similar to: mgcv::predict.gam lpmatrix for prediction outside of R

The documentation for predict.gam in library mgcv gives an example of using an "lpmatrix" to do approximate prediction via interpolation. However, the code is specific to the example wrt the number of smooth terms, df's for each,etc. (which is entirely appropriate for an example) Has anyone generalized this to directly generate code from a gam object (eg SAS or C code)? I wanted to

Dear R-Users, Dr. Wood replied to a similar topic before where confidence intervals were for a ratio of two treatments ( https://stat.ethz.ch/pipermail/r-help/2011-June/282190.html). But my question is more complicated than that one. In my case, log(E(y)) = s(x) where y is a smooth function of x. What I want is the confidence interval of a ratio of log[(E(y2))/E(y1)] given two fixed x values of

Dear Simon, Thank you for your quick reply! I used to perform the GAMsetup in the following manner: GAMsetup sintax: x.summer: vector used for construct the spline knots<-14 N<-length(x.summer) x<-array(x.summer,dim=c(1,N)) G<-list(m=1,n=N,nsdf=0,df=knots+1,dim=1,s.type=0,by=0,by.exists=FALSE,p.order=0,x=x,n.knots=knots,fit.method="mgcv") H<-GAMsetup(G) with the

Hello, Is it possible to estimate prediction interval using GAM? I looked through ?gam, ?predict.gam etc and the mgcv.pdf Simon Wood. I found it can calculate confidence interval but not clear if I can get it to calculate prediction interval. I read "Inference for GAMs is difficult and somewhat contentious." in Kuhnert and Venable An Introduction to R, and wondering why and if that

Sometimes, for specific models, I get this error from predict.gam in library mgcv: Error in complete.cases(object) : negative length vectors are not allowed Here's an example: model.calibrate <- gam(meansalesw ~ s(tscore,bs="cs",k=4), data=toplot, weights=weight, gam.method="perf.magic") > test <- predict(model.calibrate,newdata) Error in

I fitted a GAM model with Poisson distribution using the function gam() in the mgcv package. My model is of the form: mod<-gam(y~s(x0)+s(x1)+s(x2),family=poisson). To extract estimates at a specified set of covariate values I used the gam `predict' method. But I want to get estimate and standard error of the difference of two fitted values. Can someone explain what should I do? Thank

hi I'm puzzled as to the relation between the REML score computed by gam and the formula (4) on p.4 here: http://opus.bath.ac.uk/22707/1/Wood_JRSSB_2011_73_1_3.pdf I'm ok with this for poisson, or for quasipoisson when phi=1. However, when phi differs from 1, I'm stuck. #simulate some data library(mgcv) set.seed(1) x1<-runif(500) x2<-rnorm(500)

R-help, Sorry for posting again the same question (dated 26-03-2007) but all my mails have been sent to the recycle bin without possibility of recovering and thus I don't know if anyone has answer my query. Here is the original message: I'm applying a gam model (package mgcv) to predict relative abundances of a fish species. The covariates are year, month, vessel and statistical

Hi all, I am using GAM to model time trends in a logistic regression. Yet I would like to extract the the fitted spline from it to add it to another model, that cannot be fitted in GAM or GAMM. Thus I have 2 questions: 1) How can I fit a smoother over time so that I force one knot to be at a particular location while letting the model to find the other knots? 2) how can I extract the matrix

Hello I'm using function gam from package mgcv to fit splines. ?When I try to make a prediction slightly beyond the original 'x' range, I get this error: > A = runif(50,1,149) > B = sqrt(A) + rnorm(50) > range(A) [1] 3.289136 145.342961 > > > fit1 = gam(B ~ s(A, bs="ps"), outer.ok=TRUE) > predict(fit1, newdata=data.frame(A=149.9), outer.ok=TRUE) Error

In Dr. Wood's book on GAM, he suggests in section 4.1.6 that it might be useful to shrink a single smooth by adding S=S+epsilon*I to the penalty matrix S. The context was the need to be able to shrink the term to zero if appropriate. I'd like to do this in order to shrink the coefficients towards zero (irrespective of the penalty for "wiggliness") - but not necessarily all the

Dear all, I run the GAMs (generalized additive models) in gam(mgcv) using the following codes. m.gam <-gam(mark~s(x)+s(y)+s(lstday2004)+s(ndvi2004)+s(slope)+s(elevation)+disbinary,family=binomial(logit),data=point) And two repeated warnings appeared. Warnings$B!'(B 1: In gam.fit(G, family = G$family, control = control, gamma = gamma, ... : Algorithm did not converge 2: In gam.fit(G,

I would like to fit a logistic regression using a smothing spline, where the spline is a piecewise cubic polynomial. Is the knots option used to define the subintervals for each piece of the cubic spline? If yes and there are k knots, then why does the coefficients field in the returned object from gam only list k coefficients? Shouldn't there be 4k -4 coefficients? Sincerely, Bill

Please help! Im trying to run a GAM: model3=gam(data2$Symptoms~as.factor(data2$txerad)+s(data2$maritalStatus),family=binomial,data=data2) But keep getting this error: Error in dl[[i]] : subscript out of bounds Can someone please tell me what this error is? Thanks -- View this message in context: http://r.789695.n4.nabble.com/Help-with-GAM-mgcv-tp3074165p3074165.html Sent from the R help

Dear listers, I'm using gam(from mgcv) for semi-parametric regression on small and noisy datasets(10 to 200 observations), and facing a problem of overfitting. According to the book(Simon N. Wood / Generalized Additive Models: An Introduction with R), it is suggested to avoid overfitting by inflating the effective degrees of freedom in GCV evaluation with increased "gamma"

Dear list, does anybody know whether there is a way to easily retrieve the so called "partial residuals" of a gam fit with package mgcv? The partial residuals are the residuals you would get if you would "leave out" a particular predictor and are the dots in the plots created by plot(gam.object,residuals=TRUE) residuals.gam() gives me whole model residuals and

For summary(GAM) in the mgcv package smooth the degrees of freedom for the F value for test of smooth terms are the rank of covariance matrix of \hat{beta} and the residuals df. I've noticed that in a lot of GAMs I've fit the rank of the covariance turns out to be 9. In Simon Wood's book, the rank of covariance matrix is usually either 9 or 99 (pages 239-230 and 259). Can anyone

Hi R-users, I'm using R 2.11.1, mgcv 1.6-2 to fit a generalized additive mixed model. I'm new to this package...and just got more and more problems... 1. Can I include correlation and/or random effect into gam( ) also? or only gamm( ) could be used? 2. I want to estimate the smoothing function s(x) under each level of treatment. i.e. different s(x) in each level of treatment. shall I

Dear all, I'm trying to fit a pure additive model of the following formula : fit <- gam(y~x1+te(x2, x3, bs="cr")) ,with the smoothing parameter estimation method "magic"(default). Regarding this, I have two questions : Question 1 : In some cases the value of "mgcv.conv$fully.converged" becomes "FALSE", which tells me that the method stopped with a

Dear all, Following the help from gam(mgcv) help page, i tried to analyze my dataset with all the default arguments. Unfortunately, it can't be run successfully. I list the errors below. #m.gam<-gam(mark~s(x,y)+s(lstday2004)+s(slope)+s(ndvi2004)+s(elevation)+s(disbinary),family=binomial(logit),data=point)