I have
a = [1, 2, 3, 4, 5]
b = ['a', 'b', 'c', 'd']
where a has one more element.
zip(a,b) returns [(1, 'a'), (2, 'b'), (3, 'c'), (4, 'd')]. However, I want
[1, 'a', 2, 'b', 3, 'c', 4, 'd']
What is the most elegant way?
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Friedrich -- Слава Україні
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I use now `sum(zip(a, b), ())`. It returns a flat tuple. It is a modification of this controverse [answer](https://stackoverflow.com/a/952946/11769765). – Friedrich -- Слава Україні May 21 '20 at 21:42
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Mapping the `+` operator to all elements, I come across [reduce](https://book.pythontips.com/en/latest/map_filter.html#reduce), thus `reduce(lambda x,y: x+y, zip(a,b))`. But this is also given in https://stackoverflow.com/a/952943/11769765. – Friedrich -- Слава Україні Jun 07 '20 at 13:49
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Another duplicate: [Python: Intertwining two lists](https://stackoverflow.com/q/6356041/7851470). – Georgy Jun 19 '20 at 23:07
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len_a = len(list_a) ; result = [ list_a[ii//2] if ii%2==0 else list_b[ii//2] for ii in range(2*len_a) ] – Daniol Dan Aug 11 '22 at 21:19
2 Answers
5
itertools has a function for this.
from itertools import chain
a = [1, 2, 3, 4, 5]
b = ['a', 'b', 'c', 'd']
result = list(chain.from_iterable(zip(a, b)))
Adam Smith
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Using list comprehensions, one can use the following:
a = [1, 2, 3, 4, 5]
b = ['a', 'b', 'c', 'd']
result = [item for sublist in zip(a, b) for item in sublist]
# result is [1, 'a', 2, 'b', 3, 'c', 4, 'd']
This uses the list-flattening comprehension in https://stackoverflow.com/a/952952/5666087.
jkr
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but those are two for loops, I think this is not as clean, but might be faster ` xx = [] , _ = [xx.extend([x,y]) for x, y in zip(a,b)] ` now xx is ` [1, 'a', 2, 'b', 3, 'c', 4, 'd'] ` – Jose Angel Sanchez May 21 '20 at 21:22
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Maybe, but I would choose clean code over fast code. *Premature optimization is the root of all evil* – jkr May 21 '20 at 21:27