如何检查子集是否包含在列表中？

Question

我有以下列表：

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']
]

我想删除所有不符合以下约束的列表：列表必须包含['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']
]

我该怎么做？

我尝试使用set(l1).issubset(t)和set(l2).issubset(l) ，但set只返回唯一值。

Answer 1

您可以使用filter和set来检查条件。

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

我想你可以做一个会员检查：

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']]

Answer 3

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

result = []
for a in l:
    found = True

    for x in l1:
        if x not in a:
            found = False
    for x in l2:
        if x not in a:
            found = False
    if found:
        result.append(a)

print(result)