这段代码不像之前发布的答案那么复杂，但它可以工作:

m = max(a)
n = 0    # frequency of max (a)
for number in a :
    if number == m :
        n = n + 1
ilist = [None] * n  # a list containing index values of maximum number in list a.
ilistindex = 0
aindex = 0  # required index value.    
for number in a :
    if number == m :
        ilist[ilistindex] = aindex
        ilistindex = ilistindex + 1
    aindex = aindex + 1

print ilist

上面代码中的Ilist将包含列表中最大数目的所有位置。

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]

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]

import pandas as pd

pd.Series(a).idxmax()

9

我通常都是这么做的。

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]

@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