如果字符串包含指定字符，则从列表中删除字符串

Question

有以下几点：

list1 = ['something',"somet'hing",'somet"hing','some;thing','']
list2 = [';','"',"'"]

如果列表中的字符串包含来自 list2 的任何字符或字符串为空白，我想获得过滤的 list1。期望的输出：

list3 = ['something']

目前我是这样手动操作的：

list1withoutEmptyLines= list(filter(None, list1))
list1withoutQuote = [x for x in list1withoutEmptyLines if "'" not in x]
list1withoutDoublequotes = [x for x in list1withoutQuote if "\"" not in x]
list1withoutSemicolon = [x for x in list1withoutDoublequotes if ";" not in x]

而且它工作得非常好。我还尝试通过创建这样的禁用字符列表来自动化它：

forbiddenCharacters = ['"', ';', '\'']
filteredLines = []

for character in forbiddenCharacters:
    filteredLines = [x for x in uniqueLinesInFile if character not in x]

但名为filteredLines 的列表仍然包含带有分号“;”的字符串。任何建议将不胜感激。

Answer 1

您可以使用list comprehension 结合内置函数any 来做到这一点：

list1 = ['something', "somet'hing", 'somet"hing', 'some;thing', '']
list2 = [';', '"', "'"]

result = [s for s in list1 if s and not any(c in s for c in list2)]
print(result)

输出

['something']

列表推导等价于：

result = []
for s in list1:
    if s and not any(c in s for c in list2):
        result.append(s)