python 字符串子集的所有组合

Question

我需要字符串子集的所有组合。此外，长度为 1 的子集后面只能跟着长度 > 1 的子集。例如对于字符串 4824，结果应该是：

 [ [4, 824], [4, 82, 4], [48, 24], [482, 4], [4824] ]

到目前为止，我设法检索了所有可能的子集：

    length = len(number)
    ss = []
    for i in xrange(length):
        for j in xrange(i,length):
            ss.append(number[i:j + 1])

这给了我：

  ['4', '48', '482', '4824', '8', '82', '824', '2', '24', '4']

但我现在不知道如何将它们结合起来。

Answer 1

首先，编写一个生成所有字符串分区的函数：

def partitions(s):
    if s:
        for i in range(1, len(s) + 1):
            for p in partitions(s[i:]):
                yield [s[:i]] + p
    else:
        yield []

这将迭代所有可能的第一段（一个字符、两个字符等），并将它们与字符串的相应剩余部分的所有分区组合。

>>> list(partitions("4824"))
[['4', '8', '2', '4'], ['4', '8', '24'], ['4', '82', '4'], ['4', '824'], ['48', '2', '4'], ['48', '24'], ['482', '4'], ['4824']]

现在，您可以只过滤符合您条件的那些，即没有两个长度为 1 的连续子串的那些。

>>> [p for p in partitions("4824") if not any(len(x) == len(y) == 1 for x, y in zip(p, p[1:]))]
[['4', '82', '4'], ['4', '824'], ['48', '24'], ['482', '4'], ['4824']]

此处，zip(p, p[1:]) 是迭代所有连续项目对的常用方法。

更新：实际上，将您的约束直接合并到 partition 函数中也不难。只需跟踪最后一段并相应地设置最小长度。

def partitions(s, minLength=1):
    if len(s) >= minLength:
        for i in range(minLength, len(s) + 1):
            for p in partitions(s[i:], 1 if i > 1 else 2):
                yield [s[:i]] + p
    elif not s:
        yield []

演示：

>>> print list(partitions("4824"))
[['4', '82', '4'], ['4', '824'], ['48', '24'], ['482', '4'], ['4824']]

Answer 2

看到更多测试用例会很有趣，下面的算法按照你说的做：

s="4824"

def partitions(s):
  yield [s]
  if(len(s)>2):
    for i in range(len(s)-1, 0, -1):
      for g in partitions(s[i:]):
        out = [s[:i]] + g
        if not any([len(out[i]) == len(out[i+1]) and len(out[i])==1 for i in range(len(out)-1)]):
          yield out

list(partitions(s))

你得到：

[['4824'], ['482', '4'], ['48', '24'], ['4', '824'], ['4', '82', '4']]

说明

我基于以下算法：

s="4824"

def partitions_original(s):
  #yield original string
  yield [s]
  if(len(s)>2):
    for i in range(len(s)-1, 0, -1):
      #divide string in two parts
      #iteration 1: a="482", b="4"
      #iteration 2: a="48", b="24"
      #iteration 3: a="4", b="824"
      a = s[:i]
      b = s[i:]
      #recursive call of b
      for g in partitions_original(b):
        #iteration 1: b="4", g=[['4']]
        #iteration 2: b="24", g=[['24']]
        #iteration 3: b="824", g=[['824'], ['82', '4'], ['8', '24']]
        yield [a] + g

list(partitions_original(s))

你得到：

[['4824'], ['482', '4'], ['48', '24'], ['4', '824'], 
['4', '82', '4'], ['4', '8', '24']]

问题是['4', '8', '24'].....那么我必须在代码中添加if，因为"a subset of length 1 can only be followed by a subset with length > 1"

[len(out[i]) == len(out[i+1]) and len(out[i])==1 for i in range(len(out)-1)] return for ['4', '8', '24'] -> [True, False] .... any Return 如果 iterable 的任何元素是真

注意

也可以用:

if all([len(out[i]) != len(out[i+1]) or len(out[i])!=1 for i in range(len(out)-1)]):

Answer 3

我这里做的是获取字符串所有可能的分割位置，去掉最后一个。

例如，在一些包含 5 个数字“12345”的字符串中，有 4 个可能的位置来拆分字符串，将其命名为 possibility = (0,0,0,0),(1,0,1,0)... 用 (0,0,1,0) mean (don't separate 1 and 2345,don't separate 12 and 345,separate 123 and 45,don't separate 1234 and 5) 这样你就可以由于我们消除了 (1,1,1,1) 情况，因此在验证您的条件时获得所有可能性。

import itertools
from math import factorial
from itertools import product

def get_comb(string):
    L = len(string_)
    combinisation = []

    for possibility in product([0,1], repeat=len(string_)-1):
        s = []
        indexes = [i for i in range(len(string_)-1) if list(possibility)[i]!=0]
        if sum(indexes) != 0:
            if sum(indexes) != len(string_)-1:
                for index in indexes:
                    s.append(string_[:index+1])
                s.append(string_[indexes[-1:][0]+1:])
                combinisation.append(s)
            else:
                combinisation.append(string_)
    return combinisation



string_ = '4824'
print "%s combinations:"%string_
print get_comb(string_)



string_ = '478952'
print "%s combinations:"%string_
print get_comb(string_)



string_ = '1234'
print "%s combinations:"%string_
print get_comb(string_)


>> 
4824 combinations:
[['482', '4'], ['48', '24'], '4824', ['4', '482', '4'], ['4', '48', '24'], '4824
']
478952 combinations:

[['47895', '2'], ['4789', '52'], ['4789', '47895', '2'], ['478', '952'], ['478',
 '47895', '2'], '478952', ['478', '4789', '47895', '2'], ['47', '8952'], '478952
', ['47', '4789', '52'], ['47', '4789', '47895', '2'], ['47', '478', '952'], ['4
7', '478', '47895', '2'], ['47', '478', '4789', '52'], ['47', '478', '4789', '47
895', '2'], ['4', '47895', '2'], ['4', '4789', '52'], ['4', '4789', '47895', '2'
], ['4', '478', '952'], ['4', '478', '47895', '2'], '478952', ['4', '478', '4789
', '47895', '2'], ['4', '47', '8952'], '478952', ['4', '47', '4789', '52'], ['4'
, '47', '4789', '47895', '2'], ['4', '47', '478', '952'], ['4', '47', '478', '47
895', '2'], ['4', '47', '478', '4789', '52'], ['4', '47', '478', '4789', '47895'
, '2']]

1234 combinations:

[['123', '4'], ['12', '34'], '1234', ['1', '123', '4'], ['1', '12', '34'], '1234
']

Answer 4

一个普通的代码可以这样写：

s=raw_input('enter the string:')
word=[]
for i in range(len(s)):
    for j in range(i,len(s)):
        word.append(s[i:j+1])

print word
print 'no of possible combinations:',len(word)

并输出：输入字符串： 4824 ['4', '48', '482', '4824', '8', '82', '824', '2', '24', '4'] 没有可能 combinations:10

python 字符串子集的所有组合

python all combinations of subsets of a string

python

subset