@shash在其他地方回答了这个问题

找到最大列表元素的索引的python方法是 Position = max(enumerate(a)， key=lambda x: x[1])[0]

一个通过。然而，它比@Silent_Ghost的解决方案慢，甚至比@nmichaels的解决方案更慢:

for i in s m j n; do echo $i;  python -mtimeit -s"import maxelements as me" "me.maxelements_${i}(me.a)"; done
s
100000 loops, best of 3: 3.13 usec per loop
m
100000 loops, best of 3: 4.99 usec per loop
j
100000 loops, best of 3: 3.71 usec per loop
n
1000000 loops, best of 3: 1.31 usec per loop

a.index(max(a))

会告诉你列表a中值最大的元素的第一个实例的索引。

>>> m = max(a)
>>> [i for i, j in enumerate(a) if j == m]
[9, 12]

如果你想获取一个名为data的列表中最大n个数字的索引，你可以使用Pandas sort_values:

pd.Series(data).sort_values(ascending=False).index[0:n]

这里是最大值和它出现的索引:

>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> a = [32, 37, 28, 30, 37, 25, 27, 24, 35, 55, 23, 31, 55, 21, 40, 18, 50, 35, 41, 49, 37, 19, 40, 41, 31]
>>> for i, x in enumerate(a):
...     d[x].append(i)
... 
>>> k = max(d.keys())
>>> print k, d[k]
55 [9, 12]

后来:为了满足@SilentGhost

>>> from itertools import takewhile
>>> import heapq
>>> 
>>> def popper(heap):
...     while heap:
...         yield heapq.heappop(heap)
... 
>>> a = [32, 37, 28, 30, 37, 25, 27, 24, 35, 55, 23, 31, 55, 21, 40, 18, 50, 35, 41, 49, 37, 19, 40, 41, 31]
>>> h = [(-x, i) for i, x in enumerate(a)]
>>> heapq.heapify(h)
>>> 
>>> largest = heapq.heappop(h)
>>> indexes = [largest[1]] + [x[1] for x in takewhile(lambda large: large[0] == largest[0], popper(h))]
>>> print -largest[0], indexes
55 [9, 12]

这里有一个简单的单步解决方案。

import math
nums = [32, 37, 28, 30, 37, 25, 55, 27, 24, 35, 55, 23, 31]

max_val = -math.inf
res = []

for i, val in enumerate(nums):
    if(max_val < val):
        max_val = val
        res = [i]
    elif(max_val == val):
        res.append(i)
print(res)