Python 确定子列表的一部分是否在主列表中

我有2d列表

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

---

如何确定子列表的子列表是否存在于一个列表中，其中，如果根据mainlist测试slist1，则返回True，而slist2将返回False

我在想这样的事情（源代码）

有没有更“蟒蛇式”的方法？谢谢

编辑：

根据mainlist测试的slist1将返回True

slist2将返回False

slist3将返回False

slist4将返回False

---

因此，基本上，我只是测试slist是否位于mainlist[x]的前2个索引中。

如果我理解了您的问题，我想您可以使用

set（）

和集合之间的交集，如以下示例：

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缺点。。。它并没有涵盖所有的案例。

由于OP没有回应，我将涵盖这两个案例

如果顺序不重要；

['John'，'Doe']

和

['Doe'，'John']

都假定包含在

主列表中：

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

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


如果秩序确实重要<代码>['John'，'Doe']

已包含，但

['Doe'，'John']

不在

主列表中

：

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

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

---

似乎要检查

mainlist

中每个列表中的子列表是否有

slist1

或

slist2

。在这种情况下，您可以执行以下简单操作：

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

或一个较短的解决方案，包括：

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：这还假设您正在检查相邻的子列表，其中顺序很重要

您想测试mainlist列表中的所有项目还是只测试前两个项目？在这种情况下，我会做

if slist1==sublist[：2]：

而不是

all

“如何确定子列表的子列表是否存在于一个列表中，其中如果slist1根据mainlist进行测试，则返回True，而slist2将返回False”这让我头疼。你能更清楚地定义你想要什么吗？让所有

设置
sDo成为mainlist中每个子列表的索引2中的布尔值怎么样？