由 TensorFlow 计算的均方误差

Question

I am trying to figure out how the mean squared error (MSE) is calculated by tensorflow and was reading the post at https://www.tensorflow.org/api_docs/python/tf/keras/metrics/mean_squared_error .

First of all, MSE is defined as (see https://en.wikipedia.org/wiki/Mean_squared_error ):

Suppose I have a single output and create true and predicted values.

import numpy as np
import random

y_true = np.random.randint(0, 10, size=(2, 1))
print(y_true,"\n")
y_pred = np.random.randint(0,5,size=(2, 1))
print(y_pred)

[[7]
 [5]]

[[2]
 [2]]

When I call tf.keras.losses.mean_squared_error(y_true, y_pred) , what I expect to see is that [(7-2)^2 + (5-2)^2]/2 = 17 , however, it returns me array([25, 9]) . Why doesn't tensorflow compute the mean?

Then, I increase the column numbers.

y_true = np.random.randint(0, 10, size=(2, 3))
print(y_true,"\n")
y_pred = np.random.randint(0,5,size=(2, 3))
print(y_pred)

[[2 6 0]
 [3 3 4]] 

[[4 2 4]
 [3 4 2]]

The answer returned by tensorflow is array([12, 1]) . I'm not able to understand how these values are computed. What I was expecting was [(2-4)^2+ (6-2)^2+(0-4)^2]/2 + [(3-3)^2 + (3-4)^2+ (4-2)^2]/2 .

Answer 1

The second value is computed as

a = np.array([2,6,0])
b = np.array([4,2,4])
((b - a)**2).mean()

Or [(2-4)^2+ (6-2)^2+(0-4)^2]/3

According to their doc it is equivalent to np.mean(np.square(y_true - y_pred), axis=-1)

So it is computing the mse row-wise.