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。

你也可以使用numpy包:

import numpy as np
A = np.array(a)
maximum_indices = np.where(A==max(a))

这将返回包含max值的所有下标的numpy数组

如果你想把它变成一个列表:

maximum_indices_list = maximum_indices.tolist()

>>> max(enumerate([1,2,3,32,1,5,7,9]),key=lambda x: x[1])
>>> (3, 32)

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)

还有一个解决方案，只给出第一个外观，可以通过使用numpy实现:

>>> import numpy as np
>>> a_np = np.array(a)
>>> np.argmax(a_np)
9

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]