本文介绍了在 Python 中使用多个 NOT IN 语句的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我需要从一个循环中取出带有三个特定子字符串的 URL.以下代码有效，但我确信有一种更优雅的方法:
I need to URLs with three specific specific substrings out of a loop. The following code worked, but I am sure there's a more elegant way to do it:
for node in soup.findAll('loc'):
url = node.text.encode("utf-8")
if "/store/" not in url and "/cell-phones/" not in url and "/accessories/" not in url:
objlist.loc.append(url)
else:
continue
谢谢！
推荐答案
url = node.text.encode("utf-8")
sub_strings = ['/store','/cell-phones/','accessories']
if not any(x in url for x in sub_strings):
objlist.loc.append(url)
else:
continue
来自 docs:
any 返回 True.如果可迭代对象为空，则返回 False.相当于:
any returns True if any element of the iterable is true. If the iterable is empty, return False. Equivalent to:
def any(iterable):
for element in iterable:
if element:
return True
return False
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