I am using statsmodels.api.OLS to fit a linear regression model with 4 input-features.

The shape of the data is:

X_train.shape, y_train.shape  

Out[]: ((350, 4), (350,))

Then I fit the model and compute the r-squared value in 3 different ways:

import statsmodels.api as sm
import sklearn

ols = sm.OLS(y_train, X_train).fit()

y_pred = ols.predict(X_train)
res = y_train - y_pred

ss_tot = np.sum( (y_train - y_train.mean())**2 )
ss_res = np.sum( (y_train - y_pred)**2 )

(1 - ss_res/ss_tot), sklearn.metrics.r2_score(y_train, y_pred), ols.rsquared

Out[]: (0.91923900248372292, 0.91923900248372292, 0.99795455683297096)

The manually computed r-squared value and the value from sklearn.metrics.r2_score match exactly.
However, the ols.rsquared value seems to be highly over-estimated.

Why is this the case? How does statsmodels compute the rsquared value?


1 Answer 1


This is not technically an error in statsmodels, rather it is because statsmodels.OLS does not add the intercept/constant term to the right-hand-side of the regression equation by default -- you have to explicitly add it. In contrast, sklearn (and the vast majority of other regression programs) add the constant/intercept term by default unless it is explicitly suppressed.

To add the intercept term to statsmodels, use something like:

ols = sm.OLS(y_train, sm.add_constant(X_train)).fit()

The reason that omitting the intercept changes the $R^2$ is that a different definition of $R^2$ is used when there is no intercept.

We can view the usual $R^2$ as the proportional reduction in sum of squared errors between two models, A and B. $$ \text{A:} \space Y_i = \beta_0 + \beta_1X_i + e_i $$ $$ \text{B:} \space Y_i = \beta_0 + e_i $$ In words, we compare the performance of the model that includes $X$ as a predictor vs. a model that just predicts a constant value (the sample mean) for all observations.

When the intercept $\beta_0$ is omitted from model A to form a new model -- call it model C -- it no longer makes sense to compare this to the reduced model B (B is nested in A but it is not nested in C). So instead we adjust the computation of $R^2$ so that it can be viewed as the comparison between C and a new model D $$ \text{C:} \space Y_i = \beta_1X_i + e_i $$ $$ \text{D:} \space Y_i = 0 + e_i $$ In other words, we compare the slope-only model to a model that simply makes a constant prediction of 0 for all observations. This often paradoxically causes the $R^2$ to be even higher than before, but it's just because the reduced reference model D is absurd in most applications.

This and related issues are discussed a bit further in the following threads:

Removal of statistically significant intercept term increases $R^2$ in linear model

When forcing intercept of 0 in linear regression is acceptable/advisable

When is it ok to remove the intercept in a linear regression model?


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