有没有其他更快的方法来查找索引位置上的项目。
items = ['aaa','sss','ddd','fff','gggg','hhhh']
indices = [1,3,4]
My way:
[items[i] for i in indices]
operator.itemgetter
:getter = itemgetter(1,3,4)
desired = getter(items)
itemgetter
大约快 2.5 倍(但我没有计算生成 getter
函数的实际时间花费)。>>> items = ['aaa','sss','ddd','fff','gggg','hhhh']
>>> indices = [1,3,4]
>>> from operator import itemgetter
>>> import timeit
>>> getter = itemgetter(*indices)
>>> def list_comp(items=items,indices=indices):
... return [items[i] for i in indices]
...
>>> timeit.timeit('getter(items)','from __main__ import getter,items')
0.2926821708679199
>>> timeit.timeit('list_comp()','from __main__ import list_comp')
0.7736802101135254
>>> getter(items)
('sss', 'fff', 'gggg')
>>> list_comp()
['sss', 'fff', 'gggg']