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]

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()

类似的想法与列表理解，但没有枚举

m = max(a)
[i for i in range(len(a)) if a[i] == m]

@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

import operator

def max_positions(iterable, key=None, reverse=False):
  if key is None:
    def key(x):
      return x
  if reverse:
    better = operator.lt
  else:
    better = operator.gt

  it = enumerate(iterable)
  for pos, item in it:
    break
  else:
    raise ValueError("max_positions: empty iterable")
    # note this is the same exception type raised by max([])
  cur_max = key(item)
  cur_pos = [pos]

  for pos, item in it:
    k = key(item)
    if better(k, cur_max):
      cur_max = k
      cur_pos = [pos]
    elif k == cur_max:
      cur_pos.append(pos)

  return cur_max, cur_pos

def min_positions(iterable, key=None, reverse=False):
  return max_positions(iterable, key, not reverse)

>>> L = range(10) * 2
>>> L
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> max_positions(L)
(9, [9, 19])
>>> min_positions(L)
(0, [0, 10])
>>> max_positions(L, key=lambda x: x // 2, reverse=True)
(0, [0, 1, 10, 11])

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