Formulating a nested list comprising elements in the same position from another nested list

Question

How can I take a nested list and re-formulate it into a new nested list in which:

the first list of the new nested list, B, comprises the first element of all of the lists within the nested list A.
the second list of the new nested list, B, comprises the second element of all of the lists within the nested list A.
the nth list of the new nested list, B, comprises the nth element of all of the lists within the nested list A.

Examples are:

A = [[100,200,300], [400,500,600], [1000,1500,300]]

such that B becomes:

B = [[100,400,1000], [200,500,1500], [300,600,300]]

Of course, in reality A contains hundreds of lists not just three. So i need a method to automate this for large numbers of lists.

if the inner lists equal length, `B = list(zip(*(a for a in A)))`. — richard, Jun 26 '23 at 21:16
or just `B = list(zip(*A))` will also work. Alternatively `B = [*zip(*A)]` but I know some people don't love `[* ]`. — slothrop, Jun 26 '23 at 21:27

score 1 · Accepted Answer · answered Jun 26 '23 at 21:12

If you are sure that all the inner lists have the same length, then

B = [[inner_list[i] for inner_list in A] for i in range(len(A))]

Otherwise you can additionnaly check if index i is not out of bounds for current inner list:

B = [
    [inner_list[i] if i < len(inner_list) else None for inner_list in A]
    for i in range(len(A))
]

1 Answers1