similar to: Pspline smoothing

To whom it may concern, I don't know whether this is really a bug with the Pspline package or only a problem with my installation. Things work fine in Linux but not in Mac OS X (Darwin). Both system run the latest public versions of R and Pspline. predict.smooth.Pspline produces only NaN instead of predicted values when norder>2: > library (Pspline) > tt <- seq

I have a simple question that irritatingly I haven't been able to figure out on my own. It seems that some functions from the "Pspline" package are successfully installed while others are not. The code with which I'm working is more complicated, but the following highlights my problem. If I run the following code > tt <- seq (0,1,length=20) > xt <- tt^3 > fit

Hey all, I seem to be having trouble fitting a spline to a large set of data using PSpline. It seems to work fine for a data set of size n=4476, but not for anything larger (say, n=4477). For example: THIS WORKS: ----------------------------- random = array(0,c(4476,2)) random[,1] = runif(4476,0,1) random[,2] = runif(4476,0,1) random = random[order(random[,1]),] plot(random[,1],random[,2])

Hi All, I want to fit a piecewise spline of degree 1, i.e. a spline consisting of a straight line over each piece. I downloaded the R package pspline, then I have following questions: 1. in the program, the degree of the spline is specified by 2*norder-1. Why do they adopt such scheme that we can only fit a spline with odd degree? 2. norder cannot be set to 1. Is there any specific reason

Hello, I am using R and the smooth.Pspline function in the pspline package to smooth some data by using natural cubic splines. After fitting a sufficiently smooth spline using the following call: (ps=smooth.Pspline(x,y,norder=2,spar=0.8,method=1) [the values of x are age in years from 1 to 100] I tried to check that R in fact had fitted a natural cubic spline by checking that the resulting

I'm a Beginner on R-language and I' have tried to use some packages but the following message have appeared: "Error in .Fortran("pspline", as.integer(n), as.integer(nvar), as.integer(norder), : C/Fortran function name not in load table" I'm using a Windows version and I'd like to know if it is a installation problem or problems with paths? Thanks,

Dear R users: I am not entirely convinced that clogit gives me the correct result when I use pspline() and maybe you could help correct me here. When I add a constant to my covariate I expect only the intercept to change, but not the coefficients. This is true (in clogit) when I assume a linear in the logit model, but the same does not happen when I use pspline(). If I did something similar

Hi there, I have a question on how to extract the linear term in the penalized spline in coxph. Here is a sample code: n=100 set.seed(1) x=runif(100) f1 = cos(2*pi*x) hazard = exp(f1) T = 0 for (i in 1:100) { T[i] = rexp(1,hazard[i]) } C = runif(n)*4 cen = T<=C y = T*(cen) + C*(1-cen) data.tr=cbind(y,cen,x) fit=coxph(Surv(data.tr[,1],

Hi All, I am trying to figure out how to get the position of the knots in a pspline used in a cox model. my.model = coxph(Surv(agein, ageout, status) ~ pspline(x), mydata) # x being continuous How do I find out where the knot of the spline are? I would like to know to figure out how many cases are there between each knot. Best, Federico -- Federico C. F. Calboli Department of Epidemiology

Hello. I get a bit confused by the output from the predict function when used on an object from coxph in combination with p-spline, e.g. fit <- coxph(Surv(time1, time2, status)~pspline(x), Data) predict(fit, newdata=data.frame(x=1:2)) It seems like the output is somewhat independent of the x-values to predict at. For example x=1:2 gives the same result as x=21:22. Does the result span the

help Hello, I?m a statistic student in Austria and I have to do a survival analysis in R by using psplines as regressor. My problem is that I sometimes (I think it depends on the choose of the parameters) get a error message, but I do not know what it means. After that I tried the procedure with an example dataset R is providing. Although using the cancer dataset I also get this message. Input:

Hello, coxph does not use any information that are in the dataset between event times (or "death times") , since computation only occurs at event times. For instance, removing observations when there is no event at that time in the whole dataset does not change the results: > set.seed(1) > data <- as.data.frame(cbind(start=c(1:5,1:5,1:4),stop=c(2:6,2:6,2:5),status=c(rep(

Hello, I'm using R code that includes a residual permutation that was written as a supplement to the paper: Turner et al. 2010. A general hypothesis-testing framework for stable isotopes ratios in ecological studies. Ecology 91:2227-2233. The supplemental code is available at: http://www.esapubs.org/archive/ecol/E091/157/suppl-1.htm When I execute the function, no warnings are given

Hi the list, I am writing a package in S4 and I do not manage to understand a bug. The "R CMD check" and the "R CMD build" both work. Here is links to the package (not on CRAN yet for the raison that I explain bellow): http://christophe.genolini.free.fr/aTelecharger/kml_0.5.zip http://christophe.genolini.free.fr/aTelecharger/kml_0.5.tar.gz Then I install the package and I

Hi the list, One of my package has a memory issue that I do not manage to understand. The Memtest notes is here: <http://www.stats.ox.ac.uk/pub/bdr/memtests/valgrind/kml-Ex.Rout> Here is the message that I get from Memtest --- 8< ---------------- ~ Fast KmL ~ ==27283== Invalid read of size 4 ==27283== at 0x10C5DF28: kml1 (kml.c:183) ... ==27283== by 0x10C5DE4F: kml1 (kml.c:151)

Hi, I'm trying to using pspline in bic.surv{BMA}. ############################# library(BMA) library(survival) data(veteran) test.bic.surv<- bic.surv(Surv(time,status) ~ karno+pspline(age,df=3)+diagtime+prior, data = veteran, factor.type = TRUE) summary(test.bic.surv, conditional=FALSE, digits=2) ############################# The results are:

Hi the list, I am defining S4 objet. Is it possbile to define a method that change the slot of an object without using <- ? My object contain a numeric and a matrix. At some point, I would like to impute the missing value in the matrix. So I would like to use something like : ----------------- setClass("MyObj",representation(time="numeric",traj="matrix")) a

Hi, I would like like to compute first and second derivative from numerical data. I hoped I could compute a spline object from the data (that works), and then compute the derivative from that spline objects, but I couldn't find anything like that. Does somebody know how to get numerical derivatives? Oliver -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Hi, I'm using a Cox-Regression to estimate hazard rates on prepayments. I'm using the "pspline" function to face non-linearity, but I have no clue how to interpret the result. Unfortunately I did not find enough information on the "pspline" function wether in the survival package nor using google.. I got following output: * library(survival)* > > > >

Hi everybody. I'm trying to fit a weibull survival model with a spline basis for the predictor, using the survival library. I've noticed that it doesn't seem to be possible to use the aic method to choose the degrees of freedom for the spline basis in a parametric regression (although it's fine with the cox model, or if the degrees of freedom are specified directly by the user),