search for: ord2

...me=data1 ## Calculate Cox's regression coefficients' standard error and bias using the ## ordinary Bootstrap # set.seed(1234)#set the random start cox.ord1 = censboot(data=dataname,statistic=beta.fun1,R=r,F.surv=model1.surv, cox=model1.cox,sim="ordinary") cox.ord1 cox.ord2 = censboot(data=dataname,statistic=beta.fun2,R=r,F.surv=model2.surv, cox=model2.cox,sim="ordinary") cox.ord2 join=list(model1, model2) #return(join) #} #for (i in 1:2){ #result[i]=mymodels} I run the above code but for the following part: cox.ord2 = censboot(data=d...

Anybody else having problems accessing this site? Haven''t had much luck over the past couple of days. -- Posted via http://www.ruby-forum.com/.

...t with no ties > y <- sample(500000) > system.time(ord1 <- order(y)) [1] 1.18 0.00 1.18 0.00 0.00 it is very fast! This gave me the following idea: Since I don't care about keeping the order within tied values, why not add some small disturbance to x, and indeed, > unix.time(ord2 <- order(x + runif(length(x), -0.1, 0.1))) [1] 1.66 0.00 1.66 0.00 0.00 > identical(x[ord], x[ord2]) [1] TRUE it works! Is there an obvious (=better) solution to this problem that I have overlooked? In any case, I think that the problem with order and many ties is worth mentioning in th...

...imple linear model (lm), I don't have this problem, and the line is smooth: lm.1<-lm(LnFlux~Temp, data=resp) fit.2<-exp(fitted(lm.1)) plot(Flux$Temp, Flux$Flux, xlab="Temperature", ylab=expression("CO"[2]*"Flux"), xlim=c(-10, 25), ylim=c(0,250), pch=16) ord2<-order(Flux$Temp) lines(Flux$Temp[ord2], fit.2[ord2], lty=1, lwd=2) The only difference I can find between the two is the structure of the fitted objects. The fit.1 object from lme is "atomic", and lacks individual data labels. Instead, the labels are: attr(*, "label")...

Suppose one has x <- c(1, 2, 7, 9, 14) y <- c(71, 72, 77) How would one write an R function which alternates between elements of one vector and the next? In other words, one wants z <- c(x[1], y[1], x[2], y[2], x[3], y[3], x[4], y[4], x[5], y[5]) I couldn't think of a clever and general way to write this. I am aware of gdata::interleave() but it deals

Dear R user, I am a biochemist/bioinformatician, at the moment working on protein clusterings by conformation similarity. I only started seriously working with R about a couple of months ago. I have been able so far to read my way through tutorials and set-up my hierarchical clusterings. My problem is that I cannot find a way to obtain information on the rooting of specific nodes, i.e. of

...et [64.125.27.149] 13 276 ms 280 ms 283 ms so-0-1-0.mpr1.dca2.us.above.net [64.125.27.57] 14 269 ms 264 ms 260 ms so-0-1-0.mpr1.lga5.us.above.net [64.125.26.98] 15 282 ms 291 ms 294 ms xe-0-3-0.cr1.lga5.us.above.net [64.125.29.49] 16 323 ms 341 ms 295 ms xe-0-2-0.cr1.ord2.us.above.net [64.125.27.169] 17 307 ms 292 ms 293 ms xe-1-1-0.er1.ord7.above.net [64.125.26.250] 18 314 ms 308 ms 314 ms 64.124.65.218.allocated.above.net [64.124.65.218] 19 321 ms 315 ms 438 ms core1-ed2-edge3.ord1.rackspace.net [173.203.0.109] 20 310 ms 302 ms 294 m...

...at mayin.org Cc: r-help <R-help at stat.math.ethz.ch> Message-ID: <8d5a36350603070538k42884d53k8c4ff23242c1d51f at mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1 For a general solution without warnings try interleave <- function(v1,v2) { ord1 <- 2*(1:length(v1))-1 ord2 <- 2*(1:length(v2)) c(v1,v2)[order(c(ord1,ord2))] } interleave(rep(1,5),rep(3,8)) > -----Original Message----- > From: r-help-bounces at stat.math.ethz.ch > [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Gabor > Grothendieck > Sent: Monday, March 06, 2006 12:12 AM &gt...