在set中查找值比在list中查找值更快:

[item for item in x if item not in set(y)]

我相信这将会比:

[item for item in x if item not in y]

两者都保持了列表的顺序。

如果重复和订购项目是问题:

[i为a中的i，如果不是b中的i或b中的i，删除(i)]

a = [1,2,3,3,3,3,4]
b = [1,3]
result: [2, 3, 3, 3, 4]

Let:

>>> xs = [1, 2, 3, 4, 3, 2, 1]
>>> ys = [1, 3, 3]  

每一项只保留一次xs - ys == {2,4}

取集合差值:

>>> set(xs) - set(ys)
{2, 4}

删除所有xs - ys == [2,4,2]

>>> [x for x in xs if x not in ys]
[2, 4, 2]

如果ys很大，为了获得更好的性能，只将1个ys转换为一个set:

>>> ys_set = set(ys)
>>> [x for x in xs if x not in ys_set]
[2, 4, 2]

只删除相同数量的出现xs - ys == [2,4,2,1]

from collections import Counter, defaultdict

def diff(xs, ys):
    counter = Counter(ys)
    for x in xs:
        if counter[x] > 0:
            counter[x] -= 1
            continue
        yield x

>>> list(diff(xs, ys))
[2, 4, 2, 1]

1 .将xs转换为set并获取set的差异是不必要的(并且更慢，并且破坏顺序)，因为我们只需要在xs上迭代一次。

试试这个。

def subtract_lists(a, b):
    """ Subtracts two lists. Throws ValueError if b contains items not in a """
    # Terminate if b is empty, otherwise remove b[0] from a and recurse
    return a if len(b) == 0 else [a[:i] + subtract_lists(a[i+1:], b[1:]) 
                                  for i in [a.index(b[0])]][0]

>>> x = [1,2,3,4,5,6,7,8,9,0]
>>> y = [1,3,5,7,9]
>>> subtract_lists(x,y)
[2, 4, 6, 8, 0]
>>> x = [1,2,3,4,5,6,7,8,9,0,9]
>>> subtract_lists(x,y)
[2, 4, 6, 8, 0, 9]     #9 is only deleted once
>>>

在set中查找值比在list中查找值更快:

[item for item in x if item not in set(y)]

我相信这将会比:

[item for item in x if item not in y]

两者都保持了列表的顺序。

@aaronasterling提供的答案看起来不错，但是，它与列表的默认接口不兼容:x = MyList(1,2,3,4) vs x = MyList([1,2,3,4])。因此，下面的代码可以用作更友好的python列表:

class MyList(list):
    def __init__(self, *args):
        super(MyList, self).__init__(*args)

    def __sub__(self, other):
        return self.__class__([item for item in self if item not in other])

例子:

x = MyList([1, 2, 3, 4])
y = MyList([2, 5, 2])
z = x - y