Find All List Permutations Of Splitting A String In Python

June 16, 2023 Post a Comment

I have a string of letters that I'd like to split into all possible combinations (the order of letters must remain fixed), so that: s = 'monkey' becomes: combinations = [['m', 'on

Solution 1:

defsplitter(str):
    for i inrange(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.

Solution 2:

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-pythonfrom itertools import chain, combinations

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

defpartition(iterable):
    s = sliceable(iterable)
    n = len(s)
    b, mid, e = [0], list(range(1, n)), [n]
    getslice = s.__getitem__
    splits = (d for i inrange(n) for d in combinations(mid, i))
    return [[s[sl] for sl inmap(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']

Solution 3:

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'):

{'key'} # 'key' itself
{'k', 'ey'} # substring 'k' union 1st permutation of 'ey' = {'e, 'y'}
{'k', 'e', 'y'} # substring 'k' union 2nd permutation of 'ey' = {'ey'}
{'ke', 'y'} # substring 'ke' union 1st and only permutation of 'y' = {'y'}
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']

Solution 4:

Consider more_itertools.partitions:

Given

import more_itertools as mits="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:

Baca Juga

[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.

Solution 5:

A string (as opposed to list) oriented approach is to think of the each adjacent pair of characters being separated by either a space or empty string. That can be mapped to 1 and 0, and the number of possible splits are a power of 2:

2 ^ (len(s)-1)

for example, "key" can have '' or ' ' separating 'ke' and a '' or ' ' separating 'ey' which leads to 4 possibilities:

key ('' between 'k' and 'e', '' between 'e' and 'y')
k ey (' ' between 'k' and 'e', '' between 'e' and 'y')
k e y (' ' between 'k' and 'e', ' ' between 'e' and 'y')
ke y ('' between 'k' and 'e', ' ' between 'e' and 'y')

An unreadable python one liner that gives you a generator in string form:

operator_positions = (''.join([str(a >> i & 1).replace('0', '').replace('1', ' ') + s[len(s)-1-i] for i inrange(len(s)-1, -1, -1)]) for a inrange(pow(2, len(s)-1)))

A readable version of this generator with comments and sample:

s = 'monkey'
s_length = len(s)-1# represents the number of ' ' or '' that can split digits

operator_positions = (
    ''.join(
        [str(a >> i & 1).replace('0', '').replace('1', ' ') + s[s_length-i]
         for i inrange(s_length, -1, -1)])   # extra digit is for blank string to always precede first digitfor a inrange(pow(2, s_length))   # binary number loop
)
for i in operator_positions:
    print i

str(a >> i & 1) converts a into a binary string, which then has it's 0's and 1's replaced by '' and ' ', respectively. The binary string is an extra digit long so that the first digit is always ''. That way, as the digit splitter is combined with the first character, it always results in just the first character.

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