Is sklearn.metrics.mean_squared_error the larger the better (negated)?


In general, the mean_squared_error is the smaller the better.

When I am using the sklearn metrics package, it says in the document pages:

All scorer objects follow the convention that higher return values are
better than lower return values. Thus metrics which measure the
distance between the model and the data, like
metrics.mean_squared_error, are available as neg_mean_squared_error
which return the negated value of the metric.

enter image description here

However, if I go to:

It says it is the Mean squared error regression loss, didn’t say it is negated.

And if I looked at the source code and checked the example there: it is doing the normal mean squared error, i.e. the smaller the better.

So I am wondering if I missed anything about the negated part in the document. Thanks!

Asked By: Edamame



It’s a convention for implementing your own scoring object [1]. And it must be positive, because you can create a non-loss function to compute a custom positive score. That means that by using a loss function (for a score object) you have to the negative value.

The range of a loss function is: (optimum) [0. ... +] (e.g. unequal values between y and y'). For instance check the formula of the mean squared error, it’s always positive:

mean squared error

Image source:

Answered By: Darius

The actual function "mean_squared_error" doesn’t have anything about the negative part. But the function implemented when you try ‘neg_mean_squared_error’ will return a negated version of the score.

Please check the source code as to how its defined in the source code:

neg_mean_squared_error_scorer = make_scorer(mean_squared_error,

Observe how the param greater_is_better is set to False.

Now all these scores/losses are used in various other things like cross_val_score, cross_val_predict, GridSearchCV etc. For example, in cases of ‘accuracy_score’ or ‘f1_score’, the higher score is better, but in case of losses (errors), lower score is better. To handle them both in same way, it returns the negative.

So this utility is made for handling the scores and losses in same way without changing the source code for the specific loss or score.

So, you did not miss anything. You just need to take care of the scenario where you want to use the loss function. If you only want to calculate the mean_squared_error you can use mean_squared_error only. But if you want to use it to tune your models, or cross_validate using the utilities present in Scikit, use 'neg_mean_squared_error'.

Maybe add some details about that and I will explain more.

Answered By: Vivek Kumar

This is exactly what I am looking for in my code that I am trying to decipher and clarify the rmse reports to make sense of my data.

in my case, I am using this approach to calculate the rmse. How should I read the reports? Is higher better or is it the opposite?

def rmsle_cv(model):
    kf = KFold(n_folds, random_state=42).get_n_splits(train)
    rmse= np.sqrt(-cross_val_score(model, train, y_train, scoring="neg_mean_squared_error", cv = kf))

def rmsle(y, y_pred):
    return np.sqrt(mean_squared_error(y, y_pred))

In my case, I am getting these results

Lasso score(cv): 0.1176 (0.0068)
ElasticNet score(cv): 0.1177 (0.0068)
Ridge(01): 0.1270 (0.0097)
Gradient Boosting score(cv): 0.1443 (0.0109)
BayRidge(01): 0.1239 (0.0079)
Kernel Ridge score(cv): 0.1210 (0.0068)
Xgboost score(cv): 0.1177 (0.0060)
LGBM score(cv): 0.1158 (0.0064)
Answered By: Johnny

It’s a semantics kind of issue. When you use the term "loss" it is heavily implied that you want to minimize it. However when you talk about "score" it’s implied that you want to maximize it (e.g. accuracy of a classifier). Therefore, when scoring a regression model, a maximal score would mean a minimal loss, i.e. maximize the negated loss.

Answered By: Adrian