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]
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
使用集合差
>>> z = list(set(x) - set(y))
>>> z
[0, 8, 2, 4, 6]
或者你可以让x和y是集合所以你不需要做任何转换。
其他回答
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
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。
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])
from collections import Counter
y = Counter(y)
x = Counter(x)
print(list(x-y))
使用集合差
>>> z = list(set(x) - set(y))
>>> z
[0, 8, 2, 4, 6]
或者你可以让x和y是集合所以你不需要做任何转换。