python：为带有连字符的字符串创建所有可能的变体

Question

我有一个连字符字符串列表：

（样本）

myList = ['mother-in-law', 'co-operation', 'sixty-nine-eighty-ninths']

对于此列表中的每个元素，我希望能够创建连字符位于每个元素的两个或多个标记之间的所有变体：

mother-in law
mother in-law
sixty-nine eighty ninths
sixty-nine-eighty ninths
sixty nine-eighty-ninths
sixty-nine eighty-ninths
sixty nine-eighty ninths
sixty nine eighty-ninths
...

我尝试了这个问题的解决方案（Create variations of a string） 但我不知道如何适应它：

from itertools import combinations
myList = ['mother-in-law', 'co-operation', 'sixty-nine-eighty-ninths']

for e in myList :
    for i in range(len(e.split("-"))):
        for indices in combinations(range(len(e.split("-"))), i):
            print(''.join([e.split("-")[x] if x in indices else '-' for x in range(len(e))]))

这就是我得到的：

-------------
mother------------
-in-----------
--law----------
motherin-----------
mother-law----------
-inlaw----------
------------
co-----------
-operation----------
------------------------
sixty-----------------------
-nine----------------------
--eighty---------------------
---ninths--------------------
sixtynine----------------------
sixty-eighty---------------------
sixty--ninths--------------------
-nineeighty---------------------
-nine-ninths--------------------
--eightyninths--------------------
sixtynineeighty---------------------
sixtynine-ninths--------------------
sixty-eightyninths--------------------
-nineeightyninths--------------------

谢谢

Answer 1

稍微浏览一下itertools提供的工具，我发现这个产品在这里可能最有用。它让我们了解在两个单词之间有空格或破折号的所有可能性

from itertools import product, zip_longest

my_list = ['mother-in-law', 'co-operation', 'sixty-nine-eighty-ninths']
symbols = ' ', '-'

for string in my_list:
    string_split = string.split('-')
    for symbols_product in product(symbols, repeat=len(string_split)-1):
        if '-' not in symbols_product:
            continue
        rtn = ""
        for word, symbol in zip_longest(string_split, symbols_product, fillvalue=''):
            rtn += word + symbol
        print(rtn)
    print()

另外，根据您的要求，我将跳过任何两个单词之间没有破折号的迭代

输出：

mother in-law
mother-in law
mother-in-law

co-operation

sixty nine eighty-ninths
sixty nine-eighty ninths
sixty nine-eighty-ninths
sixty-nine eighty ninths
sixty-nine eighty-ninths
sixty-nine-eighty ninths
sixty-nine-eighty-ninths

Answer 2

只需制作自己的生成器来生成组合，可能会更容易一些。这可以通过递归生成器以非常可读的方式完成，只要字符串不足以运行到堆栈限制：

def hyphenCombos(s):
    head, _, rest = s.partition('-')
    if len(rest) == 0:
        yield head
    else:
        for c in hyphenCombos(rest):
            yield f'{head}-{c}'
            yield f'{head} {c}'

s = 'sixty-nine-eighty-ninths'
list(hyphenCombos(s))

结果:

['sixty-nine-eighty-ninths',
 'sixty nine-eighty-ninths',
 'sixty-nine eighty-ninths',
 'sixty nine eighty-ninths',
 'sixty-nine-eighty ninths',
 'sixty nine-eighty ninths',
 'sixty-nine eighty ninths',
 'sixty nine eighty ninths']

有了它，您可以在理解中使用它，或者将它传递给其他itertools函数来执行您需要的任何操作：

myList = ['mother-in-law', 'co-operation', 'sixty-nine-eighty-ninths']
chain.from_iterable(hyphenCombos(s) for s in myList))
# or variations...
# [list(hyphenCombos(s)) for s in myList]