similar to: predicting from coxph with pspline

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:

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 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

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

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 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),

Dear all, I'm trying to use the Pspline add-on package to fit a quintic spline (norder =3), but I keep running into a Singularity error. > traj.spl <- smooth.Pspline(time, x, norder=3 ) Error in smooth.Pspline(time, x, norder = 3) : Singularity error in solving equations > Playing around with the other parameters produces an "unused arguments" error: > traj.spl

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

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 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

Full_Name: Suharto Anggono Version: 2.8.1 OS: Windows Submission from: (NULL) (125.165.84.118) PR#14076 inspired me to write this. > t1 <- as.POSIXct("1970-01-01 00:00:00", tz="GMT") > t2 <- as.POSIXlt("1970-01-01 00:00:00", tz="GMT") > t1 - t2 Time difference of 7 hours Above, t1 and t2 represent the same time in the same specified

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:

Hi, I would like to merge two data frames. It is just that I want the merging to be done with some kind of a fuzzy criterion. Let me explain. My first data frame looks like this : ID1 time1 dt 1 2008-01-02 13:11 10 2 2008-01-02 14:20 20 3

Further pursuing my curiosity to measure the efficiency of R/C++ interface, I conducted a simple matrix-vector multiplication test using .C and .Call functions in R. In each case, I measured the execution time in R, as well as inside the C++ function. Subtracting the two, I came up with a measure of overhead associated with each call. I assume that this overhead would be non-existent of the entire

Hi r users, I have two data sets (X1, X2). For example, time1<-c( 0, 8, 15, 22, 43, 64, 85, 106, 127, 148, 169, 190 ,211 ) outpue1<-c(171 ,164 ,150 ,141 ,109 , 73 , 47 ,26 ,15 ,12 ,6 ,2 ,1 ) X1<-cbind(time1,outpue1) time2<-c( 0 ,8 ,15 , 22 ,43 , 64 ,85 ,106 ,148) output2<-c( 5 ,5 ,4 ,5 ,5 ,4 ,1 ,2 , 1 ) X2<-cbind(time2,output2) I want to

Greetings listeRs - Given a data frame such as times time1 time2 time3 time4 1 70.408543 48.92378 7.399605 95.93050 2 17.231940 27.48530 82.962916 10.20619 3 20.279220 10.33575 66.209290 30.71846 4 NA 53.31993 12.398237 35.65782 5 9.295965 NA 48.929201 NA 6 63.966518 42.16304 1.777342 NA one can use "transform" to

Dear All,   #I have the following code   Dose<-1000 Tinf <-0.5 INTERVAL <-8 TIME8 <-matrix(c((0*INTERVAL):(1*INTERVAL))) TIME7 <-matrix(c((0*INTERVAL):(2*INTERVAL))) TIME6 <-matrix(c((0*INTERVAL):(3*INTERVAL))) TIME5 <-matrix(c((0*INTERVAL):(4*INTERVAL))) TIME4 <-matrix(c((0*INTERVAL):(5*INTERVAL))) TIME3 <-matrix(c((0*INTERVAL):(6*INTERVAL))) TIME2

To Whomever is Interested, I have spent several days searching the web, help files, the R wiki and the archives of this mailing list for a solution to this problem, but nonetheless I apologize in advance if I have missed something obvious. The problem is this; I have a 5-column data frame with about 4.2 million rows, and want to create a new (and hopefully much smaller) data frame that

Hello all, I've 2 strings that representing the start and end values of a date and time. For example, time1 <- c("21/04/2005","23/05/2005","11/04/2005") time2 <- c("15/07/2009", "03/06/2008", "15/10/2005") as.difftime(time1,time2) Time differences in secs [1] NA NA NA attr(,"tzone") [1] "" How can i

Dear all, I would like to have all unique combinations in the following matrix TimeIndex<- rbind (c(1,"Week_of_21_07-29_03"),           c(2,"Thursday_21_03"),           c(3,"Friday_22_03"),           c(4,"Saturday_23_03"),           c(5,"Sunday_24_03"),           c(6,"Monday_25_03"),           c(7,"Tuesday_26_03"),