Nested For Loops Using List Comprehension
PythonFor LoopList ComprehensionPython 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