Nested For Loops Using List Comprehension

PythonFor LoopList Comprehension

Python Problem Overview


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?

Python Solutions


Solution 1 - Python

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   
       ]

Solution 2 - Python

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')

Solution 3 - Python

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.

Solution 4 - Python

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

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionJohn HowardView Question on Stackoverflow
Solution 1 - PythonaaronasterlingView Answer on Stackoverflow
Solution 2 - Pythonmiles82View Answer on Stackoverflow
Solution 3 - PythonmirekphdView Answer on Stackoverflow
Solution 4 - PythonStefan GruenwaldView Answer on Stackoverflow