Is there a zip-like function that pads to longest length?

user_116 0
Viewed 109259

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)]
zip list python
8 Answers

In addition to the accepted answer, if you're working with iterables that might be different lengths but shouldn't be, it's recommended to pass strict=True to zip() (supported since Python 3.10).

To quote the documentation:

zip() is often used in cases where the iterables are assumed to be of equal length. In such cases, it’s recommended to use the strict=True option. Its output is the same as regular zip():

>>> list(zip(('a', 'b', 'c'), (1, 2, 3), strict=True))
[('a', 1), ('b', 2), ('c', 3)]

Unlike the default behavior, it checks that the lengths of iterables are identical, raising a ValueError if they aren’t:

>>> list(zip(range(3), ['fee', 'fi', 'fo', 'fum'], strict=True))
Traceback (most recent call last):
...
ValueError: zip() argument 2 is longer than argument 1

Without the strict=True argument, any bug that results in iterables of different lengths will be silenced, possibly manifesting as a hard-to-find bug in another part of the program.

user_2442970

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)))
user_57094980

To add to the answers already given, the following works for any iterable and does not use itertools, answering @ProdIssue's question:

def zip_longest(*iterables, default_value):
    iterators = tuple(iter(i) for i in iterables)
    sentinel = object()
    while True:
        new = tuple(next(i, sentinel) for i in iterators)
        if all(n is sentinel for n in new):
            return
        yield tuple(default_value if n is sentinel else n for n in new)

The use of sentinel is needed so an iterator yielding default_value will not be erroneously be identified as empty.

user_106666680

Just use iterators, nothing fancy.

def zip_longest(*iterables):
    items = 0
    for iterable in iterables:
        items = max(items, len(iterable))

    iters = [iter(iterable) for iterable in iterables]
    while items:
        yield (*[next(i, None) for i in iters],)
        items -= 1
user_40462350

Im using a 2d array but the concept is the similar using python 2.x:

if len(set([len(p) for p in printer])) > 1:
    printer = [column+['']*(max([len(p) for p in printer])-len(column)) for column in printer]
user_122040580
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