I had never expected that this answer could eventually be so long when I posted my initial answer 2 years ago. Tss <- sum((actual - mean(actual)) ^ 2) # total sum of squaresīut there is another: regss <- sum((preds - mean(preds)) ^ 2) # regression sum of squaresĪlso, your formula can give a negative value (the proper value should be 1 as mentioned above in the Warning section). Rss <- sum((preds - actual) ^ 2) # residual sum of squares That is false, since the partition of sum of squares does not hold and you can't compute R squared in a consistent way.Īs you demonstrated, this is just one way for computing R squared: preds <- c(1, 2, 3) It looks like you skip this regression step and go straight to the sum of square computation. Given two vectors x and y, we first fit a regression line y ~ x then compute regression sum of squares and total sum of squares. You probably misunderstood the procedure. Thanks for your comments 1, 2 and your answer of details. But, do you really think that the preds is a good prediction on actual? Yes of course, one is just a linear rescaling of the other so they have a perfect linear relationship. The R squared between those two vectors is 1. Here is very extreme example: preds <- 1:4/4 Some people did this, but I don't agree with it. If you split your data into training and testing parts and fit a regression model on the training one, you can get a valid R squared value on training part, but you can't legitimately compute an R squared on the test part. But there is no justification that it can measure the goodness of out-of-sample prediction. R squared can be a (but not the best) measure of "goodness of fit". But only when such an estimate is statistically justified. The R squared is reported by summary functions associated with regression functions. R - Calculate Test MSE given a trained model from a training set and a test set.So it is a weak or even useless measure on "goodness of prediction". Think twice!! R squared between x + a and y + b are identical for any constant shift a and b. R squared between two arbitrary vectors x and y (of the same length) is just a goodness measure of their linear relationship. Lemma 1: a regression y ~ x is equivalent to y - mean(y) ~ x - mean(x) Sandipan's answer will return you exactly the same result (see the following proof), but as it stands it appears more readable (due to the evident $r.squared).īasically we fit a linear regression of y over x, and compute the ratio of regression sum of squares to total sum of squares. So you can define you function as: rsq <- function (x, y) cor(x, y) ^ 2 R squared between two vectors is just the square of their correlation. You need a little statistical knowledge to see this.
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