Python 如何让 numpy.argmax 返回所有出现的最大值？
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How to make numpy.argmax return all occurrences of the maximum?
提问by Marieke_W
I'm trying to find a function that returns alloccurrences of the maximum in a given list.
我正在尝试找到一个函数，该函数返回给定列表中所有出现的最大值。
numpy.argmaxhowever only returns the first occurrence that it finds. For instance:
numpy.argmax但是只返回它找到的第一次出现。例如：
from numpy import argmax
list = [7, 6, 5, 7, 6, 7, 6, 6, 6, 4, 5, 6]
winner = argmax(list)
print winner
gives only index 0. But I want it to give all indices: 0, 3, 5.
只给出索引0。但我希望它提供所有索引：0, 3, 5.
回答by jabaldonedo
As documentation of np.argmaxsays: "In case of multiple occurrences of the maximum values, the indices corresponding to the first occurrence are returned.", so you will need another strategy.
正如文档所述np.argmax：“如果多次出现最大值，则返回与第一次出现相对应的索引。” ，因此您将需要另一种策略。
One option you have is using np.argwherein combination with np.amax:
您拥有的一种选择是np.argwhere结合使用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]
回答by user5838456
Much simpler...
简单多了...
list[list == np.max(list)]
列表[列表== np.max(列表)]
回答by EricRobertBrewer
In case it matters, the following algorithm runs in O(n) instead of O(2n) (i.e., using np.argmaxand then np.argwhere):
万一重要，以下算法以 O(n) 而不是 O(2n) 运行（即，使用np.argmax然后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_
