Return all possible combinations of a string when splitted into n strings

Question

I made a search for stackoverflow about this but couldn\'t find a way to do it. It probably involves itertools.

I want to find all the possible results of splitting

Answer 1

Here is a recipe for partitioning a sequence into n groups, based on code by Raymond Hettinger:

import itertools as IT

def partition_into_n(iterable, n, chain=IT.chain, map=map):
    """
    Based on http://code.activestate.com/recipes/576795/ (Raymond Hettinger)
    Modified to include empty partitions, and restricted to partitions of length n
    """
    s = iterable if hasattr(iterable, '__getslice__') else tuple(iterable)
    m = len(s)
    first, middle, last = [0], range(m + 1), [m]
    getslice = s.__getslice__
    return (map(getslice, chain(first, div), chain(div, last))
            for div in IT.combinations_with_replacement(middle, n - 1))

---

In [149]: list(partition_into_n(s, 3))
Out[149]: 
[['', '', 'thisisateststring'],
 ['', 't', 'hisisateststring'],
 ['', 'th', 'isisateststring'],
 ['', 'thi', 'sisateststring'],
 ...
 ['thisisateststrin', '', 'g'],
 ['thisisateststrin', 'g', ''],
 ['thisisateststring', '', '']]

---

It's slower than the recursive solution for small n,

def partitions_recursive(s, n):
    if not n>1:
        yield [s]
        return
    for i in range(len(s) + 1):
        for tail in partitions_recursive(s[i:], n - 1):
            yield [s[:i]] + tail

s = "thisisateststring"
In [150]: %timeit list(partition_into_n(s, 3))
1000 loops, best of 3: 354 µs per loop

In [151]: %timeit list(partitions_recursive(s, 3))
10000 loops, best of 3: 180 µs per loop

but as you might expect, it is faster for large n (as the recursion depth increases):

In [152]: %timeit list(partition_into_n(s, 10))
1 loops, best of 3: 9.2 s per loop

In [153]: %timeit list(partitions_recursive(s, 10))
1 loops, best of 3: 10.2 s per loop