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我需要对列表中的每个元素应用一些正则表达式替换。我写了一个函数来少重复自己。反正还是有太多的重复。我该如何优化呢?

def cleanlist(mylist, regex, substitution):
    tmp_list = mylist
    cleaned_list = [re.sub(regex, substitution, line) for line in tmp_list]
    return cleaned_list


create_table_parts = cleanlist(create_table_parts, "(SET).+?(\n)", "\n") 
create_table_parts = cleanlist(create_table_parts, "(__|\(__).*?\n|(^\)|(?<=\n)(\n))", "")
create_table_parts = cleanlist(create_table_parts, "\"", "")
create_table_parts = cleanlist(create_table_parts, "(?<=CREATE\sTABLE\s).+?(\.)", "") 
create_table_parts = cleanlist(create_table_parts, "(PRIMARY\sKEY\s).+?(\n)|(FOREIGN\sKEY\s).+?(\n)|", "")
create_table_parts = cleanlist(create_table_parts, "(CREATE_INDEX\s).+?(\n)", "") 
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1 回答 1

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将您的模式放入列表并循环:

patterns = [
    ("(SET).+?(\n)", "\n"),
    ("(__|\(__).*?\n|(^\)|(?<=\n)(\n))", ""),
    ("\"", ""), ("(?<=CREATE\sTABLE\s).+?(\.)", ""),
    ("(PRIMARY\sKEY\s).+?(\n)|(FOREIGN\sKEY\s).+?(\n)|", ""),
    ("(CREATE_INDEX\s).+?(\n)", "")
]

for patt, sub in patterns:
    create_table_parts = cleanlist(create_table_parts, patt, sub)

你甚至可以使用reduce()它,而不是for循环:

create_table_parts = reduce(lambda ctp, patt: cleanlist(ctp, *patt),
                            patterns, create_table_parts)

但无论是否更具可读性,这都是个人要求。

于 2014-05-28T10:02:24.050 回答