Predicted R-Squared (r2, r^2) Calculation in `python`

stats.py

def press_statistic(y_true, y_pred, xs):
    """
    Calculation of the `Press Statistics <https://www.otexts.org/1580>`_
    """
    res = y_pred - y_true
    hat = xs.dot(np.linalg.pinv(xs))
    den = (1 - np.diagonal(hat))
    sqr = np.square(res/den)
    return sqr.sum()

def predicted_r2(y_true, y_pred, xs):
    """
    Calculation of the `Predicted R-squared <https://rpubs.com/RatherBit/102428>`_
    """
    press = press_statistic(y_true=y_true,
                            y_pred=y_pred,
                            xs=xs
    )

    sst  = np.square( y_true - y_true.mean() ).sum()
    return 1 - press / sst
 
def r2(y_true, y_pred):
    """
    Calculation of the unadjusted r-squared, goodness of fit metric
    """
    sse  = np.square( y_pred - y_true ).sum()
    sst  = np.square( y_true - y_true.mean() ).sum()
    return 1 - sse/sst

However, you should make it clear that your code works only for models with an intercept. If you force the intercept to 0 (no intercept), your code will not work due to incorrect SST calculation:

You calculated the centered total sum of squares (SST):

sst  = np.square( y_true - y_true.mean() ).sum()

But the original R implementation calculates the uncentered total sum of squares:

+     #' Use anova() to get the sum of squares for the linear model
+     lm.anova <- anova(linear.model)
+     #' Calculate the total sum of squares
+     sst <- sum(lm.anova$'Sum Sq')

If you use the sample data in my comment above:

• Your sst (centered) = 640.9
• Original R implementation sst (uncentered) = 5437