Sorting 3 lists in python into 1 formatted list

Question

I have a set of lists:

A = [[A1,A2,A3],[A4,A5,A6]...,[A(n-2),A(n-1),A(n)]] #A has length n
B = [[B1,B2,B3],[B4,B5,B6]...,[B(n-2),B(n-1),B(n)]] #B has length n
C = [[C1,C2,C3],[C4,C5,C6]...,[C(n-2),C(n-1),C(n)]] #C has length n

and I want to sort it into the following format:

f = [(A1,A2,A3,B1,B2,B3,C1,C2,C3),(A4,A5,A6,B4,B5,B6,C4,C5,C6),...,(A(n-2),A(n-1),A(n),B(n-2),B(n-1),B(n),C(n-2),C(n-1),C(n))]

I'm pretty new to python and I cant think of a way to do this.

Any input will be greatly appreciated.

Ive started by using:

for item in range(len(A)):
   f[item][0] = A[item][0]
   f[item][1] = A[item][1]
   f[item][2] = A[item][2]

for item in range(len(B)):
   f[item][3] = B[item][0]
   f[item][4] = B[item][1]
   f[item][5] = B[item][2]

for item in range(len(C)):
   f[item][6] = C[item][0]
   f[item][7] = C[item][1]
   f[item][8] = C[item][2]

But this just sets all items in the list f to be equal to the last item in f for some reason.

Jean-François Fabre · Accepted Answer · 2016-11-24 22:51:35Z

1

interleave sublists using zip, and flatten the resulting sublists with itertools.chain in a list comprehension with this nice one-liner:

import itertools

A = [["A1","A2","A3"],["A4","A5","A6"]] #A has length n
B = [["B1","B2","B3"],["B4","B5","B6"]] #B has length n
C = [["C1","C2","C3"],["C4","C5","C6"]] #C has length n

print([tuple(itertools.chain(*l)) for l in zip(A,B,C)])

result:

[('A1', 'A2', 'A3', 'B1', 'B2', 'B3', 'C1', 'C2', 'C3'), ('A4', 'A5', 'A6', 'B4', 'B5', 'B6', 'C4', 'C5', 'C6')]

General case if you have a variable amount of lists, stored in a list of lists:

list_of_lists = [A,B,C]

print([tuple(itertools.chain(*l)) for l in zip(*list_of_lists)])

(use * operator to expand list items to arguments for zip)

note: works well if sublists have different lengths, as long as there are as many sublists in each list (else zip will drop the last one(s)):

A = [["A1","A2","A3"],["A4","A5","A6","A7"],["I will be discarded"]] #A has length n+1, last element will be lost
B = [["B1","B2","B3","B3bis"],["B4","B5","B6"]] #B has length n
C = [["C0","C1","C2","C3"],["C4","C5","C6"]] #C has length n

yields:

[('A1', 'A2', 'A3', 'B1', 'B2', 'B3', 'B3bis', 'C0', 'C1', 'C2', 'C3'), ('A4', 'A5', 'A6', 'A7', 'B4', 'B5', 'B6', 'C4', 'C5', 'C6')]

edited Nov 24, 2016 at 22:51

answered Nov 24, 2016 at 22:43

Jean-François Fabre♦

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1 Comment

user3208454 Over a year ago

Thanks very much, could I get this to work if the lists were of different lengths? Would it be easier to make all lists equal to the length of the largest list?

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