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])

这可能比马克的清单理解还要快:

list(itertools.filterfalse(set(temp2).__contains__, temp1))

def diffList(list1, list2):     # returns the difference between two lists.
    if len(list1) > len(list2):
        return (list(set(list1) - set(list2)))
    else:
        return (list(set(list2) - set(list1)))

例如，如果list1 =[10, 15, 20, 25, 30, 35, 40]和list2 =[25, 40, 35]，那么返回的列表将输出= [10,20,30,15]

如果散差表的元素是排序和集合的，你可以使用朴素方法。

list1=[1,2,3,4,5]
list2=[1,2,3]

print list1[len(list2):]

或者使用本机set方法:

subset=set(list1).difference(list2)

print subset

import timeit
init = 'temp1 = list(range(100)); temp2 = [i * 2 for i in range(50)]'
print "Naive solution: ", timeit.timeit('temp1[len(temp2):]', init, number = 100000)
print "Native set solution: ", timeit.timeit('set(temp1).difference(temp2)', init, number = 100000)

朴素解:0.0787101593292

本机集解决方案:0.998837615564

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']

两个列表(例如list1和list2)之间的差异可以使用下面的简单函数找到。

def diff(list1, list2):
    c = set(list1).union(set(list2))  # or c = set(list1) | set(list2)
    d = set(list1).intersection(set(list2))  # or d = set(list1) & set(list2)
    return list(c - d)

or

def diff(list1, list2):
    return list(set(list1).symmetric_difference(set(list2)))  # or return list(set(list1) ^ set(list2))

通过使用上述函数，可以使用diff(temp2, temp1)或diff(temp1, temp2)找到差异。两者都会给出结果['Four'， 'Three']。你不需要担心列表的顺序或者先给出哪个列表。

Python文档参考