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Predicted R-Squared (r2, r^2) Calculation in `python`
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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
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Okay, I compared your code:
with the original R implementation:
And can confirm the results are the same:
Python:
R:
Thanks.