我想取列表x和y的差值:

>>> x = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> y = [1, 3, 5, 7, 9]  
>>> x - y
# should return [0, 2, 4, 6, 8]

当前回答

@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

其他回答

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。

@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

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
>>>

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   

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]) ]