@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

这是一个“集合减法”操作。使用设定的数据结构。

在Python 2.7中:

x = {1,2,3,4,5,6,7,8,9,0}
y = {1,3,5,7,9}
print x - y

输出:

>>> print x - y
set([0, 8, 2, 4, 6])

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
>>>

如果列表允许重复元素，你可以使用Counter from collections:

from collections import Counter
result = list((Counter(x)-Counter(y)).elements())

如果你需要保留x中元素的顺序:

result = [ v for c in [Counter(y)] for v in x if not c[v] or c.subtract([v]) ]

from collections import Counter

y = Counter(y)
x = Counter(x)

print(list(x-y))

@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