be create numpy array with zeros and fill with list values.

import numpy as npdef pad_list(lst):    inner_max_len = max(map(len, lst))    map(lambda x: x.extend([0]*(inner_max_len-len(x))), lst)    return np.array(lst)def apply_to_zeros(lst, dtype=np.int64):    inner_max_len = max(map(len, lst))    result = np.zeros([len(lst), inner_max_len], dtype)    for i, row in enumerate(lst):        for j, val in enumerate(row):            result[i][j] = val    return result

Test case:

>>> pad_list([[ 1, 2, 3], [2], [2, 4]])array([[1, 2, 3],       [2, 0, 0],       [2, 4, 0]])>>> apply_to_zeros([[ 1, 2, 3], [2], [2, 4]])array([[1, 2, 3],       [2, 0, 0],       [2, 4, 0]])

Performance:

>>> timeit.timeit('from __main__ import pad_list as f; f([[ 1, 2, 3], [2], [2, 4]])', number = 10000)0.3937079906463623>>> timeit.timeit('from __main__ import apply_to_zeros as f; f([[ 1, 2, 3], [2], [2, 4]])', number = 10000)0.1344289779663086

Not strictly a function from numpy, but you could do something like this

from itertools import izip, izip_longestimport numpya=[[1,2,3], [4], [5,6]]res1 = numpy.array(list(izip(*izip_longest(*a, fillvalue=0))))

or, alternatively:

res2=numpy.array(list(izip_longest(*a, fillvalue=0))).transpose()

If you use python 3, use zip, and itertools.zip_longest.