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Estimados miembros de la lista, Estoy haciendo una análisis de supervivencia con R. Adjunto mis datos. Quiero analizar la supervivencia de 5 grupos diferentes y compararla. Para ello estoy utilizando el paquete survival. > s = Surv(c$tiempo, c$estado) > f = survfit(s ~ tratamiento, data = c) > d = survdiff(s ~ tratamiento, data = c) > d Call: survdiff(formula = s ~ tratamiento,

Hi all, I was trying to convert a data frame to a zoo object so I can use some time series functions like lag(). But it seems then everything became a factor, so I have to convert it back to numeric to run the correct regressions. Is there a way to avoid it? Here is an example: ############################# a <- data.frame(nn =as.character(c("a", "b", "c",

Dear R users, I have a response variable in a csv file called "y" and a matrix of predictor variables in a csv file called "mat". I have used the function "nls" I have specified the nonlinear relation between these variable.The code I have witten is called Rprog which begins with the phrase: L.minor.m1<-nls(Y~a ....etc.. The program when I execute the program, I

As I am reading ?arima, only NA entries in the argument fixed= imports. The following seems to indicate otherwise: x <- arima.sim(model=list(ar=0.8), n=100) + (1:100)/50 > t <- 1:100 > mod1 <- lm(x ~ t) > > init1 <- c(0, coef(mod1)[2]) > fixed1 <- c(as.numeric(NA), 0) > > arima(x, order=c(1,0,0), xreg=t, include.mean=FALSE, init=init1, fixed=fixed1)

I am trying to recreate an analysis that has been done by another group (in SAS I believe). I'm stuck on one part, I think because my stats knowledge is lacking, and while it's OT, I'm hoping someone here can help. Given this dataframe; foo*<-*structure(list(OBS = structure(1:18, .Label = c("1", "2", "3", "4", "5",

Hola, Manuel, No entiendo tu pregunta (la repito aqui para que sea mas explicito): hay alguna forma de comparar la probabilidad de supervivencias (en este caso anual) entre grupos sin utilizar un chi-cuadrado y un valor de P. Entiendo que lo que hace survdiff es comparar las curvas de supervivencia, pero yo quiero comparar la probabilidad de supervivencia entre grupos al final del estudio. Con

Hi Experts, I am new to R, and was wondering how to do 3D linear regression in R. In other words, I need to Fit a 3-Dimensional Line to Data Points (input). I googled before posting this, and found that it is possible in Matlab and other commercial packages. For example, see the Matlab link:

Target: FreeBSD 6.2-STABLE on i386 autoconf says: configure:2122: checking build system type configure:2140: result: i386-unknown-freebsd6.2 [...] configure:2721: gcc -v >&5 Using built-in specs. Configured with: FreeBSD/i386 system compiler Thread model: posix gcc version 3.4.6 [FreeBSD] 20060305 [...] objdir != srcdir, for both llvm and gcc. Release build. llvm-gcc 4.2 from source.

LLVMers, The 2.2 prerelease is now available for testing: http://llvm.org/prereleases/2.2/ If anyone can help test this release, I ask that you do the following: 1) Build llvm and llvm-gcc (or use a binary). You may build release (default) or debug. You may pick llvm-gcc-4.0, llvm-gcc-4.2, or both. 2) Run 'make check'. 3) In llvm-test, run 'make TEST=nightly report'. 4) When