使用Python set类型。这是最Pythonic的。：）

此外，由于它是原生的，它也应该是最优化的方法。

See:

http://docs.python.org/library/stdtypes.html#set

http://docs.python.org/library/sets.htm(适用于较旧的python)

# Using Python 2.7 set literal format.
# Otherwise, use: l1 = set([1,2,6,8])
#
l1 = {1,2,6,8}
l2 = {2,3,5,8}
l3 = l1 - l2

使用Set推导式{x for x in l2}或Set (l2)来获取Set，然后使用List推导式来获取List

l2set = set(l2)
l3 = [x for x in l1 if x not in l2set]

基准测试代码:

import time

l1 = list(range(1000*10 * 3))
l2 = list(range(1000*10 * 2))

l2set = {x for x in l2}

tic = time.time()
l3 = [x for x in l1 if x not in l2set]
toc = time.time()
diffset = toc-tic
print(diffset)

tic = time.time()
l3 = [x for x in l1 if x not in l2]
toc = time.time()
difflist = toc-tic
print(difflist)

print("speedup %fx"%(difflist/diffset))

基准测试结果:

0.0015058517456054688
3.968189239501953
speedup 2635.179227x    

如果你想要那种行为，集合方法是最好的。如果您不想删除列表l1中仅在l2中存在过一次的元素的所有实例，那么这些set操作将导致错误的结果。假设你在l1中有重复的元素，甚至在l2中也有重复的元素，并且想要两个列表l1 - l2的实际差值，同时保持其余元素的顺序:

l1 = [1, 2, 3, 4, 5, 5, 6, 5, 5, 2]
l2 = [1, 2, 2, 5]
_ = [l1.remove(item) for item in l2 if item in l1] # discard return value
print(l1) # [3, 4, 5, 6, 5, 5]

注意，这将明显比设置操作慢，只在用例需要时使用它 如果你不想修改原来的列表-先创建一个列表的副本

Python 3.8上的集合和列表理解基准

(加起来就是Moinuddin Quadri的基准)

tldr:使用Arkku的集合解决方案，相比之下比承诺的还要快!

根据列表检查现有文件

在我的例子中，我发现使用Arkku的集合解决方案要比python的列表理解方法快40倍(!)。

列表理解:

%%time
import glob
existing = [int(os.path.basename(x).split(".")[0]) for x in glob.glob("*.txt")]
wanted = list(range(1, 100000))
[i for i in wanted if i not in existing]

墙壁时间:28.2秒

Sets

%%time
import glob
existing = [int(os.path.basename(x).split(".")[0]) for x in glob.glob("*.txt")]
wanted = list(range(1, 100000))
set(wanted) - set(existing)

壁时间:689毫秒

使用Python set类型。这是最Pythonic的。：）

此外，由于它是原生的，它也应该是最优化的方法。

See:

http://docs.python.org/library/stdtypes.html#set

http://docs.python.org/library/sets.htm(适用于较旧的python)

# Using Python 2.7 set literal format.
# Otherwise, use: l1 = set([1,2,6,8])
#
l1 = {1,2,6,8}
l2 = {2,3,5,8}
l3 = l1 - l2

使用filterfalse而不使用lambda-expression

When using functions like filter or filterfalse and similar from itertools you can usually save performance by avoiding lambda-expressions and using already existing functions. Instances of list and set defines a __contains__-method to use for containment checks. The in-operator calls this method under the hood, so using x in l2 can be replaced by l2.__contains__(x). Usually this replacement is not really prettier but in this specific case it allows us to gain better performance than using a lambda-expression, when used in combination with filterfalse:

>>> from itertools import filterfalse
>>> l1 = [1, 2, 6, 8]
>>> l2 = [2, 3, 5, 8]
>>> list(filterfalse(l2.__contains__, l1))
[1, 6]

Filterfalse创建一个迭代器，该迭代器生成的所有元素在用作12.2 .__contains__的参数时返回false。

Sets有一个更快的__contains__实现，所以更好的是:

>>> from itertools import filterfalse
>>> l1 = [1, 2, 6, 8]
>>> l2 = set([2, 3, 5, 8])
>>> list(filterfalse(l2.__contains__, l1))
[1, 6]

性能

使用列表:

$  python3 -m timeit -s "from itertools import filterfalse; l1 = [1,2,6,8]; l2 = set([2,3,5,8]);" "list(filterfalse(l2.__contains__, l1))"
500000 loops, best of 5: 522 nsec per loop

使用设置:

$ python3 -m timeit -s "from itertools import filterfalse; l1 = [1,2,6,8]; l2 = set([2,3,5,8]);" "list(filterfalse(l2.__contains__, l1))"
1000000 loops, best of 5: 359 nsec per loop