单线版arulmr解决方案

def diff(listA, listB):
    return set(listA) - set(listB) | set(listB) -set(listA)

你可以使用列表推导式:

temp3 = [item for item in temp1 if item not in temp2]

以下是@SuperNova的回答的修改版本

def get_diff(a: list, b: list) -> list:
    return list(set(a) ^ set(b))

这里有一些简单的、保持顺序的方法来区分两个字符串列表。

Code

使用pathlib的一个不寻常的方法:

import pathlib


temp1 = ["One", "Two", "Three", "Four"]
temp2 = ["One", "Two"]

p = pathlib.Path(*temp1)
r = p.relative_to(*temp2)
list(r.parts)
# ['Three', 'Four']

这假设两个列表包含相同开头的字符串。更多细节请参阅文档。注意，与set操作相比，它不是特别快。

使用itertools.zip_longest的直接实现:

import itertools as it


[x for x, y in it.zip_longest(temp1, temp2) if x != y]
# ['Three', 'Four']

如果要删除列表a中存在于列表b中的所有值。

def list_diff(a, b):
    r = []

    for i in a:
        if i not in b:
            r.append(i)
    return r

list_diff([1、2、2],[1])

结果(2,2):

or

def list_diff(a, b):
    return [x for x in a if x not in b]

如果您想要更类似于变更集的东西……可以使用Counter

from collections import Counter

def diff(a, b):
  """ more verbose than needs to be, for clarity """
  ca, cb = Counter(a), Counter(b)
  to_add = cb - ca
  to_remove = ca - cb
  changes = Counter(to_add)
  changes.subtract(to_remove)
  return changes

lista = ['one', 'three', 'four', 'four', 'one']
listb = ['one', 'two', 'three']

In [127]: diff(lista, listb)
Out[127]: Counter({'two': 1, 'one': -1, 'four': -2})
# in order to go from lista to list b, you need to add a "two", remove a "one", and remove two "four"s

In [128]: diff(listb, lista)
Out[128]: Counter({'four': 2, 'one': 1, 'two': -1})
# in order to go from listb to lista, you must add two "four"s, add a "one", and remove a "two"