Is there a zip-like function that pads to longest length?
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Chapters
00:00 Is There A Zip-Like Function That Pads To Longest Length?
00:19 Answer 1 Score 92
00:38 Accepted Answer Score 399
01:14 Answer 3 Score 9
01:28 Answer 4 Score 4
01:39 Thank you
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Full question
https://stackoverflow.com/questions/1277...
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Content licensed under CC BY-SA
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Tags
#python #list #zip
#avk47
Hire the world's top talent on demand or became one of them at Toptal: https://topt.al/25cXVn
and get $2,000 discount on your first invoice
--------------------------------------------------
Music by Eric Matyas
https://www.soundimage.org
Track title: Unforgiving Himalayas Looping
--
Chapters
00:00 Is There A Zip-Like Function That Pads To Longest Length?
00:19 Answer 1 Score 92
00:38 Accepted Answer Score 399
01:14 Answer 3 Score 9
01:28 Answer 4 Score 4
01:39 Thank you
--
Full question
https://stackoverflow.com/questions/1277...
--
Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...
--
Tags
#python #list #zip
#avk47
ACCEPTED ANSWER
Score 399
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)]
ANSWER 2
Score 92
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)]
ANSWER 3
Score 9
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)
ANSWER 4
Score 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)))