[R] "NaN" answer don't understand why
varin sacha
v@r|n@@ch@ @end|ng |rom y@hoo@|r
Sat Nov 14 21:44:38 CET 2020
Dear Bill,
Many thanks for your response. I got it.
Best,
Le samedi 14 novembre 2020 à 00:17:10 UTC+1, Bill Dunlap <williamwdunlap using gmail.com> a écrit :
fit <- robustgam::robustgam(...) produces a list, with no class attached, so residuals(fit) invokes the default method for residuals(), which essentially returns the 'residuals' component of 'fit'. There is no such component so it returns NULL, an object of length zero. The mean of a length-zero object is NaN.
It would make sense for mean(NULL) or sum(NULL) to give an error since they are only meant to work on numbers. However this would probably break some existing code, since there are various functions that return NULL instead of numeric(0).
-Bill
On Fri, Nov 13, 2020 at 2:05 PM varin sacha via R-help <r-help using r-project.org> wrote:
> Dear R-experts,
>
> Here below my reproducible example. No error message but I can not get a result. I get "NaN" as a result. I don't understand what is going on. Many thanks for your precious help, as usual.
>
>
> # # # # # # # # # # # # # # # # # # # # # # # # #
> x<-c(499,491,500,517,438,495,501,525,516,494,500,453,479,481,505,465,477,520,520,480,477,416,502,503,497,513,492,469,504,482,502,498,463,504,495)
> y<-c(499,496,424,537,480,484,503,575,540,436,486,506,496,481,508,425,501,519,546,507,452,498,471,495,499,522,509,474,502,534,504,466,527,485,525)
>
> library(robustgam)
> true.family <- poisson()
>
> #Robust GAM
> fit=robustgam(x,y,sp=0,family=true.family,smooth.basis='ps',K=3)
>
> #OLS
> fit1 <- lm(y~x)
>
> #Huber-M
> library(robustbase)
> library(MASS)
> fit2=rlm(y~x)
>
> #GAM
> library(mgcv)
> fit3=gam(y~s(x))
>
> # MSE of OLS linear model
> mean(residuals(fit1)^2)
>
> # MSE of Huber-M linear model
> mean(residuals(fit2)^2)
>
> # MSE of GAM
> mean(residuals(fit3)^2)
>
> # MSE of robust GAM
> mean(residuals(fit)^2)
> # # # # # # # # # # # # # # # # # # # # # # # # #
>
>
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