如何解读np.argmax()的输出结果？相关案例疑问咨询

Hey there! Let's clear up these two confusion points about np.argmax() with your examples.

First, let's recap your 2D array:

import numpy as np
a = np.arange(6).reshape(2,3)
# a looks like:
# array([[0, 1, 2],
#        [3, 4, 5]])

When you specify axis=0, np.argmax() looks for the index of the maximum value along the vertical axis (columns). Let's break down each column:

• First column values: [0, 3] → max is 3, which is at index 1 (the second row)
• Second column values: [1, 4] → max is 4, at index 1
• Third column values: [2, 5] → max is 5, at index 1

That's why the result is array([1, 1, 1]) — it's not pointing to the global maximum's position, but rather giving the row index of the maximum in each individual column.

Your 3D array b has shape (2, 2, 3):

b = np.array([[[2,3,4],[4,5,6]],[[3,7,1],[2,5,9]]])

When you don't specify an axis parameter, np.argmax() automatically flattens the entire array into a 1D sequence first, then finds the index of the maximum value in this flattened array.

Let's flatten b manually:

Flattened b: [2, 3, 4, 4, 5, 6, 3, 7, 1, 2, 5, 9]
Indices (0-based):  0  1  2  3  4  5  6  7  8  9 10 11

The maximum value 9 is at index 11 in this flattened list — that's exactly what np.argmax(b) returns. If you want to map this 1D index back to the original 3D coordinates, you can use np.unravel_index():

np.unravel_index(11, b.shape)  # returns (1, 1, 2)

And indeed, b[1][1][2] gives you the value 9, which is the global maximum.

内容的提问来源于stack exchange，提问作者FaCoffee