当你有本地方法可用时，不确定为什么上面的解释如此复杂:

main_list = list(set(list_2)-set(list_1))

使用这样的列表理解:

main_list = [item for item in list_2 if item not in list_1]

输出:

>>> list_1 = ["a", "b", "c", "d", "e"]
>>> list_2 = ["a", "f", "c", "m"] 
>>> 
>>> main_list = [item for item in list_2 if item not in list_1]
>>> main_list
['f', 'm']

编辑:

就像下面评论中提到的，对于大列表，上面不是理想的解决方案。在这种情况下，更好的选择是先将list_1转换为set:

set_1 = set(list_1)  # this reduces the lookup time from O(n) to O(1)
main_list = [item for item in list_2 if item not in set_1]

main_list=[]
list_1=["a", "b", "c", "d", "e"]
list_2=["a", "f", "c", "m"]

for i in list_2:
    if i not in list_1:
        main_list.append(i)

print(main_list)

输出:

['f', 'm']

你可以使用集合:

main_list = list(set(list_2) - set(list_1))

输出:

>>> list_1=["a", "b", "c", "d", "e"]
>>> list_2=["a", "f", "c", "m"]
>>> set(list_2) - set(list_1)
set(['m', 'f'])
>>> list(set(list_2) - set(list_1))
['m', 'f']

根据@JonClements的评论，这里是一个更整洁的版本:

>>> list_1=["a", "b", "c", "d", "e"]
>>> list_2=["a", "f", "c", "m"]
>>> list(set(list_2).difference(list_1))
['m', 'f']

我将这些列表压缩在一起，逐个元素进行比较。

main_list = [b for a, b in zip(list1, list2) if a!= b]

当你有本地方法可用时，不确定为什么上面的解释如此复杂:

main_list = list(set(list_2)-set(list_1))