PattuX - 1 year ago 68

Python Question

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`

`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`

`facs`

`facs`

`l1`

`l1`

`l2`

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

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Answer Source

`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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