Nested For Loops Using List Comprehension

Question

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?

score 173 · Accepted Answer · edited Aug 20 '21 at 12:37

173

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   
       ]

edited Aug 20 '21 at 12:37

Alex

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answered Sep 03 '10 at 04:58

aaronasterling

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3

What if you want to do the nested loop to iterate over a nested list? Something like: [print('a') for ax in axs for axs in axes] is printing a bunch on [None, None...] up to len(axes) – Pablo Ruiz Ruiz Nov 14 '18 at 16:09
@Pablo I think you had your loops reversed. `L1 = [[[e1, e2, ...], ...], ...]` -> `[ e for L2 in L1 for L3 in L2 for e in L3 ]` – AnOccasionalCashew Oct 19 '19 at 04:25
1

The order of the `for` statements is the same as if you had written it as two `for` loops on two separate lines. – Boris Verkhovskiy Jan 08 '20 at 17:07
aaronasterling Can i use conditional statements in the following? – Salik Malik Jul 24 '20 at 10:04
`lst = [j+k if BLAHBLAHBLAH for j in s1 for k in s2]` or something – Salik Malik Jul 24 '20 at 10:06
This behavior is quite unexpected. I would prefer `for k` (the latter one) to be the outer loop which is consistent with math convention/natural language. – Shen Zhuoran Jul 26 '21 at 18:42

score 42 · Answer 2 · answered Sep 03 '10 at 07:53

42

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')

answered Sep 03 '10 at 07:53

miles82

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The Cartesian product is only a subset of the more general answer provided by @aaronasterling in which the inner loop values can be a function of each outer loop values. For example: for x in range(5) for y in range(x). – jpcgt Jun 01 '22 at 21:58

mirekphd · Answer 3 · 2022-03-13T10:46:49.813

1

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(list - itertools.product returns an iterator.

edited Mar 13 '22 at 10:46

answered Mar 13 '22 at 10:41

mirekphd

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score 0 · Answer 4 · edited Mar 14 '17 at 09:19

0

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

edited Mar 14 '17 at 09:19

Tshilidzi Mudau

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answered Mar 24 '14 at 20:53

Stefan Gruenwald

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Accourding to OP's question, I think the output should give couples of letters, and there should be 9 combinations. – Mattia Jul 30 '15 at 08:42
1

What happened to: abd, abe, acd, ace, acf, adb, adc, ade, adf, aeb, aed, afb, afc, afd, afe, and all the ones starting with c, e, or f? Even if order is not important, omitted are: bda, ade, etc. – Harry Binswanger Mar 26 '17 at 02:38
The way this works is, that the left most position can only be "a" or "d", the middle position can only be "b" or "e", and the right position can only be "c" or "f". – Stefan Gruenwald Mar 27 '17 at 22:31

Nested For Loops Using List Comprehension

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