> tstDf <- data.frame(y = 1:11, x=1:11) > fit <- nls(y~a/x, data=tstDf, start=list(a=1)) > predict(fit, se.fit=TRUE) [1] 7.0601879 3.5300939 2.3533960 1.7650470 1.4120376 1.1766980 1.0085983 [8] 0.8825235 0.7844653 0.7060188 0.6418353
The same code in S-Plus 6.1 produces the following:
> predict(fit, se.fit = TRUE)
$fit:
[1] 7.0601876 3.5300938 2.3533959 1.7650469 1.4120375 1.1766979 1.0085982 0.8825234
[9] 0.7844653 0.7060188 0.6418352
$se.fit:
[1] 5.2433042 2.6216521 1.7477681 1.3108261 1.0486608 0.8738840 0.7490435 0.6554130
[9] 0.5825894 0.5243304 0.4766640
$residual.scale: [1] 6.544753
$df: [1] 10
Unfortunately, I'm not in a position to fix the problem, but this toy example might make it easier for someone else to fix it.
spencer graves
p.s. The following command in S-Plus 6.1 seems to work fine but produces an error in R 1.7.1:
nls(y~a, data=tstDf, start=list(a=1))
Error in nlsModel(formula, mf, start) : singular gradient matrix at initial parameter estimates
#############################################################
Enrique Portilla wrote:
Hi, Does anyone know how to compute the confidence prediction intervals for
a nonlinear least squares models (nls)?
I was trying to use the function 'predict' as I usually do for other models fitting (glm, lm, gams...), but it seems that se.fit, and interval computation is not implemented for the nls...
Cheers
Enrique
~~~~~~~~~~~~~~~~~~~~~~~~~~~ Fisheries Research Services, Marine Laboratory, Victoria Road, Torry, Aberdeen, UK. Tel. 44 (0) 1224 295314 ~~~~~~~~~~~~~~~~~~~~~~~~~~~
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