如何返回可迭代的第 n 个最大项目的原始列表中的索引
heapq.nlargest(2, [100, 2, 400, 500, 400])
output = [(3,500), (2, 400)]
这已经花了我几个小时。我想不通。
>>> seq = [100, 2, 400, 500, 400]
>>> heapq.nlargest(2, enumerate(seq), key=lambda x: x[1])
[(3, 500), (2, 400)]
您可以list.index
与 结合使用map
,这对于小型快速n
(注意返回值为 x的第一list.index
项的列表中的索引):
>>> iterable = [100, 2, 400, 500, 400]
>>> map(iterable.index, heapq.nlargest(2, iterable))
[3, 2]
要查看关联的值...
>>> map(lambda n: (n, iterable.index(n)), heapq.nlargest(2, iterable))
[(500, 3), (400, 2)]
更大的n
请参阅@SilentGhost 的帖子。
编辑:基准一些解决方案:
#!/usr/bin/env python
import heapq
from timeit import Timer
seq = [100, 2, 400, 500, 400]
def a(seq):
"""returns [(3, 500), (2, 400)]"""
return heapq.nlargest(2, enumerate(seq), key=lambda x: x[1])
def b(seq):
"""returns [3, 2]"""
return map(seq.index, heapq.nlargest(2, seq))
def c(seq):
"""returns [(500, 3), (400, 2)]"""
map(lambda n: (n, seq.index(n)), heapq.nlargest(2, seq))
if __name__ == '__main__':
_a = Timer("a(seq)", "from __main__ import a, seq")
_b = Timer("b(seq)", "from __main__ import b, seq")
_c = Timer("c(seq)", "from __main__ import c, seq")
loops = 1000000
print _a.timeit(number=loops)
print _b.timeit(number=loops)
print _c.timeit(number=loops)
# Core i5, 2.4GHz, Python 2.6, Darwin
# 8.92712688446
# 5.64332985878
# 6.50824809074