为什么MSE sklearn库给我的平方误差不同于l2范数平方误差？

Question

I was trying to compute the mean squared error as in:

in python. I saw that scipt/sklearn had an implementation for it already. However, when I tried comparing it to my own implementation they did NOT agree. Why is that? My implementation simply uses the norm 2 (or the Frobenius norm not matching) and nothing else fancy.

To test this I wrote the following script:

import sklearn
from sklearn.decomposition import PCA
from sklearn.metrics import mean_squared_error
import numpy as np
from numpy import linalg as LA

X_truth = np.ones((5,6))
X_pred = 1.7*np.ones((5,6))

print 'LA error: ', (1.0/5)*LA.norm(X_truth - X_pred)**2
print 'LA error: ', (1.0/X_truth.shape[0])*LA.norm(X_truth - X_pred)**2
print 'LA error:: ', (1.0/5)*LA.norm(X_truth - X_pred, 'fro')**2
print 'LA error: ', LA.norm(X_truth - X_pred)**2
print 'LA error: ', LA.norm(X_truth - X_pred)
print 'LA error: ', (1.0/X_truth.shape[0])*LA.norm(X_truth - X_pred)
print 'sklearn MSE error: ', mean_squared_error(X_truth, X_pred)

I literally tested every combination I could think of and I still can't have them to match. Any ideas?

Answer 1

The formula that's used is a little unusual in that it doesn't take the square root of the sum of squares, whereas LA.norm does.

If you look carefully at the docs, you can recreate the formula

np.sum((X_truth-X_pred)**2)/X_truth.size

gives 0.49 just like

mean_squared_error(X_truth, X_pred)

or

LA.norm(X_truth - X_pred)**2/X_truth.size