Question

我有2d列表主列表

mainlist = [['John','Doe',True],['Mary','Jane',False],['James','Smith',False]]
slist1 = ['John', 'Doe']
slist2 = ['John', 'Smith']
slist3 = ['Doe', 'John']
slist4 = ['John', True]

如何确定列表中是否存在子列表的子列表，如果slist1针对mainlist进行测试，则返回True，而slist2将返回False

我在考虑这样的事情（代码来自here）

for sublist in mainlist:
    if all(i in sublist for i in slist1):
        return True
        break

有更多的“pythonic”方式吗？感谢

编辑：

针对主列表测试的slist1将返回True
slist2将返回False
slist3将返回False
slist4将返回False

所以基本上，我只是测试slist是否在主列表[x]

Answer 1

如果我确实理解了你的问题，我认为你可以使用set()和集合之间的交集，就像这个例子：

def list_intersection(a, b):
   for sub in a:
       condition = set(sub) & set(b)
       condition2 = len(set(b)) == len(condition)
       if condition and condition2:
           return True
   return False


mainlist = [['John','Doe',True],['Mary','Jane',False],['James','Smith',False]]
slist1 = ['John', 'Doe']
slist2 = ['John', 'Smith']

list_intersection(mainlist, slist1)
# True
list_intersection(mainlist, slist2)
# False

PS：这个解决方案有很多弱点......并不能涵盖所有案例。

Answer 2

由于OP没有回复，我将介绍两种情况。

如果订单不重要; ['John', 'Doe']中的['Doe', 'John']和mainlist 假设包含：

def list_intersection_no_order(a, b):
    b = set(b)
    if any(b.difference(sublist) == set() for sublist in a):
        return True
    return False


mainlist = [['John','Doe',True],['Mary','Jane',False],['James','Smith',False]]
slist1 = ['John', 'Doe']
slist2 = ['John', 'Smith']
slist3 = ['Doe', 'John']

print(list_intersection_no_order(mainlist, slist1))
# True
print(list_intersection_no_order(mainlist, slist2))
# False
print(list_intersection_no_order(mainlist, slist3))
# True

如果订单确实重要; ['John', 'Doe']已包含但['Doe', 'John']不在mainlist：

def list_intersection_with_order(a, b):
    if any(b == sublist[:2] for sublist in a):
        return True
    return False


mainlist = [['John','Doe',True],['Mary','Jane',False],['James','Smith',False]]
slist1 = ['John', 'Doe']
slist2 = ['John', 'Smith']
slist3 = ['Doe', 'John']

print(list_intersection_with_order(mainlist, slist1))
# True
print(list_intersection_with_order(mainlist, slist2))
# False
print(list_intersection_with_order(mainlist, slist3))
# False

Answer 3

您似乎想要检查mainlist中每个列表中的子列表是slist1还是slist2。在这种情况下，你可以做一些简单的事情：

def sublist_intersection(lst, sub):
    sub_len = len(sub)

    for l in lst:
        for i in range(0, len(l), sub_len):
            if l[i:i+sub_len] == sub:
                return True

    return False

或使用any()的简短解决方案：

def sublist_intersection(lst, sub):
    sub_len = len(sub)
    return any(l[i:i+sub_len] == sub for l in lst for i in range(0, len(l), sub_len))

其工作原理如下：

>>> mainlist = [['John','Doe',True],['Mary','Jane',False],['James','Smith',False]]
>>> slist1 = ['John', 'Doe']
>>> slist2 = ['John', 'Smith']
>>> slist3 = ['Doe', 'John']
>>> sublist_intersection(mainlist, slist1)
True
>>> sublist_intersection(mainlist, slist2)
False
>>> sublist_intersection(mainlist, slist3)
False

注意：这也假设您正在检查连续的子列表，其中顺序很重要。

确定子列表的一部分是否在主列表中

3 个答案: