Transpose/Unzip Function (inverse of zip)?
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Chapters
00:00 Transpose/Unzip Function (Inverse Of Zip)?
00:24 Accepted Answer Score 918
01:01 Answer 2 Score 29
01:40 Answer 3 Score 21
02:15 Answer 4 Score 24
02:29 Thank you
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Full question
https://stackoverflow.com/questions/1933...
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Tags
#python #list #matrix #transpose
#avk47
ACCEPTED ANSWER
Score 918
In 2.x, zip is its own inverse! Provided you use the special * operator.
>>> zip(*[('a', 1), ('b', 2), ('c', 3), ('d', 4)])
[('a', 'b', 'c', 'd'), (1, 2, 3, 4)]
This is equivalent to calling zip with each element of the list as a separate argument:
zip(('a', 1), ('b', 2), ('c', 3), ('d', 4))
except the arguments are passed to zip directly (after being converted to a tuple), so there's no need to worry about the number of arguments getting too big.
In 3.x, zip returns a lazy iterator, but this is trivially converted:
>>> list(zip(*[('a', 1), ('b', 2), ('c', 3), ('d', 4)]))
[('a', 'b', 'c', 'd'), (1, 2, 3, 4)]
ANSWER 2
Score 29
You could also do
result = ([ a for a,b in original ], [ b for a,b in original ])
It should scale better. Especially if Python makes good on not expanding the list comprehensions unless needed.
(Incidentally, it makes a 2-tuple (pair) of lists, rather than a list of tuples, like zip does.)
If generators instead of actual lists are ok, this would do that:
result = (( a for a,b in original ), ( b for a,b in original ))
The generators don't munch through the list until you ask for each element, but on the other hand, they do keep references to the original list.
ANSWER 3
Score 24
I like to use zip(*iterable) (which is the piece of code you're looking for) in my programs as so:
def unzip(iterable):
return zip(*iterable)
I find unzip more readable.
ANSWER 4
Score 21
If you have lists that are not the same length, you may not want to use zip as per Patricks answer. This works:
>>> zip(*[('a', 1), ('b', 2), ('c', 3), ('d', 4)])
[('a', 'b', 'c', 'd'), (1, 2, 3, 4)]
But with different length lists, zip truncates each item to the length of the shortest list:
>>> zip(*[('a', 1), ('b', 2), ('c', 3), ('d', 4), ('e', )])
[('a', 'b', 'c', 'd', 'e')]
You can use map with no function to fill empty results with None:
>>> map(None, *[('a', 1), ('b', 2), ('c', 3), ('d', 4), ('e', )])
[('a', 'b', 'c', 'd', 'e'), (1, 2, 3, 4, None)]
zip() is marginally faster though.