Separating a String

问题

Given a string, I want to generate all possible combinations. In other words, all possible ways of putting a comma somewhere in the string.

For example:

input:  ["abcd"]
output: ["abcd"]
        ["abc","d"]
        ["ab","cd"]
        ["ab","c","d"]
        ["a","bc","d"]
        ["a","b","cd"]
        ["a","bcd"]
        ["a","b","c","d"]

I am a bit stuck on how to generate all the possible lists. Combinations will just give me lists with length of subset of the set of strings, permutations will give all possible ways to order.

I can make all the cases with only one comma in the list because of iterating through the slices, but I can't make cases with two commas like "ab","c","d" and "a","b","cd"

My attempt w/slice:

test="abcd"

for x in range(len(test)):
     print test[:x],test[x:]

回答1:

How about something like:

from itertools import combinations

def all_splits(s):
    for numsplits in range(len(s)):
        for c in combinations(range(1,len(s)), numsplits):
            split = [s[i:j] for i,j in zip((0,)+c, c+(None,))]
            yield split

after which:

>>> for x in all_splits("abcd"):
...     print(x)
...     
['abcd']
['a', 'bcd']
['ab', 'cd']
['abc', 'd']
['a', 'b', 'cd']
['a', 'bc', 'd']
['ab', 'c', 'd']
['a', 'b', 'c', 'd']

回答2:

You can certainly use itertools for this, but I think it's easier to write a recursive generator directly:

def gen_commas(s):
    yield s
    for prefix_len in range(1, len(s)):
        prefix = s[:prefix_len]
        for tail in gen_commas(s[prefix_len:]):
            yield prefix + "," + tail

Then

print list(gen_commas("abcd"))

prints

['abcd', 'a,bcd', 'a,b,cd', 'a,b,c,d', 'a,bc,d', 'ab,cd', 'ab,c,d', 'abc,d']

I'm not sure why I find this easier. Maybe just because it's dead easy to do it directly ;-)

回答3:

You could generate the power set of the n - 1 places that you could put commas:

what's a good way to combinate through a set?

and then insert commas in each position.

回答4:

Using itertools:

import itertools
input_str =  "abcd"
for k in range(1,len(input_str)):
    for subset in itertools.combinations(range(1,len(input_str)), k): 
        s = list(input_str)
        for i,x in enumerate(subset): s.insert(x+i, ",")
        print "".join(s)

Gives:

a,bcd
ab,cd
abc,d
a,b,cd
a,bc,d
ab,c,d
a,b,c,d

Also a recursive version:

def commatoze(s,p=1):
    if p == len(s):
        print s
        return
    commatoze(s[:p] + ',' + s[p:], p + 2)
    commatoze(s, p + 1)

input_str =  "abcd"
commatoze(input_str)

回答5:

You can solve the integer composition problem and use the compositions to guide where to split the list. Integer composition can be solved fairly easily with a little bit of dynamic programming.

def composition(n):
    if n == 1: 
        return [[1]] 
    comp = composition (n - 1) 
    return [x + [1] for x in comp] + [y[:-1] + [y[-1]+1] for y in comp]

def split(lst, guide):
    ret = []
    total = 0
    for g in guide:
        ret.append(lst[total:total+g])
        total += g
    return ret

lst = list('abcd')
for guide in composition(len(lst)):
    print split(lst, guide)

Another way to generate integer composition:

from itertools import groupby
def composition(n):
    for i in xrange(2**(n-1)):
        yield [len(list(group)) for _, group in groupby('{0:0{1}b}'.format(i, n))]

回答6:

Given

import more_itertools as mit

Code

list(mit.partitions("abcd"))

Output

[[['a', 'b', 'c', 'd']],
 [['a'], ['b', 'c', 'd']],
 [['a', 'b'], ['c', 'd']],
 [['a', 'b', 'c'], ['d']],
 [['a'], ['b'], ['c', 'd']],
 [['a'], ['b', 'c'], ['d']],
 [['a', 'b'], ['c'], ['d']],
 [['a'], ['b'], ['c'], ['d']]]

Install more_itertools via > pip install more-itertools.

来源：https://stackoverflow.com/questions/20754857/separating-a-string