Using R Squared with Mean Very Close to 0 I am using $R^2$ to evaluate the quality of fitted model. The mean of the Y values in my model is very close to $0$, around $0.003$. This makes the denominator of the $R^2$ calculation very small, less than $1$, so I often get massively negative $R^2$ values, think $-2000$. Is there a way I can adjust $R^2$ so it is less biased by the very small denominator?
 A: $R^2 = 1 - \dfrac{\sum_{i=1}^N (y_i - \hat{y_i})^2}{\sum_{i=1}^N (y_i - \bar{y})^2}$
$y_i$ is the $i^{th}$ observation of $y$.
$\hat{y_i}$ is predicted value of the the $i^{th}$ observation of $y$.
$\bar{y}$ is the mean value of all $y_i$.
$N$ is the number of observationf of $y$.
You're saying that $\bar{y}$ is tiny and that makes your denominator small. A small denominator could, conceivable, cause some issues when you divide on a computer, but short of that, a small denominator means an even smaller numerator, since the model will fit the data better than always guessing the average value of the response variable.$^{\dagger}$. Consequently, your fraction will still be in $(0,1)$, meaning that your $R^2$ is in $(0,1)$ like $R^2$ is supposed to be.
$^{\dagger}$I am assuming you're doing a linear regression, OLS-style model. If you're not, there might be bigger issues to discuss about using $R^2$. You also can wreck this idea if you do OLS without an intercept, but, almost always, that is a poor strategy.
EDIT
In R, $R^2$ can be calculated like this.
R2 <- 1 - (sum((y-predict(model))^2))/(sum((y-mean(y))^2))

