python - 检查list1在list2内后返回输出

list1 = [[['a','b'],['c','d']],[['f','g'],['h','i']],[['j','k','l'], ['a','b']]]
list2 = [['a','b'],['c','d'],['f','g'],['h','i']]

所以在列表1中，有3个列表。我想检查list1中的列表是否是list 2的子集。但列表列表中的所有列表都必须在列表2中，才能使其为True/Correct。如果所有内容都在list2中，则为true；如果不在list2中，则为false。

[['a','b'],['c','d']]
[['f','g'],['h','i']]
[['j','k','l'], ['a','b']]

所以情况如下

These are from list1, and we're checking against list2

Both [a, b] and [c, d] should be in list 2 -> Both are in list 2, so Return True
Both [f, g] and [h, i] should be in list 1 -> Both are in list 2, so return true
Both [j, k, l] and [a, b] should be in list 1 -> f, k, l is not in list 2, so return False even though a, b are in list 2

Here is my desired output for above results

[True, True, False]

或

val1 = True
val2 = True
val3 = False

代码

def xlist(list1, list2):
    if all(letter in list1 for letter in list2):
        print('True')

print xlist(list1, list2)

final = []
"""I am checking i in list1. In actual, I should be checking all lists within the list of list1."""
    for i in list1:
        print(xlist(list1, list2))
        final.append(xlist(list1, list2))
        print(final)

最佳答案

您的问题是您没有从xlist函数返回任何值（print与return不同）。更改为：

def xlist(list1, list2):
    return all(letter in list1 for letter in list2)

然后：

final = []
for i in list1:
    final.append(xlist(list2, i))
print(final)

结果是：

[True, True, False]

作为另一种较短的方法，您可以将all函数与嵌套的list comprehension一起使用：

>>> list1 = [[['a','b'],['c','d']],[['f','g'],['h','i']],[['j','k','l'], ['a','b']]]
>>> list2 = [['a','b'],['c','d'],['f','g'],['h','i']]
>>> [all(item in list2 for item in sublist) for sublist in list1]
[True, True, False]

关于python - 检查list1在list2内后返回输出，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/57154748/