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
>>>

from collections import Counter

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

print(list(x-y))

使用一个列表推导式来计算差值，同时保持x的原始顺序:

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

如果你不需要列表属性(例如，排序)，使用一个集差异，正如其他答案所建议的:

list(set(x) - set(y))

为了允许x - y中缀语法，在从list继承的类上重写__sub__:

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])

用法:

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])

这个例子减去了两个列表:

# List of pairs of points
list = []
list.append([(602, 336), (624, 365)])
list.append([(635, 336), (654, 365)])
list.append([(642, 342), (648, 358)])
list.append([(644, 344), (646, 356)])
list.append([(653, 337), (671, 365)])
list.append([(728, 13), (739, 32)])
list.append([(756, 59), (767, 79)])

itens_to_remove = []
itens_to_remove.append([(642, 342), (648, 358)])
itens_to_remove.append([(644, 344), (646, 356)])

print("Initial List Size: ", len(list))

for a in itens_to_remove:
    for b in list:
        if a == b :
            list.remove(b)

print("Final List Size: ", len(list))

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。