python - 从字符串 python 生成所有的字谜

Question

我今天在想这个问题，我来了以下伪代码(Python 3.2):

def anagrams( string ):

    for c in string:
      anagram = c + anagram( string - {c} ) # remove the char from its position in the string
      print(anagram)

    return

def main():

    word = "abcd"
    anagrams( word )

    return

但是，我想知道一种执行此操作的 pythonic 方法: anagram = c + anagram( 字符串 - {c} )

如何从字符串中删除该字符？例如:

"abc" -> 'a' + "bc" -> 'a' + 'b' + "c" -> 'a' + 'b' + 'c' = 'abc'
             + "cb" -> 'a' + 'c' + "b" -> 'a' + 'c' + 'b' = 'acb'
      -> 'b' + "ac" -> 'b' + 'a' + "c" -> 'b' + 'a' + 'c' = 'bac'
             + "ca" -> 'b' + 'c' + "a" -> 'b' + 'c' + 'a' = 'bca'
      -> 'c' + "ba" -> 'c' + 'b' + "a" -> 'c' + 'b' + 'a' = 'cba'
             + "ab" -> 'c' + 'a' + "b" -> 'c' + 'a' + 'b' = 'cab'

谢谢

Answer 1

为什么不直接使用 itertools ？

>>> import itertools
>>> ["".join(perm) for perm in itertools.permutations("abc")]
['abc', 'acb', 'bac', 'bca', 'cab', 'cba']

documentation还包含如何完成排列的代码。

---

编辑:

Without itertools :

def all_perms(elements):
    if len(elements) <=1:
        yield elements
    else:
        for perm in all_perms(elements[1:]):
            for i in range(len(elements)):
                yield perm[:i] + elements[0:1] + perm[i:]


word = "abc"
print list(all_perms(word))

没有 itertools 也没有 generators:

def all_perms(elements):
    if len(elements) <=1:
        return elements
    else:
        tmp = []
        for perm in all_perms(elements[1:]):
            for i in range(len(elements)):
                tmp.append(perm[:i] + elements[0:1] + perm[i:])
        return tmp

结果:

['abc', 'bac', 'bca', 'acb', 'cab', 'cba']