python - Reshape list of lists based on position of the element -

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this question has answer here:

i have list of lists looks this;

[[4, 0, 1], [0, 0, 1], [0, 1, 2], [1, 1, 0], [2, 0, 0]]

which efficient way 3 lists above list based on position of element?

result:

[4,0,0,1,2] [0,0,1,1,0] [1,1,2,0,0]

you can use zip , list comprehension:

>>> lst = [[4, 0, 1], [0, 0, 1], [0, 1, 2], [1, 1, 0], [2, 0, 0]] >>> [list(x) x in zip(*lst)] [[4, 0, 0, 1, 2], [0, 0, 1, 1, 0], [1, 1, 2, 0, 0]] >>>

placing * before lst unpacks list arguments zip function. in case:

zip(*lst)

is equivalent writing:

zip([4, 0, 1], [0, 0, 1], [0, 1, 2], [1, 1, 0], [2, 0, 0])

zip zips these lists returning iterator of tuples n-th tuple contains n-th item each of lists. in case, returns:1

>>> list(zip(*lst)) [(4, 0, 0, 1, 2), (0, 0, 1, 1, 0), (1, 1, 2, 0, 0)] >>>

finally, list comprehension converts these tuples lists , collects them new list.

1you need call list() on zip(*lst) view items because zip returns iterator in python 3.x.


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