如何检查一个子集是否包含在列表中？

Question

我有以下清单：

l =
[
 ['s1', 's5', 's6', 's8', 's10', 's5', 's15', 's23'],
 ['s1', 's5', 's8', 's10', 's5', 's6', 's8', 's15', 's23'],
 ['s1', 's5', 's6', 's10', 's14', 's15', 's23']
]

我想删除不符合以下约束的所有列表:列表必须包括['s5','s6','s8']和['s15', 's23']

l1 = ['s5','s6','s8']
l2 = ['s15', 's23']

预期结果是：

[
 ['s1', 's5', 's6', 's8', 's10', 's5', 's15', 's23'],
 ['s1', 's5', 's8', 's10', 's5', 's6', 's8', 's15', 's23']
]

我怎样才能做到呢？

我尝试使用set(l1).issubset(t)和set(l2).issubset(l)，但是set只返回唯一的值。

Answer 1

您可以使用filter和set检查条件。

使用filter，您也可以维护订单。

>>> list(filter(lambda x : not(set(l1+l2)-set(x)), l))
>>> [['s1', 's5', 's6', 's8', 's10', 's5', 's15', 's23'], 
     ['s1', 's5', 's8', 's10', 's5', 's6', 's8', 's15', 's23']]

Answer 2

我想你可以做一次会员检查：

l = [
 ['s1', 's5', 's6', 's8', 's10', 's5', 's15', 's23'],
 ['s1', 's5', 's8', 's10', 's5', 's6', 's8', 's15', 's23'],
 ['s1', 's5', 's6', 's10', 's14', 's15', 's23']
]
l1 = ['s5','s6','s8']
l2 = ['s15', 's23']

l1_s = ''.join(l1)
l2_s = ''.join(l2)

print([x for x in l if l1_s in ''.join(x) and l2_s in ''.join(x)])

# [['s1', 's5', 's6', 's8', 's10', 's5', 's15', 's23'],
#  ['s1', 's5', 's8', 's10', 's5', 's6', 's8', 's15', 's23']]