Find all list permutations of splitting a string in Python

The idea is to realize that the permutation of a string s is equal to a set containing s itself, and a set union of each substring X of s with the permutation of s\X. For example, permute('key'):

1. {'key'} # 'key' itself
2. {'k', 'ey'} # substring 'k' union 1st permutation of 'ey' = {'e, 'y'}
3. {'k', 'e', 'y'} # substring 'k' union 2nd permutation of 'ey' = {'ey'}
4. {'ke', 'y'} # substring 'ke' union 1st and only permutation of 'y' = {'y'}
5. Union of 1, 2, 3, and 4, yield all permutations of the string key.

With this in mind, a simple algorithm can be implemented:

>>> def permute(s):
    result = [[s]]
    for i in range(1, len(s)):
        first = [s[:i]]
        rest = s[i:]
        for p in permute(rest):
            result.append(first + p)
    return result

>>> for p in permute('monkey'):
        print(p)    

['monkey']
['m', 'onkey']
['m', 'o', 'nkey']
['m', 'o', 'n', 'key']
['m', 'o', 'n', 'k', 'ey']
['m', 'o', 'n', 'k', 'e', 'y']
['m', 'o', 'n', 'ke', 'y']
['m', 'o', 'nk', 'ey']
['m', 'o', 'nk', 'e', 'y']
['m', 'o', 'nke', 'y']
['m', 'on', 'key']
['m', 'on', 'k', 'ey']
['m', 'on', 'k', 'e', 'y']
['m', 'on', 'ke', 'y']
['m', 'onk', 'ey']
['m', 'onk', 'e', 'y']
['m', 'onke', 'y']
['mo', 'nkey']
['mo', 'n', 'key']
['mo', 'n', 'k', 'ey']
['mo', 'n', 'k', 'e', 'y']
['mo', 'n', 'ke', 'y']
['mo', 'nk', 'ey']
['mo', 'nk', 'e', 'y']
['mo', 'nke', 'y']
['mon', 'key']
['mon', 'k', 'ey']
['mon', 'k', 'e', 'y']
['mon', 'ke', 'y']
['monk', 'ey']
['monk', 'e', 'y']
['monke', 'y']

Consider more_itertools.partitions:

Given

import more_itertools as mit


s = "monkey"

Demo

As-is:

list(mit.partitions(s))
#[[['m', 'o', 'n', 'k', 'e', 'y']],
# [['m'], ['o', 'n', 'k', 'e', 'y']],
# [['m', 'o'], ['n', 'k', 'e', 'y']],
# [['m', 'o', 'n'], ['k', 'e', 'y']],
# [['m', 'o', 'n', 'k'], ['e', 'y']],
# [['m', 'o', 'n', 'k', 'e'], ['y']],
# ...]

After some string joining:

[list(map("".join, x)) for x in mit.partitions(s)]

Output

[['monkey'],
 ['m', 'onkey'],
 ['mo', 'nkey'],
 ['mon', 'key'],
 ['monk', 'ey'],
 ['monke', 'y'],
 ['m', 'o', 'nkey'],
 ['m', 'on', 'key'],
 ['m', 'onk', 'ey'],
 ['m', 'onke', 'y'],
 ['mo', 'n', 'key'],
 ['mo', 'nk', 'ey'],
 ['mo', 'nke', 'y'],
 ['mon', 'k', 'ey'],
 ['mon', 'ke', 'y'],
 ['monk', 'e', 'y'],
 ['m', 'o', 'n', 'key'],
 ['m', 'o', 'nk', 'ey'],
 ['m', 'o', 'nke', 'y'],
 ['m', 'on', 'k', 'ey'],
 ['m', 'on', 'ke', 'y'],
 ['m', 'onk', 'e', 'y'],
 ['mo', 'n', 'k', 'ey'],
 ['mo', 'n', 'ke', 'y'],
 ['mo', 'nk', 'e', 'y'],
 ['mon', 'k', 'e', 'y'],
 ['m', 'o', 'n', 'k', 'ey'],
 ['m', 'o', 'n', 'ke', 'y'],
 ['m', 'o', 'nk', 'e', 'y'],
 ['m', 'on', 'k', 'e', 'y'],
 ['mo', 'n', 'k', 'e', 'y'],
 ['m', 'o', 'n', 'k', 'e', 'y']]

Install via > pip install more_itertools.

http://wordaligned.org/articles/partitioning-with-python contains an interesting post about sequence partitioning, here is the implementation they use:

#!/usr/bin/env python

# From http://wordaligned.org/articles/partitioning-with-python

from itertools import chain, combinations

def sliceable(xs):
    '''Return a sliceable version of the iterable xs.'''
    try:
        xs[:0]
        return xs
    except TypeError:
        return tuple(xs)

def partition(iterable):
    s = sliceable(iterable)
    n = len(s)
    b, mid, e = [0], list(range(1, n)), [n]
    getslice = s.__getitem__
    splits = (d for i in range(n) for d in combinations(mid, i))
    return [[s[sl] for sl in map(slice, chain(b, d), chain(d, e))]
            for d in splits]

if __name__ == '__main__':
    s = "monkey"
    for i in partition(s):
        print i

Which would print:

['monkey']
['m', 'onkey']
['mo', 'nkey']
['mon', 'key']
['monk', 'ey']
['monke', 'y']
['m', 'o', 'nkey']
['m', 'on', 'key']
['m', 'onk', 'ey']
['m', 'onke', 'y']
['mo', 'n', 'key']
['mo', 'nk', 'ey']
['mo', 'nke', 'y']
['mon', 'k', 'ey']
['mon', 'ke', 'y']
['monk', 'e', 'y']
...
['mo', 'n', 'k', 'e', 'y']
['m', 'o', 'n', 'k', 'e', 'y']

def splitter(str):
    for i in range(1, len(str)):
        start = str[0:i]
        end = str[i:]
        yield (start, end)
        for split in splitter(end):
            result = [start]
            result.extend(split)
            yield result

combinations = list(splitter(str))

Note that I defaulted to a generator to save you from running out of memory with long strings.