如何使 numpy.argmax 返回最大值的所有匹配项？

我试图找到一个函数，它返回给定列表中最大值的 所有出现次数。

然而，numpy.argmax 只返回它找到的第一个匹配项。例如:

from numpy import argmax


list = [7, 6, 5, 7, 6, 7, 6, 6, 6, 4, 5, 6]
winner = argmax(list)


print winner

只给出 0的索引，但是我希望它给出所有的索引: 0, 3, 5。

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小开

As documentation of np.argmax says: "In case of multiple occurrences of the maximum values, the indices corresponding to the first occurrence are returned.", so you will need another strategy.

One option you have is using np.argwhere in combination with np.amax:

>>> import numpy as np
>>> listy = [7, 6, 5, 7, 6, 7, 6, 6, 6, 4, 5, 6]
>>> winner = np.argwhere(listy == np.amax(listy))
>>> print(winner)
[[0]
[3]
[5]]
>>> print(winner.flatten().tolist()) # if you want it as a list
[0, 3, 5]
小开

Much simpler...

list[list == np.max(list)]

小开

In case it matters, the following algorithm runs in O(n) instead of O(2n) (i.e., using np.argmax and then np.argwhere):

def allmax(a):
if len(a) == 0:
return []
all_ = [0]
max_ = a[0]
for i in range(1, len(a)):
if a[i] > max_:
all_ = [i]
max_ = a[i]
elif a[i] == max_:
all_.append(i)
return all_
小开

It is even easier, when compared to other answers, if you use np.flatnonzero:

>>> import numpy as np
>>> your_list = np.asarray([7, 6, 5, 7, 6, 7, 6, 6, 6, 4, 5, 6])
>>> winners = np.flatnonzero(your_list == np.max(your_list))
>>> winners
array([0, 3, 5])

If you want a list:

>>> winners.tolist()
[0, 3, 5]

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