如果你只是想要一个测试,将目标列表加入一个字符串并测试bad的每个元素,如下所示:
>>> my_list = ['abc-123', 'def-456', 'ghi-789', 'abc-456', 'def-111', 'qwe-111']
>>> bad = ['abc', 'def']
>>> [e for e in bad if e in '\n'.join(my_list)]
['abc', 'def']
根据您的问题,您可以通过这种方式将每个元素作为子字符串与另一个元素的每个元素进行测试:
>>> [i for e in bad for i in my_list if e in i]
['abc-123', 'abc-456', 'def-456', 'def-111']
它很快(与其他方法之一相比):
>>> def f1():
... [item for item in my_list if any(x in item for x in bad)]
...
>>> def f2():
... [i for e in bad for i in my_list if e in i]
...
>>> timeit.Timer(f1).timeit()
5.062238931655884
>>> timeit.Timer(f2).timeit()
1.35371994972229
根据您的评论,您可以通过以下方式获取不匹配的元素:
>>> set(my_list)-{i for e in bad for i in my_list if e in i}
{'ghi-789', 'qwe-111'}