英文:
How to save previous "output" or a state of the previous output in a recursive function call in python?
问题
我正在使用递归函数来生成文本，使用正则表达式匹配，在其中找到一个由方括号内的同义词组合组成的模式（pattern = '[.*?]'），由字符串分隔符分隔（我定义了一个 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
函数的结构如下：
def combinations(self, sentence, master_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 regex_match in re.finditer(pattern, sentence, flags=re.IGNORECASE):repl = regex_match.group(0)[1:-1]start_span = regex_match.span()[0]end_span = regex_match.span()[1]for word in repl.split(self.SEPARATOR):tmpsentence = (sentence[0: start_span] +word +sentence[end_span:])new_sentence = deepcopy(tmpsentence)self.combinations(new_sentence, master_sentence_list)
因此，master_sentence_list 变量像DFS树一样不断追加句子。
我想避免在同一句子中重复使用相同的单词 - 例如，如果我使用了单词“decline”，那么在递归调用后的内部for循环中选择下一组单词时不应再次使用它。是否有一种方法可以在解析第二个方括号模式中的单词时“存储”由第一个方括号内的单词使用的方式，以此类推？
就像DFS树，每个节点都必须存储其每个父节点的状态一样。
如何修改函数以在 sentence_list 中的单个句子中不再使用相同的单词？
我尝试使用一个名为 avoid_words: list 的参数来存储父节点单词的列表。但是，当我需要移动到第一个方括号中的下一个单词时（或者从不同的“根”开始时），如何擦除它？
英文:
I am using a recursive function to generate text using a RegEx match, where it finds a pattern of words according to a combination of synonyms inside square brackets (pattern = '\[.*?\]') separated by a string separator (I have defined a SEPARATOR =#lkmkmksdmf###. )
The initial sentence argument to the function is something like:
[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
and
The function reads like:
def combinations(self,sentence,master_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 regex_match in re.finditer(pattern, sentence, flags = re.IGNORECASE):repl=regex_match.group(0)[1:-1]start_span = regex_match.span()[0]end_span = regex_match.span()[1]for word in repl.split(self.SEPARATOR):tmpsentence = (sentence[0: start_span] +word +sentence[end_span:])new_sentence = deepcopy(tmpsentence)self.combinations(new_sentence,master_sentence_list)
Thus, the master_sentence_list variable keeps appending the sentences like a DFS tree
I want to avoid using the same words twice - for example if I used the word "decline" then it should not be used again while choosing the next set of words in the inner for loop after the recursive call. Is there a way of "storing" the word used by the words inside the first square bracket when a word from the second square brackets pattern is parsed and so on?
*It is like a DFS tree where each node has to store the state of each of its parent node.
*
How can I modify the function to not use the same words again in a single sentence of the sentence_list?
I tried using an argument called "avoid_words: list" to which would store the list of the parent node words. But how do I erase it when I have to move over to the next word in from the first square bracket (or starts from a different "root")?
答案1
得分: 3
Tim指出，如果确实没有其他输入字符串及其参数的方法（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。
下面是我会使用的已注释代码，如果我不得不解决这种情况。
def all_combinations(sentence) -> list:pattern = r'\[(.*?)\]'synonyms = []resulting_sentences = []# 将所有同义词放入同义词列表中list_of_synonyms = re.findall(pattern, sentence, flags=re.IGNORECASE)# 从原始句子中删除同义词sentence = re.sub(pattern, '[]', sentence)# 将同义词拆分为包含元组和时钟的字典for i, x in enumerate(list_of_synonyms):synonyms.append(tuple(x.split('#lkmkmksdmf###')))# 创建组合并将其放入集合列表中。# 集合只能容纳唯一的元素，因此在重复性的情况下，它们将更短。# 如果长度小于3，则将删除该集合。synonym_combinations = list(set(combinations) for combinations in itertools.product(*synonyms) if len(set(combinations)) == 3)# 遍历组合for combination in synonym_combinations:# 遍历组合中的单词formatted_sentence = sentencefor synonym in combination:formatted_sentence = formatted_sentence.replace('[]', synonym, 1)# 将格式化后的句子添加到结果句子中resulting_sentences.append(formatted_sentence)return resulting_sentences
英文:
As Tim pointed out, if there really is no other way to input the string and it's arguments (which I doubt) you should use split() function to separate initial sentence into words (synonyms) and pure sentence.
Bellow is the commented code I would use, have I had to solve a situation like this.
def all_combinations(sentence) -> list:pattern = r'\[(.*?)\]'synonyms = []resulting_sentences = []#Put all of the synonyms into synonyms listlist_of_synonyms = re.findall(pattern, sentence, flags = re.IGNORECASE)#Remove synonyms from the origingal sentencesentence = re.sub(pattern, '[]', sentence)#split sinynonyms into dictionaries containing tuple and clockfor 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 combinationsfor combination in synonym_combinations:#iterate over words in combinationsformatted_sentence = sentencefor synonym in combination: formatted_sentence = formatted_sentence.replace('[]',synonym,1)#append formatted sentence to resulting sentecesresulting_sentences.append(formatted_sentence)return resulting_sentences
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