试试这个:

temp3 = set(temp1) - set(temp2)

如果您真的关注性能，那么使用numpy!

以下是github上的完整笔记本，并对list、numpy和pandas进行了比较。

https://gist.github.com/denfromufa/2821ff59b02e9482be15d27f2bbd4451

试试这个:

temp3 = set(temp1) - set(temp2)

您可以循环遍历第一个列表，对于不在第二个列表中但在第一个列表中的每一项，将其添加到第三个列表中。例句:

temp3 = []
for i in temp1:
    if i not in temp2:
        temp3.append(i)
print(temp3)

你可以使用列表推导式:

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

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"