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)

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

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))

现有的解决方案都提供了其中的一个或另一个:

比O(n*m)性能快。 保持输入列表的顺序。

但到目前为止，还没有一种解决方案兼而有之。如果你两者都想要，试试这个:

s = set(temp2)
temp3 = [x for x in temp1 if x not in s]

性能测试

import timeit
init = 'temp1 = list(range(100)); temp2 = [i * 2 for i in range(50)]'
print timeit.timeit('list(set(temp1) - set(temp2))', init, number = 100000)
print timeit.timeit('s = set(temp2);[x for x in temp1 if x not in s]', init, number = 100000)
print timeit.timeit('[item for item in temp1 if item not in temp2]', init, number = 100000)

结果:

4.34620224079 # ars' answer
4.2770634955  # This answer
30.7715615392 # matt b's answer

我提出的方法以及保持顺序也比集合减法(稍微)快，因为它不需要构造一个不必要的集合。如果第一个列表比第二个列表长得多，并且散列代价昂贵，那么性能差异将更加明显。下面是第二个测试:

init = '''
temp1 = [str(i) for i in range(100000)]
temp2 = [str(i * 2) for i in range(50)]
'''

结果:

11.3836875916 # ars' answer
3.63890368748 # this answer (3 times faster!)
37.7445402279 # matt b's answer

我在这个游戏中有点晚了，但你可以做一个性能的比较，上面提到的一些代码和这个，两个最快的竞争者是，

list(set(x).symmetric_difference(set(y)))
list(set(x) ^ set(y))

我为我的初级编码水平道歉。

import time
import random
from itertools import filterfalse

# 1 - performance (time taken)
# 2 - correctness (answer - 1,4,5,6)
# set performance
performance = 1
numberoftests = 7

def answer(x,y,z):
    if z == 0:
        start = time.clock()
        lists = (str(list(set(x)-set(y))+list(set(y)-set(y))))
        times = ("1 = " + str(time.clock() - start))
        return (lists,times)

    elif z == 1:
        start = time.clock()
        lists = (str(list(set(x).symmetric_difference(set(y)))))
        times = ("2 = " + str(time.clock() - start))
        return (lists,times)

    elif z == 2:
        start = time.clock()
        lists = (str(list(set(x) ^ set(y))))
        times = ("3 = " + str(time.clock() - start))
        return (lists,times)

    elif z == 3:
        start = time.clock()
        lists = (filterfalse(set(y).__contains__, x))
        times = ("4 = " + str(time.clock() - start))
        return (lists,times)

    elif z == 4:
        start = time.clock()
        lists = (tuple(set(x) - set(y)))
        times = ("5 = " + str(time.clock() - start))
        return (lists,times)

    elif z == 5:
        start = time.clock()
        lists = ([tt for tt in x if tt not in y])
        times = ("6 = " + str(time.clock() - start))
        return (lists,times)

    else:    
        start = time.clock()
        Xarray = [iDa for iDa in x if iDa not in y]
        Yarray = [iDb for iDb in y if iDb not in x]
        lists = (str(Xarray + Yarray))
        times = ("7 = " + str(time.clock() - start))
        return (lists,times)

n = numberoftests

if performance == 2:
    a = [1,2,3,4,5]
    b = [3,2,6]
    for c in range(0,n):
        d = answer(a,b,c)
        print(d[0])

elif performance == 1:
    for tests in range(0,10):
        print("Test Number" + str(tests + 1))
        a = random.sample(range(1, 900000), 9999)
        b = random.sample(range(1, 900000), 9999)
        for c in range(0,n):
            #if c not in (1,4,5,6):
            d = answer(a,b,c)
            print(d[1])