查找在Python中拆分字符串的所有列表排列

Question

I have a string of letters that I'd like to split into all possible combinations (the order of letters must be remain fixed), so that:

s = 'monkey'

becomes:

combinations = [['m', 'onkey'], ['mo', 'nkey'], ['m', 'o', 'nkey'] ... etc]

Any ideas?

Answer 1

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']

Answer 2

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.

Answer 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') :

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']

Answer 4

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')
• key (' ' 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 in range(len(s)-1, -1, -1)]) for a in range(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 in range(s_length, -1, -1)])   # extra digit is for blank string to always precede first digit
    for a in range(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.

Answer 5

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 .

Answer 6

My solution allows you to also set a threshold for the minimum size of sub string

This is my code:

def split_string (s, min_str_length = 2, root_string=[], results=[] ):
    """
    :param s: word to split, string
    :param min_str_length: the minimum character for a sub string
    :param root_string:  leave empty
    :param results: leave empty
    :return: nested list of all possible combinations of word split according to the minimum substring length
    """
    for i in range(min_str_length,len(s)):
        if i == min_str_length:
            primary_root_string=root_string
        else:
            root_string = primary_root_string
        if len(s[i:])>= min_str_length :
            results.append(list(chain(*[root_string,[s[:i]],[s[i:]]])))
            root_string = list(chain(*[root_string,[s[:i]]]))
            split_string(s[i:], min_str_length, root_string, results)
    return results

Examples of use:

Input: split_string ('monkey', min_str_length = 1, root_string=[], results=[] )
Output: 
[['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']]

or

Input: split_string ('monkey', min_str_length = 2, root_string=[], results=[] )
Output: [['mo', 'nkey'], ['mo', 'nk', 'ey'], ['mon', 'key'], ['monk', 'ey']]