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

cross-validationpythonregressionscikit-learn

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.

Best Answer

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.

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I dont have the reputation to comment but I want to provide this link for you and/or a passersby where the negative output of the MSE in scikit learn is discussed - https://github.com/scikit-learn/scikit-learn/issues/2439

In addition (to make this a real answer) your first option is correct in that not only is MSE the metric you want to use to compare models but R^2 cannot be calculated depending (I think) on the type of cross-val you are using.

If you choose MSE as a scorer, it outputs a list of errors which you can then take the mean of, like so:

# Doing linear regression with leave one out cross val

from sklearn import cross_validation, linear_model
import numpy as np

# Including this to remind you that it is necessary to use numpy arrays rather 
# than lists otherwise you will get an error
X_digits = np.array(x)
Y_digits = np.array(y)

loo = cross_validation.LeaveOneOut(len(Y_digits))

regr = linear_model.LinearRegression()

scores = cross_validation.cross_val_score(regr, X_digits, Y_digits, scoring='mean_squared_error', cv=loo,)

# This will print the mean of the list of errors that were output and 
# provide your metric for evaluation
print scores.mean()

Python – K-fold cross validation implementation python

The reason why your validation score is low is subtle.

The issue is how you have partitioned the dataset. Remember, when doing cross-validation you should randomly split the dataset. It is the randomness that you are missing.

Your data is loaded category by category, which means in your input dataset, class labels and examples follow one after the other. By not doing the random split, you have completely removed a class which your model never sees during the training phase and hence you get a bad result on your test/validation phase.

You can solve this by doing a random shuffle. So, do this:

from sklearn.utils import shuffle    

processor = Data_Preprocessor()
td, tl = processor.raw_to_vector(path="C:/Users/Pankaj/Downloads/ng/")
vectorizer = CountVectorizer(stop_words='english', lowercase=True, min_df=2, analyzer="word")
data = vectorizer.fit_transform(td)
# Shuffle the data and labels
data, tl = shuffle(data, tl, random_state=0)
clfMNB = MultinomialNB(alpha=.0001)
score = Cross_Validation.Cross_Validation(clfMNB, 10, data, tl)

print("Train score" + str(score[0]))
print("Test score" + str(score[1]))

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