如何用尾随零填充多个列表？

Question

Suppose I have two lists containing the same number of elements which are lists of integers. For instance:

a = [[1, 7, 3, 10, 4], [1, 3, 8], ..., [2, 5, 10, 91, 54, 0]]
b = [[5, 4, 23], [1, 2, 0, 4], ..., [5, 15, 11]]

For each index, I want to pad the shorter list with trailing zeros. The example above should look like:

a = [[1, 7, 3, 10, 4], [1, 3, 8, 0], ..., [2, 5, 10, 91, 54, 0]]
b = [[5, 4, 23, 0, 0], [1, 2, 0, 4], ..., [51, 15, 11, 0, 0, 0]]

Is there an elegant way to perform this comparison and padding build into Python lists or perhaps numpy? I am aware that numpy.pad can perform the padding, but its the iteration and comparison over the lists that has got me stuck.

Answer 1

I'm sure there's an elegant Python one-liner for this sort of thing, but sometimes a straightforward imperative solution will get the job done:

for i in xrange(0, len(a)):
    x = len(a[i])
    y = len(b[i])
    diff = max(x, y)
    a[i].extend([0] * (diff - x))
    b[i].extend([0] * (diff - y))

print a, b

Be careful with "elegant" solutions too, because they can be very difficult to comprehend (I can't count the number of times I've come back to a piece of code I wrote using reduce() and had to struggle to figure out how it worked).

Answer 2

One line? Yes. Elegant? No.

In [2]: from itertools import izip_longest
In [3]: A, B = map(list, zip(*[map(list, zip(*izip_longest(l1,l2, fillvalue=0)))
                               for l1,l2 in zip(a,b)]))

In [4]: A
Out[4]: [[1, 7, 3, 10, 4], [1, 3, 8, 0], [2, 5, 10, 91, 54, 0]]

In [5]: B
Out[5]: [[5, 4, 23, 0, 0], [1, 2, 0, 4], [5, 15, 11, 0, 0, 0]]

Answer 3

Note: Creates 2 new lists. Preserves the old lists.

from itertools import repeat

>>> b = [[5, 4, 23], [1, 2, 0, 4],[5, 15, 11]]
>>> a = [[1, 7, 3, 10, 4], [1, 3, 8],[2, 5, 10, 91, 54, 0]]

>>> [y+list(repeat(0, len(x)-len(y))) for x,y in zip(a,b)]
[[5, 4, 23, 0, 0], [1, 2, 0, 4], [5, 15, 11, 0, 0, 0]]

>>> [x+list(repeat(0, len(y)-len(x))) for x,y in zip(a,b)]
[[1, 7, 3, 10, 4], [1, 3, 8, 0], [2, 5, 10, 91, 54, 0]]

Answer 4

a = [[1, 7, 3, 10, 4], [1, 3, 8], [2, 5, 10, 91, 54, 0]]
b = [[5, 4, 23], [1, 2, 0, 4], [5, 15, 11]]

for idx in range(len(a)):
    size_diff = len(a[idx]) - len(b[idx])
    if size_diff < 0:
        a[idx].extend( [0] * abs(size_diff) )
    elif size_diff > 0:
        b[idx].extend( [0] * size_diff )