PattuX - 1 year ago 75
Python Question

# All possibilities to split a list into two lists

I have a list with some elements and want to iterate over all possible ways to divide this list into two lists. By that I mean all permutations, so the order doesn't matter (i.e. Element 1 and 3 could be in the one list and Element 2 in the other). Currently I do it like this, where

`facs`
is my initial list:

``````patterns = []
for i in range(2**(len(facs)-1)):
pattern = []
for j in range((len(facs)-1)):
pattern.append(i//(2**j)%2)
patterns.append(pattern)

for pattern in patterns:
l1 = [facs[-1]]
l2 = []
for i in range(len(pattern)):
if pattern[i] == 1:
l1.append(facs[i])
else:
l2.append(facs[i])
``````

So I basically create a list of length
`len(facs)-1`
and fill it with every possible combination of ones and zeros. I then 'overlay' every pattern with
`facs`
, except for the last element of
`facs`
which is always in
`l1`
, as I'd otherwise get every result twice, as I handle two lists the same, no matter what lists is
`l1`
or
`l2`
.

Is there a faster and more elegant (shorter/more pythonic) way to do this?

`itertools` has `product()` which could be used to generate the masks and `izip()` which could combine the lists for easy filtering. As a bonus, since they return iterators, they don't use much memory.

``````from itertools import *

facs = ['one','two','three']

l1 = []
l2 = []
for pattern in product([True,False],repeat=len(facs)):
l1.append([x[1] for x in izip(pattern,facs) if x[0]])
l2.append([x[1] for x in izip(pattern,facs) if not x[0]])
``````
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