我正在寻求帮助，以查找允许我获取字符串列表（例如["I like ", " and ", " because "]和单个目标字符串，例如）的Python函数"I like lettuce and carrots and onions because I do"，并找到可以对目标字符串中的字符进行分组的所有方式，以便每个字符串列表中的顺序。
例如：
solution(["I like ", " and ", " because ", "do"],
"I like lettuce and carrots and onions because I do")
应该返回：
[("I like ", "lettuce", " and ", "carrots and onions", " because ", "I ", "do"),
("I like ", "lettuce and carrots", " and ", "onions", " because ", "I ", "do")]
请注意，在每个元组中，list参数中的字符串都是按顺序排列的，并且该函数返回每种可能的方法来拆分目标字符串以实现此目的。
另一个例子，这次只有一种可能的组织字符的方式：
solution(["take ", " to the park"], "take Alice to the park")
应该给出结果：
[("take ", "Alice", " to the park")]
这是一个无法正确组织字符的示例：
solution(["I like ", " because ", ""],
"I don't like cheese because I'm lactose-intolerant")
应该回馈：
[]
因为没有办法做到。请注意，"I like "第一个参数中的不能拆分。目标字符串中没有字符串"I like "，因此无法匹配。
这是最后一个示例，同样具有多个选项：
solution(["I", "want", "or", "done"],
"I want my sandwich or I want my pizza or salad done")
应该回来
[("I", " ", "want", " my sandwich ", "or", " I want my pizza or salad ", "done"),
("I", " ", "want", " my sandwich or I want my pizza ", "or", " salad ", "done"),
("I", " want my sandwich or I", "want", " my pizza ", "or", " salad ", "done")]`
再次注意，每个字符串["I", "want", "or", "done"]按顺序包含在每个元组中，并且其余字符以任何可能的方式围绕这些字符串重新排序。返回所有可能的重新排序列表。
请注意，还假定列表中的第一个字符串将出现在目标字符串的开头，列表中的最后一个字符串将出现在目标字符串的结尾。（如果没有，该函数应返回一个空列表。）
哪些Python函数将允许我执行此操作？
我已经尝试过使用正则表达式函数，但是在有多个选项的情况下，它似乎失败了。
我有一个解决方案，它需要大量的重构，但它似乎可以工作，我希望这会有所帮助，这是一个非常有趣的问题。
import itertools
import re
from collections import deque
def solution(search_words, search_string):
found = deque()
for search_word in search_words:
found.append([(m.start()) for m in re.compile(search_word).finditer(search_string)])
if len(found) != len(search_words) or len(found) == 0:
return [] # no search words or not all words found
word_positions_lst = [list(i) for i in itertools.product(*found) if sorted(list(i)) == list(i)]
ret_lst = []
for word_positions in word_positions_lst:
split_positions = list(itertools.chain.from_iterable(
(split_position, split_position + len(search_word))
for split_position, search_word in zip(word_positions, search_words)))
last_seach_word = search_string[split_positions[-1]:]
ret_strs = [search_string[a:b] for a, b in zip(split_positions, split_positions[1:])]
if last_seach_word:
ret_strs.append(last_seach_word)
if len(search_string) == sum(map(len,ret_strs)):
ret_lst.append(tuple(ret_strs))
return ret_lst
print(solution(["I like ", " and ", " because ", "do"],
"I like lettuce and carrots and onions because I do"))
print([("I like ", "lettuce", " and ", "carrots and onions", " because ", "I ", "do"),
("I like ", "lettuce and carrots", " and ", "onions", " because ", "I ", "do")])
print()
print(solution(["take ", " to the park"], "take Alice to the park"))
print([("take ", "Alice", " to the park")])
print()
print(solution(["I like ", " because "],
"I don't like cheese because I'm lactose-intolerant"))
print([])
print()
输出：
[('I like ', 'lettuce', ' and ', 'carrots and onions', ' because ', 'I ', 'do'), ('I like ', 'lettuce and carrots', ' and ', 'onions', ' because ', 'I ', 'do')]
[('I like ', 'lettuce', ' and ', 'carrots and onions', ' because ', 'I ', 'do'), ('I like ', 'lettuce and carrots', ' and ', 'onions', ' because ', 'I ', 'do')]
[('take ', 'Alice', ' to the park')]
[('take ', 'Alice', ' to the park')]
[]
[]
[('I', ' ', 'want', ' my sandwich ', 'or', ' I want my pizza or salad ', 'done'), ('I', ' ', 'want', ' my sandwich or I want my pizza ', 'or', ' salad ', 'done'), ('I', ' want my sandwich or I ', 'want', ' my pizza ', 'or', ' salad ', 'done')]
[('I', ' ', 'want', ' my sandwich ', 'or', ' I want my pizza or salad ', 'done'), ('I', ' ', 'want', ' my sandwich or I want my pizza ', 'or', ' salad ', 'done'), ('I', ' want my sandwich or I', 'want', ' my pizza ', 'or', ' salad ', 'done')]
编辑：重构代码以具有有意义的变量名。
Edit2：添加了我忘记的最后一种情况。
本文收集自互联网，转载请注明来源。
如有侵权，请联系 [email protected] 删除。
我来说两句