如何在Python中的递归函数调用中保存先前的“输出”或先前输出的状态？

Question

我正在使用递归函数使用正则表达式匹配来生成文本，它根据方括号 (

pattern = '\[.*?\]'

) 内的同义词组合查找单词模式，并用字符串分隔符分隔（我定义了

_STRING_SEPARATOR =#lkmkmksdmf###

. )

函数的初始句子参数类似于：

[decreasing#lkmkmksdmf###shrinking#lkmkmksdmf###falling#lkmkmksdmf###contracting#lkmkmksdmf###faltering#lkmkmksdmf###the contraction in] exports of services will drive national economy to a 0.3% real GDP [decline#lkmkmksdmf###decrease#lkmkmksdmf###contraction] in 2023 from an estimated 5.0% [decline#lkmkmksdmf###decrease#lkmkmksdmf###contraction] in 2022

和

该函数如下所示：

def combinations(self,sentence,sentence_list:list):
        pattern = '\[.*?\]'

        if not re.findall(pattern, sentence, flags = re.IGNORECASE):
            if sentence not in sentence_list:
                sentence_list.append(sentence)
        else:
            for single_match in re.finditer(pattern, sentence, flags = re.IGNORECASE):
                repl=single_match.group(0)[1:-1]
                start_span = single_match.span()[0]
                end_span = single_match.span()[1]
                for candidate_word in repl.split(self._STRING_SEPARATOR):
                    tmp_sentence = (
                        sentence[0: start_span] +
                        candidate_word +
                        sentence[end_span:]
                    )
                    new_sentence = deepcopy(tmp_sentence)
                    self.combinations(new_sentence,sentence_list)

因此，

sentence_list

变量像DFS树一样不断追加句子

我想避免两次使用相同的单词 - 例如，如果我使用了单词“decline”，那么在递归调用后在内部 for 循环中选择下一组单词时，不应再次使用它。当解析第二个方括号模式中的单词等时，是否有一种方法可以“存储”第一个方括号内的单词所使用的单词？

*它就像一个 DFS 树，其中每个节点都必须存储其每个父节点的状态。 * 如何修改该函数，以便在句子列表的单个句子中不再使用相同的单词？

我尝试使用名为“avoid_words: list”的参数来存储父节点单词的列表。但是，当我必须从第一个方括号移到下一个单词（或从不同的“根”开始）时，如何删除它？

Answer 1

正如 Tim 指出的，如果确实没有其他方法来输入字符串及其参数（我对此表示怀疑），你应该使用

split()

 函数将初始句子分成单词（同义词）和纯句子。

波纹管是我将使用的注释代码，我是否必须解决这样的情况。

def all_combinations(sentence) -> list:
    pattern = r'\[(.*?)\]'
    synonyms = []
    resulting_sentences = []

#Put all of the synonyms into synonyms list
    list_of_synonyms = re.findall(pattern, sentence, flags = re.IGNORECASE)
#Remove synonyms from the origingal sentence
    sentence = re.sub(pattern, '[]', sentence)


#split sinynonyms into dictionaries containing tuple and clock
    for i, x in enumerate(list_of_synonyms):
            synonyms.append(tuple(x.split('#lkmkmksdmf###')))
 
#Create combinations and put those into list of sets. 
# Sets can hold only unique elements, thus in case of duplicity thwy will be shorter.
# The set will be removed if it's length is <3.
    synonym_combinations = list(set(combinations) for combinations in itertools.product(*synonyms) if len(set(combinations)) == 3)

#iterate over combinations
    for combination in synonym_combinations:
#iterate over words in combinations
        formatted_sentence = sentence
        for synonym in combination: formatted_sentence = formatted_sentence.replace('[]',synonym,1)
#append formatted sentence to resulting senteces
        resulting_sentences.append(formatted_sentence)
    return resulting_sentences