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

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

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]

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。

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])

只有一句话:

idx = max(range(len(a)), key = lambda i: a[i])