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有一个称为列表推导式的语言特性，它非常适合使这类事情变得极其简单。下面的语句完全是你想要的，并将结果存储在l3中:

l3 = [x for x in l1 if x not in l2]

L3将包含[1,6]。

一种方法是使用集合:

>>> set([1,2,6,8]) - set([2,3,5,8])
set([1, 6])

但是请注意，集合不会保留元素的顺序，并且会删除任何重复的元素。元素也需要是可哈希的。如果这些限制是可以容忍的，那么这通常是最简单和性能最高的选项。

性能比较

在Python 3.9.1和Python 2.7.16上比较这里提到的所有答案的性能。

Python 3.9.1

答案按表现顺序列出:

Arkku's set difference using subtraction "-" operation - (91.3 nsec per loop) mquadri$ python3 -m timeit -s "l1 = set([1,2,6,8]); l2 = set([2,3,5,8]);" "l1 - l2" 5000000 loops, best of 5: 91.3 nsec per loop Moinuddin Quadri's using set().difference()- (133 nsec per loop) mquadri$ python3 -m timeit -s "l1 = set([1,2,6,8]); l2 = set([2,3,5,8]);" "l1.difference(l2)" 2000000 loops, best of 5: 133 nsec per loop Moinuddin Quadri's list comprehension with set based lookup- (366 nsec per loop) mquadri$ python3 -m timeit -s "l1 = [1,2,6,8]; l2 = set([2,3,5,8]);" "[x for x in l1 if x not in l2]" 1000000 loops, best of 5: 366 nsec per loop Donut's list comprehension on plain list - (489 nsec per loop) mquadri$ python3 -m timeit -s "l1 = [1,2,6,8]; l2 = [2,3,5,8];" "[x for x in l1 if x not in l2]" 500000 loops, best of 5: 489 nsec per loop Daniel Pryden's generator expression with set based lookup and type-casting to list - (583 nsec per loop) : Explicitly type-casting to list to get the final object as list, as requested by OP. If generator expression is replaced with list comprehension, it'll become same as Moinuddin Quadri's list comprehension with set based lookup. mquadri$ mquadri$ python3 -m timeit -s "l1 = [1,2,6,8]; l2 = set([2,3,5,8]);" "list(x for x in l1 if x not in l2)" 500000 loops, best of 5: 583 nsec per loop Moinuddin Quadri's using filter() and explicitly type-casting to list (need to explicitly type-cast as in Python 3.x, it returns iterator) - (681 nsec per loop) mquadri$ python3 -m timeit -s "l1 = [1,2,6,8]; l2 = set([2,3,5,8]);" "list(filter(lambda x: x not in l2, l1))" 500000 loops, best of 5: 681 nsec per loop Akshay Hazari's using combination of functools.reduce + filter -(3.36 usec per loop) : Explicitly type-casting to list as from Python 3.x it started returned returning iterator. Also we need to import functools to use reduce in Python 3.x mquadri$ python3 -m timeit "from functools import reduce; l1 = [1,2,6,8]; l2 = [2,3,5,8];" "list(reduce(lambda x,y : filter(lambda z: z!=y,x) ,l1,l2))" 100000 loops, best of 5: 3.36 usec per loop

Python 2.7.16

答案按表现顺序列出:

Arkku's set difference using subtraction "-" operation - (0.0783 usec per loop) mquadri$ python -m timeit -s "l1 = set([1,2,6,8]); l2 = set([2,3,5,8]);" "l1 - l2" 10000000 loops, best of 3: 0.0783 usec per loop Moinuddin Quadri's using set().difference()- (0.117 usec per loop) mquadri$ mquadri$ python -m timeit -s "l1 = set([1,2,6,8]); l2 = set([2,3,5,8]);" "l1.difference(l2)" 10000000 loops, best of 3: 0.117 usec per loop Moinuddin Quadri's list comprehension with set based lookup- (0.246 usec per loop) mquadri$ python -m timeit -s "l1 = [1,2,6,8]; l2 = set([2,3,5,8]);" "[x for x in l1 if x not in l2]" 1000000 loops, best of 3: 0.246 usec per loop Donut's list comprehension on plain list - (0.372 usec per loop) mquadri$ python -m timeit -s "l1 = [1,2,6,8]; l2 = [2,3,5,8];" "[x for x in l1 if x not in l2]" 1000000 loops, best of 3: 0.372 usec per loop Moinuddin Quadri's using filter() - (0.593 usec per loop) mquadri$ python -m timeit -s "l1 = [1,2,6,8]; l2 = set([2,3,5,8]);" "filter(lambda x: x not in l2, l1)" 1000000 loops, best of 3: 0.593 usec per loop Daniel Pryden's generator expression with set based lookup and type-casting to list - (0.964 per loop) : Explicitly type-casting to list to get the final object as list, as requested by OP. If generator expression is replaced with list comprehension, it'll become same as Moinuddin Quadri's list comprehension with set based lookup. mquadri$ python -m timeit -s "l1 = [1,2,6,8]; l2 = set([2,3,5,8]);" "list(x for x in l1 if x not in l2)" 1000000 loops, best of 3: 0.964 usec per loop Akshay Hazari's using combination of functools.reduce + filter -(2.78 usec per loop) mquadri$ python -m timeit "l1 = [1,2,6,8]; l2 = [2,3,5,8];" "reduce(lambda x,y : filter(lambda z: z!=y,x) ,l1,l2)" 100000 loops, best of 3: 2.78 usec per loop

试试这个:

l1=[1,2,6,8]
l2=[2,3,5,8]
r=[]
for x in l1:
    if x in l2:
        continue
    r=r+[x]
print(r)

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