python字符串子集的所有组合

时间:2015-05-22 13:19:01

标签: python subset

我需要字符串子集的所有组合。另外,长度为1的子集之后只能是长度> 1的子集。例如对于字符串4824,结果应为:

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

到目前为止,我设法检索所有可能的子集:

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

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。

4 个答案:

答案 0 :(得分:8)

首先,编写一个函数来生成 all 字符串的分区:

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

这会迭代所有可能的第一个段(一个字符,两个字符等),并将这些段与字符串的相应剩余部分的所有分区组合在一起。

>>> [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']]

现在,您可以只过滤那些与您的条件相符的内容,即那些没有两个连续长度为1的子字符串的内容。

zip(p, p[1:])

在这里,partition是迭代所有连续项对的常用方法。

更新:实际上,将约束直接纳入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 [] 函数也不是那么难。只需跟踪最后一段并相应地设置最小长度。

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

演示:

{{1}}

答案 1 :(得分:2)

看到更多测试用例会很有趣,下面的算法就是你所说的:

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']]

<强>解释

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'] .....然后我必须将if添加到代码中,因为“长度为1的子集后面只能跟一个长度> 1的子集“

[len(out[i]) == len(out[i+1]) and len(out[i])==1 for i in range(len(out)-1)]返回['4', '8', '24'] - &gt; [True, False] .... any如果iterable的任何元素为真,则返回True

注意

也可以使用:

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

答案 2 :(得分:0)

我在这里做的是获得字符串的所有可能的分割位置并消除最后一个。

例如,在一些包含5个数字的字符串中#34; 12345&#34;例如,有4个可能的位置来分割字符串,将其称为possibility = (0,0,0,0),(1,0,1,0) ...与(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),这样您可以在条件得到验证时获得所有可能性,因为我们消除了(1,1, 1,1)案例。

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
']

答案 3 :(得分:0)

普通代码可以写成:

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)

输出: 输入字符串: 4824 [&#39; 4&#39;,&#39; 48&#39;,&#39; 482&#39;,&#39; 4824&#39;,&#39; 8&#39;,&#39; 82&#39;,&#39; 824&#39;,&#39; 2&#39;,&#39; 24&#39;,&#39; 4&#39;] 没有可能的组合:10