在Python中检查set中的项目成员资格

Question

您好我已经编写了几个月了,并且知道了基础知识,但是我遇到了一个集合成员问题,我无法找到解决方案.

我有一个整数对列表的列表,我想删除其中包含"a"整数的列表.我认为使用套装是最简单的方法.贝娄是代码:

## This is the item to test against. 
a = set([3]) 
## This is the list to test.      
groups = [[3, 2], [3, 4], [1, 2], [5, 4], [4, 3]]     

## This is a list that will contain the lists present
## in groups which do not contain "a"
groups_no_a = []        

for group in groups:
    group = set(group)
    if a in group:
        groups_no_a.append(group)
    ## I thought the problem had something to do with
    ## clearing the variable so I put this in,   
    ## but to no remedy. 
    group.clear()  


print groups_no_a 

我一直在使用也尝试s.issubset(t)直到我意识到这一点,如果测试的每一个在元素s中t.

谢谢!

Answer 1

你想测试是否没有交叉点:

if not a & group:

要么

if not a.intersection(group):

或者,相反,这些集合是不相交的:

if a.isdisjoint(group):

方法形式采用任何迭代,你甚至不必group变成一个集合.以下单行也可以使用:

groups_no_a = [group for group in groups if a.isdisjoint(group)]

演示:

>>> a = set([3]) 
>>> groups = [[3, 2], [3, 4], [1, 2], [5, 4], [4, 3]]     
>>> [group for group in groups if a.isdisjoint(group)]
[[1, 2], [5, 4]]

如果您要测试的只是一个元素,那么创建集合的性能可能会比您在成员资格测试中获得的成本更高,而且只是:

3 not in group

哪里group是一个简短的清单.

您可以使用该timeit模块比较Python代码片段,以查看哪种方法最适合您的特定典型列表大小.