下面是@arkolec的回答，下面是一个用于比较列表、元组和集的实用程序类:

from difflib import SequenceMatcher

class ListDiffer:

    def __init__(self, left, right, strict:bool=False):
        assert isinstance(left, (list, tuple, set)), "left must be list, tuple or set"
        assert isinstance(right, (list, tuple, set)), "right must be list, tuple or set"
        self.l = list(left) if isinstance(left, (tuple, set)) else left
        self.r = list(right) if isinstance(left, (tuple, set)) else right

        if strict:
            assert isinstance(left, right.__class__), \
                f'left type ({left.__class__.__name__}) must equal right type ({right.__class__.__name__})'

        self.diffs = []
        self.equal = []

        for tag, i, j, k, l in SequenceMatcher(None, self.l, self.r).get_opcodes():
            if tag in ['delete', 'replace', 'insert']:
                self.diffs.append((tag, i, j, k, l))
            elif tag == 'equal':
                [self.equal.append(v) for v in left[i:j]]
                


    def has_diffs(self):
        return len(self.diffs) > 0


    def only_left(self):
        a = self.l[:]
        [a.remove(v) for v in self.equal]
        return a

    def only_right(self):
        a = self.r[:]
        [a.remove(v) for v in self.equal]
        return a


    def __str__(self, verbose:bool=False):
        iD = 0
        sb = []
        if verbose:
            sb.append(f"left: {self.l}\n")
            sb.append(f"right: {self.r}\n")
            sb.append(f"diffs: ")
        for tag, i, j, k, l in self.diffs:
            s = f"({iD})"
            if iD > 0: sb.append(' | ')
            if tag in ('delete', 'replace'): s = f'{s} l:{self.l[i:j]}'
            if tag in ('insert', 'replace'): s = f'{s} r:{self.r[k:l]}'
            sb.append(s)
            iD = iD + 1

        if verbose:
            sb.append(f"\nequal: {self.equal}")
        return ''.join(sb)

    def __repr__(self) -> str:
        return "<ListDiffer> {}".format(self.__str__())

用法:

left = ['a','b','c']
right = ['aa','b','c','d']
# right = ('aa','b','c','d')
ld = ListDiffer(left, right, strict=True)
print(f'ld.has_diffs(): {ld.has_diffs()}')
print(f'ld: {ld}')
print(f'ld.only_left(): {ld.only_left()}')
print(f'ld.only_right(): {ld.only_right()}')

输出:

ld.has_diffs(): True
ld: (0) l:['a'] r:['aa'] | (1) r:['d']
ld.only_left(): ['a']
ld.only_right(): ['aa', 'd']

我不能说性能，但你可以使用ld.only_left()来获得你正在寻找的结果。

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])

这里有一个简单的方法来区分两个列表(无论内容是什么)，你可以得到如下所示的结果:

>>> from sets import Set
>>>
>>> l1 = ['xvda', False, 'xvdbb', 12, 'xvdbc']
>>> l2 = ['xvda', 'xvdbb', 'xvdbc', 'xvdbd', None]
>>>
>>> Set(l1).symmetric_difference(Set(l2))
Set([False, 'xvdbd', None, 12])

希望这对你有所帮助。

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

由于目前的解决方案都不产生一个元组，我将抛出:

temp3 = tuple(set(temp1) - set(temp2))

另外:

#edited using @Mark Byers idea. If you accept this one as answer, just accept his instead.
temp3 = tuple(x for x in temp1 if x not in set(temp2))

像其他在这个方向上得到答案的非元组一样，它保持了顺序

最简单的方法，

使用设置().difference(设置())

list_a = [1,2,3]
list_b = [2,3]
print set(list_a).difference(set(list_b))

答案设置([1])

可以打印为列表，

print list(set(list_a).difference(set(list_b)))