Python: zip-like function that pads to longest len

问题:

Is there a built-in function that works like zip() but that will pad the results so that the length of the resultant list is the length of the longest input rather than the shortest input?

>>> a=[\'a1\']
>>> b=[\'b1\',\'b2\',\'b3\']
>>> c=[\'c1\',\'c2\']

>>> zip(a,b,c)
[(\'a1\', \'b1\', \'c1\')]

>>> What command goes here?
[(\'a1\', \'b1\', \'c1\'), (None, \'b2\', \'c2\'), (None, \'b3\', None)]

回答1:

In Python 3 you can use itertools.zip_longest

>>> list(itertools.zip_longest(a, b, c))
[(\'a1\', \'b1\', \'c1\'), (None, \'b2\', \'c2\'), (None, \'b3\', None)]

You can pad with a different value than None by using the fillvalue parameter:

>>> list(itertools.zip_longest(a, b, c, fillvalue=\'foo\'))
[(\'a1\', \'b1\', \'c1\'), (\'foo\', \'b2\', \'c2\'), (\'foo\', \'b3\', \'foo\')]

With Python 2 you can either use itertools.izip_longest (Python 2.6+), or you can use map with None. It is a little known feature of map (but map changed in Python 3.x, so this only works in Python 2.x).

>>> map(None, a, b, c)
[(\'a1\', \'b1\', \'c1\'), (None, \'b2\', \'c2\'), (None, \'b3\', None)]

回答2:

For Python 2.6x use itertools module\'s izip_longest.

For Python 3 use zip_longest instead (no leading i).

>>> list(itertools.izip_longest(a, b, c))
[(\'a1\', \'b1\', \'c1\'), (None, \'b2\', \'c2\'), (None, \'b3\', None)]

回答3:

non itertools Python 3 solution:

def zip_longest(*lists):
    def g(l):
        for item in l:
            yield item
        while True:
            yield None
    gens = [g(l) for l in lists]    
    for _ in range(max(map(len, lists))):
        yield tuple(next(g) for g in gens)

回答4:

non itertools My Python 2 solution:

if len(list1) < len(list2):
    list1.extend([None] * (len(list2) - len(list1)))
else:
    list2.extend([None] * (len(list1) - len(list2)))