Python 从不带标点符号的字符串搜索到主字符串，然后从主字符串中提取带标点符号但不带库的片段，可能吗？

我要做的作业是图书馆不允许做的，我低估了这个问题：

假设我们有一个字符串列表：str_list=[我的头，终于自由了，进入警报状态，再过一会儿，脖子]

关于这一点，我们可以肯定的是，每一个字符串都包含在master_字符串中，它们是有序的，并且没有标点符号。所有这些都要感谢我以前的控制

然后我们有了字符串：master_string=‘来吧，我的头终于自由了！’爱丽丝用一种高兴的口吻说，过了一会儿，这种口吻变成了惊慌，因为她发现自己的肩膀不见了。她往下看，只看到一条长长的脖子，它好像从她下面一片绿叶中拔出一根麦秆

这里我要做的基本上是检查主字符串中包含的str_列表中至少k的字符串序列，在这种情况下k=2，然而，我低估了一个事实，即在str_列表中，我们每个字符串中有超过1个单词，因此执行master_string.split不会带我去任何地方，因为这意味着询问我的头是否=我的，这当然是错误的

我在考虑做一些事情，比如一次连接一个字符串，然后搜索master_string.strip.，：；！？但是如果我找到了相应的序列，我绝对需要直接从master_字符串中获取它们，因为我需要结果变量中的标点符号。这基本上意味着直接从master_字符串中获取切片，但这怎么可能呢？有可能吗，或者我必须改变方法？这让我发疯，尤其是因为没有图书馆可以这么做

如果您想知道这里的预期结果是什么：

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[我的头终于自由了！在另一个时刻发出警报，]因为两者都尊重stru列表中至少k个字符串的条件，neck将保存在丢弃列表中，因为它不尊重不能丢弃的条件。pop因为我需要用丢弃的变量做其他事情

我有两个不同的版本，数字1给你脖子：，但数字2没有给你那么多，这是数字1：

master_string = "Come, my head’s free at last!’ said Alice in a tone of delight, which changed into alarm in another moment, when she found that her shoulders were nowhere to be found: all she could see, when she looked down, was an immense length of neck, which seemed to rise like a stalk out of a sea of green leaves that lay far below her."

str_list = ["my head's", "free", "at last", "into alarm", "in another moment", "neck"]

new_str = ''

for word in str_list:
    if word in master_string:
       new_str += word + ' '
            

print(new_str)

这是第二条：

master_string = "Come, my head’s free at last!’ said Alice in a tone of delight, which changed into alarm in another moment, when she found that her shoulders were nowhere to be found: all she could see, when she looked down, was an immense length of neck, which seemed to rise like a stalk out of a sea of green leaves that lay far below her."

str_list = ["my head's", "free", "at last", "into alarm", "in another moment", "neck"]

new_str = ''

for word in str_list:
    if word in master_string:
        new_word = word.split(' ')
        if len(new_word) == 2:
            new_str += word + ' '
            

print(new_str)

遵循我的解决方案：

尝试根据master_字符串和一组有限的标点字符扩展所有代码，例如my head's->my head's free at last！；自由->终于自由了！。 仅保留已扩展至少k次的子字符串。 删除多余的子字符串，例如“最终释放”！现在我的头终于自由了！。 代码如下：

str_list = ["my head’s", "free", "at last", "into alarm", "in another moment", "neck"]
master_string = "‘Come, my head’s free at last!’ said Alice in a tone of delight, which changed into alarm in another moment, when she found that her shoulders were nowhere to be found: all she could see, when she looked down, was an immense length of neck, which seemed to rise like a stalk out of a sea of green leaves that lay far below her."
punctuation_characters = ".,:;!?"  # list of punctuation characters
k = 1

def extend_string(current_str, successors_num = 0) :
    # check if the next token is a punctuation mark
    for punctuation_mark in punctuation_characters :
        if current_str + punctuation_mark in master_string :
            return extend_string(current_str + punctuation_mark, successors_num)
    
    # check if the next token is a proper successor
    for successor in str_list :
        if current_str + " " + successor in master_string :
            return extend_string(current_str + " " + successor, successors_num+1)
    
    # cannot extend the string anymore
    return current_str, successors_num

extended_strings = []
for s in str_list :
    extended_string, successors_num = extend_string(s)
    if successors_num >= k : extended_strings.append(extended_string)

extended_strings.sort(key=len)  # sorting by ascending length
result_list = []
for es in extended_strings :
    result_list = list(filter(lambda s2 : s2 not in es, result_list))
    result_list.append(es)
print(result_list)      # result: ['my head’s free at last!', 'into alarm in another moment,']

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所以neck被排除在预期输出之外，因为它前面没有其他字符串，所以k=1；是吗？问题：str_list的字符串只有在与空格连接时才能出现在master_字符串中？是的，确切地说，neck只有一个字符串，因此不符合在前后至少有一个字符串的要求。关于str_列表的字符串是的，它们可以在master_字符串中显示，但它们之间有空格，但缺少标点。那么顶点呢？在str_列表中，我们有我的头，但它与master_字符串的我的头不匹配。我们可以替换str_列表中的所有'with'吗？这只是一个例子，无论如何我不明白你的意思，它们是同一个字符串不？这可能有效，但我无法导入库来解决这个问题，这是违反规则的rules@Krist我用更好的解决方案更新了答案谢谢！现在我要检查一下，我想了解如何将它集成到我的代码中。既然k是一个可以从一次代码运行到另一次代码运行的值，那么这会起作用吗？@Krist I将代码概括为任何可能的k。现在，你将用k=1获得['my head's free！''，'into alarm in Other片刻，']，用k=2获得['my head's free at last！']。