Python list transpose and fill

Question

I have a list of lists such that the length of each inner list is either 1 or n (assume n > 1).

>>> uneven = [[1], [47, 17, 2, 3], [3], [12, 5, 75, 33]]

I want to transpose the list, but instead of truncating the longer list (as with zip) or filling the shorter lists with None, I want to fill the shorter lists with their own singular value. In other words, I'd like to get:

>>> [(1, 47, 3, 12), (1, 17, 3, 5), (1, 2, 3, 75), (1, 3, 3, 33)]

I can do this with a couple of iterations:

>>> maxlist = len(max(*uneven, key=len))
>>> maxlist
4
>>> from itertools import repeat
>>> uneven2 = [x if len(x) == maxlist else repeat(x[0], maxlist) for x in uneven]
>>> uneven2
[[1, 1, 1, 1], [47, 17, 2, 3], [3, 3, 3, 3], [12, 5, 75, 33]]
>>> zip(*uneven2)
[(1, 47, 3, 12), (1, 17, 3, 5), (1, 2, 3, 75), (1, 3, 3, 33)]

But is there a better approach? Do I really need to know maxlist in advance to accomplish this?

Answer 1

You can repeat one element list forever:

uneven = [[1], [47, 17, 2, 3], [3], [12, 5, 75, 33]]

from itertools import repeat

print zip(*(repeat(*x) if len(x)==1 else x for x in uneven))

Answer 2

You could use itertools.cycle() instead:

>>> from itertools import cycle
>>> uneven3 = [x if len(x) != 1 else cycle(x) for x in uneven]
>>> zip(*uneven3)
[(1, 47, 3, 12), (1, 17, 3, 5), (1, 2, 3, 75), (1, 3, 3, 33)]

That means you don't need to know maxlist ahead of time.