所有可能的替换版本

Question

我有一个小问题：

例如，我们有字符串'YXY00'。'Y'的每个'X'可以分别替换为'Y'或'X'。在本例中，我们有2^3=8个替换选项，例如：

YXY00
YXX00
YYY00
YYX00
XXY00
XXX00
XYY00
XYX00

我怎样才能用python3.x得到这个替代品呢？你知道吗

Answer 1

我重构了更多，现在可读性更强：

def replace_at_index(string, index, replacement):
    """
    Credit to: http://stackoverflow.com/users/95612/jochen-ritzel

    >>> replace_at_index("abc", 1, "z")
    'azc'
    """
    return string[:index] + replacement + string[index + 1:]

def possible_replaces(string):
    """
    >>> list(possible_replaces('YXY00'))
    ['XXY00', 'YXY00', 'YXY00', 'YYY00', 'YXX00', 'YXY00', 'YXY00', 'YXY00']
    >>> list(possible_replaces('XYY000000'))
    ['XYY000000', 'YYY000000', 'XXY000000', 'XYY000000', 'XYX000000', 'XYY000000', 'XYY000000', 'XYY000000', 'XYY000000', 'XYY000000', 'XYY000000', 'XYY000000']
    """
    for index, char in enumerate(string):
        if char in 'XY':
            yield replace_at_index(string, index, 'X')
            yield replace_at_index(string, index, 'Y')
        else:
            yield string

---

我还写了一个更一般的解决方案：

def possible_replaces(string, to_multipy_replace = 'XY'):
    """
    Returns a list of strings where each member of `to_multipy_replace` is replaced
    by each member of said set.

    >>> list(possible_replaces('YXY00'))
    ['XXY00', 'YXY00', 'YXY00', 'YYY00', 'YXX00', 'YXY00', 'YXY00', 'YXY00']
    >>> list(possible_replaces('XYY0'))
    ['XYY0', 'YYY0', 'XXY0', 'XYY0', 'XYX0', 'XYY0', 'XYY0']
    """
    for index, char in enumerate(string):
        if char in to_multipy_replace:
            for replacement in to_multipy_replace:
                yield replace_at_index(string, index, replacement)
        else:
            yield string

你现在不仅限于“XY”，而是任何你喜欢的字符集。你知道吗

Answer 2

可以使用itertools.permutations()获取字符串的排列。然后通过应用一些替换逻辑，您可以得到以下结果：

import itertools

def permutate(source, changeset):
    count = sum(1 for char in source if char in changeset)
    holder = ''.join('{}' if char in changeset else char for char in source)
    for perm in set(itertools.permutations(changeset * count, count)):
        print(holder.format(*perm))

permutate('XY000Y00', 'XY')

结果：

XY000X00
XX000Y00
XY000Y00
XX000X00
YX000X00
YY000Y00
YX000Y00
YY000X00