python - scikit-learn cross validation, negative values with mean squared error

59
votes

When I use the following code with Data matrix X of size (952,144) and output vector y of size (952), mean_squared_error metric returns negative values, which is unexpected. Do you have any idea?

from sklearn.svm import SVR
from sklearn import cross_validation as CV

reg = SVR(C=1., epsilon=0.1, kernel='rbf')
scores = CV.cross_val_score(reg, X, y, cv=10, scoring='mean_squared_error')

all values in scores are then negative.

3
pythonregressionscikit-learncross-validation
Yes, this is supposed to happen. I forget exactly why, but I believe it's related to them minimizing the result when performing grid searching. The actual MSE is simply the postive version of the number you're getting. - David
possible duplicate of sklearn GridSearchCV with Pipeline -- @David is right, when the unified scoring API was introduced, we decided to always maximize the score, which means scores that are actually losses need to be negated. - Fred Foo

3 Answers

72
votes

Trying to close this out, so am providing the answer that David and larsmans have eloquently described in the comments section:

Yes, this is supposed to happen. The actual MSE is simply the positive version of the number you're getting.

The unified scoring API always maximizes the score, so scores which need to be minimized are negated in order for the unified scoring API to work correctly. The score that is returned is therefore negated when it is a score that should be minimized and left positive if it is a score that should be maximized.

This is also described in sklearn GridSearchCV with Pipeline.

1
votes

To see what are available scoring keys use:

import sklearn
print(sklearn.metrics.SCORERS.keys())

You can either use 'r2' or 'neg_mean_squared_error'. There are lots of options based on your requirement.

0
votes

You can fix it by changing scoring method to "neg_mean_squared_error" as you can see below:

from sklearn.svm import SVR
from sklearn import cross_validation as CV

reg = SVR(C=1., epsilon=0.1, kernel='rbf')
scores = CV.cross_val_score(reg, X, y, cv=10, scoring='neg_mean_squared_error')