How to obtain RMSE out of lm result?

问题:

I know there is a small difference between $sigma and the concept of root mean squared error. So, i am wondering what is the wasiest way to obtain RMSE out of lm function in R?

res<-lm(randomData$price ~randomData$carat+
                     randomData$cut+randomData$color+
                     randomData$clarity+randomData$depth+
                     randomData$table+randomData$x+
                     randomData$y+randomData$z)

length(coefficients(res))

contains 24 coefficient, and i cannot make my model manually anymore. So, how can i evaluate the RMSE based on coefficients drived of lm?

回答1:

Residual sum of squares:

RSS <- c(crossprod(res$residuals))

Mean squared error:

MSE <- RSS / length(res$residuals)

Root MSE:

RMSE <- sqrt(MSE)

Pearson estimated residual variance (as returned by summary.lm):

sig2 <- RSS / res$df.residual

Statistically, MSE is the maximum likelihood estimator of residual variance, but is biased (downward). The Pearson one is the restricted maximum likelihood estimator of residual variance, which is unbiased.

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Remark

• Given two vectors x and y, c(crossprod(x, y)) is equivalent to sum(x * y) but much faster. c(crossprod(x)) is likewise faster than sum(x ^ 2).
• sum(x) / length(x) is also faster than mean(x).

回答2:

To get the RMSE in one line, with just functions from base, I would use:

sqrt(mean(res$residuals^2))

回答3:

I think the other answers might be incorrect. The MSE of regression is the SSE divided by (n - k - 1), where n is the number of data points and k is the number of model parameters.

Simply taking the mean of the residuals squared (as other answers have suggested) is the equivalent of dividing by n instead of (n - k - 1).

I would calculate RMSE by sqrt(sum(res$residuals^2) / res$df).

The quantity in the denominator res$df gives you the degrees of freedom, which is the same as (n - k - 1). Take a look at this for reference: https://www3.nd.edu/~rwilliam/stats2/l02.pdf