Is there a middle ground between `zip` and `zip_longest`

Question

Say I have these three lists:

a = [1, 2, 3, 4]
b = [5, 6, 7, 8, 9]
c = [10, 11, 12]

Is there a builtin function such that:

somezip(a, b) == [(1, 5), (2, 6), (3, 7), (4, 8)]
somezip(a, c) == [(1, 10), (2, 11), (3, 12), (4, None)]

Behaving somewhere between zip and zip_longest?

Answer 1

No there isn't, but you can easily combine the functionality of takewhile and izip_longest to achieve what you want

from itertools import takewhile, izip_longest
from operator import itemgetter
somezip = lambda *p: list(takewhile(itemgetter(0),izip_longest(*p)))

(In case the first-iterator may have items which evaluates to False, you can replace the itemgetter with a lambda expression - refer @ovgolovin's comment)

somezip = lambda *p: list(takewhile(lambda e: not e[0] is None,izip_longest(*p)))

Examples

>>> from itertools import takewhile, izip_longest
>>> from operator import itemgetter
>>> a = [1, 2, 3, 4]
>>> b = [5, 6, 7, 8, 9]
>>> c = [10, 11, 12]
>>> somezip(a,b)
[(1, 5), (2, 6), (3, 7), (4, 8)]
>>> somezip(a,c)
[(1, 10), (2, 11), (3, 12), (4, None)]
>>> somezip(b,c)
[(5, 10), (6, 11), (7, 12), (8, None), (9, None)]

Answer 2

import itertools as it

somezip = lambda *x: it.islice(it.izip_longest(*x), len(x[0]))



>>> list(somezip(a,b))
[(1, 5), (2, 6), (3, 7), (4, 8)]

>>> list(somezip(a,c))
[(1, 10), (2, 11), (3, 12), (4, None)]