如何根据包含部分值和通配符的其他列表过滤列表?以下示例是我到目前为止的例子:
l1 = ['test1', 'test2', 'test3', 'test4', 'test5']
l2 = set(['*t1*', '*t4*'])
filtered = [x for x in l1 if x not in l2]
print filtered
此示例导致:
['test1', 'test2', 'test3', 'test4', 'test5']
但是,我希望将基于l2
的结果限制为以下内容:
['test2', 'test3', 'test5']
答案 0 :(得分:10)
>>> from fnmatch import fnmatch
>>> l1 = ['test1', 'test2', 'test3', 'test4', 'test5']
>>> l2 = set(['*t1*', '*t4*'])
>>> [x for x in l1 if not any(fnmatch(x, p) for p in l2)]
['test2', 'test3', 'test5']
答案 1 :(得分:1)
你也可以使用filter()而不是list comprehension,这可能会让你轻松交换过滤函数以获得更大的灵活性:
>>> l1 = ['test1', 'test2', 'test3', 'test4', 'test5']
>>> l2 = set(['*t1*', '*t4*'])
>>> filterfunc = lambda item: not any(fnmatch(item, pattern) for pattern in l2)
>>> filter(filterfunc, l1)
Out: ['test2', 'test3', 'test5']
>>> # now we don't like our filter function no more, we assume that our l2 set should match on any partial match so we can get rid of the star signs:
>>> l2 = set(['t1', 't4'])
>>> filterfunc = lambda item: not any(pattern in item for pattern in l2)
>>> filter(filterfunc, l1)
Out: ['test2', 'test3', 'test5']
这样,您甚至可以将filterfunc概括为使用多个模式集:
>>> from functools import partial
>>> def filterfunc(item, patterns):
return not any(pattern in item for pattern in patterns)
>>> filter(partial(filterfunc, patterns=l2), l1)
Out: ['test2', 'test3', 'test5']
>>> filter(partial(filterfunc, patterns={'t1','test5'}), l1)
Out: ['test2', 'test3', 'test4']
当然,您可以轻松升级filterfunc以接受模式集中的正则表达式。例如。
答案 2 :(得分:1)
我认为用例的最简单的方法是使用Python in
来测试子字符串(虽然这意味着删除你的星号):
def remove_if_not_substring(l1, l2):
return [i for i in l1 if not any(j in i for j in l2)]
所以这是我们的数据:
l1 = ['test1', 'test2', 'test3', 'test4', 'test5']
l2 = set(['t1', 't4'])
用它调用我们的函数:
remove_if_not_substring(l1, l2)
返回:
['test2', 'test3', 'test5']