是否还有其他更快捷的方法可以在索引位置查找项目。
items = ['aaa','sss','ddd','fff','gggg','hhhh']
indices = [1,3,4]
My way:
[items[i] for i in indices]
答案 0 :(得分:7)
如果您反复使用相同的索引,那么使用operator.itemgetter
可能会做得更好:
getter = itemgetter(1,3,4)
desired = getter(items)
根据我的简单基准测试,itemgetter
的速度提高了约2.5倍(但我没有花时间实际构建getter
函数需要多长时间。)
>>> 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']