# Nested For Loops Using List Comprehension

Each Answer to this Q is separated by one/two green lines.

If I had two strings, 'abc' and 'def', I could get all combinations of them using two for loops:

for j in s1:
for k in s2:
print(j, k)

However, I would like to be able to do this using list comprehension. I’ve tried many ways, but have never managed to get it. Does anyone know how to do this?

lst = [j + k for j in s1 for k in s2]

or

lst = [(j, k) for j in s1 for k in s2]

if you want tuples.

Like in the question, for j... is the outer loop, for k... is the inner loop.

Essentially, you can have as many independent ‘for x in y’ clauses as you want in a list comprehension just by sticking one after the other.

To make it more readable, use multiple lines:

lst = [
j + k         # result
for j in s1   # for loop
for k in s2 # for loop
# condition
]

Since this is essentially a Cartesian product, you can also use itertools.product. I think it’s clearer, especially when you have more input iterables.

itertools.product('abc', 'def', 'ghi')

It’s just a ready-to-go version of @miles82 answer (please give credit where it’s due):

from itertools import product
list(map(list, product('abc', 'def') ))

Output:

[['a', 'd'],
['a', 'e'],
['a', 'f'],
['b', 'd'],
['b', 'e'],
['b', 'f'],
['c', 'd'],
['c', 'e'],
['c', 'f']]

In case you wondered why we need list(map(listitertools.product returns an iterator.

Try recursion too:

s=""
s1="abc"
s2="def"
def combinations(s,l):
if l==0:
print s
else:
combinations(s+s1[len(s1)-l],l-1)
combinations(s+s2[len(s2)-l],l-1)

combinations(s,len(s1))

Gives you the 8 combinations:

abc
abf
aec
aef
dbc
dbf
dec
def

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .