search for: survivorship

Hello, My advisor and I have been working on some survivorship analyses in R and we are hoping to get some feedback on a particular issue involving nesting. We are interested in patterns of food discovery by ant species. Our observations consist of time to discovery by an ant for three different food types, each of two different sizes. These data were collec...

...gt; MIME-version: 1.0 From: Diana Fischer <diana.fischer at yale.edu> To: maechler at stat.math.ethz.ch Subject: R-Project Date: Wed, 25 Oct 2000 11:42:36 -0500 I briefly looked at the R documentation of available statistical procedures. It appears that there are no "lifetable" or survivorship-failure model procedures. Am I correct? Thanks, Diana B. Fischer, Ph.D. Dept. Therapeutic Radiology Yale School of Medicine ------- end of forwarded message ------- -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac...

...11.5 1 4 C1 8 2 1 1 1 There are 81 cages and each 20 individuals whose survival was followed over time. The columns S,F,G,L and S are experimentally manipulated factors thought to have an influence on survival. Using survfit(Surv(deathtime,status)~cage) gives me the survivorship curves for every cage. But what I??d like to have is a mean survivorship value for every cage. Obviously, using tapply (deathtime,cage,mean) gives me mean values, but I??d like to have a better estimate of this using a proper statistical model. I??ve tried a glm with poisson errors (as suggest...

...4024 48.579639 48.111326 [16] 44.994786 65.241722 65.852035 [19] 53.254482 43.217719 30.255150 Reading in some data, the 'mydat' line extends across the console, but the output is wrapped around again? > mydat<-data.frame(ID, Year = factor(Year), Dayrelease, Agerelease, Survivorship, Entry, Exit, Fate, Gender) > mydat ID Year Dayrelease Agerelease 1 16240 1996 205 95 2 16319 1996 205 88 3 16378 1996 248 108 . . . 576 3094 2005 251 136 577 2667 2005 264 861 578 3035 2005 264...

Dear List, Is there an automated way to use the survival package to generate survival rate estimates and their standard errors? To be clear, *not *the survivorship estimates (which are cumulative), but the survival *rate * estimates... Thank you in advance for any help. Best, Brian [[alternative HTML version deleted]]

Hello, I am trying to compute the mdeian of the survival time from the function survfit: > fit <- survfit(Surv(time, status) ~ 1) > fit Call: survfit(formula = Surv(time, status) ~ 1) records n.max n.start events median 0.95LCL 0.95UCL 111 111 111 20 NA NA NA The results is NA? the fit$surv gives values between 1 and 0.749! Am I doing this correct?

...I want to plot female survival, as estimated by my model, by mating status and therefore two survival curves should be plotted. My code results in one survival curve bounded by, what I assume are confidence intervals. My code for the plot is: plot(survfit(F.cox.weight), lty=c(1,3), ylab = "Survivorship", xlab = "Day", bty="L") legend(35,0.9,c("Mated","Virgin"), lty=c(1,3), bty = "n") The plot looks like this: http://r.789695.n4.nabble.com/file/n4325120/Fig_survival_female.jpg Probably a simple bit of code I'm missing! Cheers, -- Vie...

Hi All, I am trying to do a survivorship analysis with library(survival)from a data set that looks like this: I followed a bunch of naturally germinated seedlings of an annual plant from germination to death (none made it to reproduce, and died in a period of ~60 days after germination.) I also know the size of the seed of every individua...

...gerelease + Dayrelease + strata(Year), data=mydat) coxfit.apc<-survfit(apc.coxfit1) coxfit.apc plot(survfit(apc.coxfit1), conf.int=F, log=T, lty=c(1:2), col=c(1:2), xlim=c(205, 800)) #not run--first entry for this example is day 205 for 1996, 259 for 1997 >mydat ID Year Dayrelease Agerelease Survivorship Entry Exit Fate Sex 16240 1996 205 95 164 205 369 1 0 16319 1996 205 88 140 205 345 1 0 16378 1996 248 108 100 248 348 1 0 20383 1996 241 98 204 241 445 1 0 16324 1996 219 90 227 219 446 1 0 16327 1996 219 90 497 219 716 1 0 20373 1996 241 114 413 241 654 1 0 20374 1996 241 111 211 241 452 1 0 1624...

...o work correctly. Here is my problem. I have three ages: Nage<-c(1,2,3) I have an weight matrix: Wt<-c( 0.04952867, 0.23808432, 0.34263880) I have an age schedule of maturity: Mat<-c(0,1,1) where 0 is not mature, and 1 is mature I have a vulnerability schedule: Vul<-c(0,1,1) I have an survivorship schedule: Survship<-c(1,0.4,0.16) I also have leading parameters R0<-130.66; recK<-3.068; a<-5.48; b<-0.0282; S<-0.4 I have annual catches for 100 years, ct<-runif(100,5,20) Now I want a matrix of 100 years x 3 ages yr<-c(1:100) Nt<-matrix(0,nrow=length(yr)+1),ncol=lengt...

Hello, I am using the clogit (conditional logistic regression) on a simple data set which is not related to survivorship, which I understand to be fine. I have trouble understanding the output. I would like to find parameter estimates of the logistic models I am constructing with R, and am unclear as to what the coefficients represent that are provided. The below is a copy of my output. > summary(clogit(Pl...

...0.524 -0.96 0.34 Likelihood ratio test=0.92 on 1 df, p=0.337 n= 19 > survfit(brettest1)$surv [1] 0.9502716 0.9010356 0.8508737 0.7996929 0.7473818 0.6957615 0.5872297 0.5324263 0.4756545 [10] 0.4201535 0.3547775 0.2916481 0.2189030 0.1374015 >mydat ID Year Dayrelease Agerelease Survivorship Entry Exit Fate Sex 1 16240 1996 205 95 164 205 369 1 0 2 16319 1996 205 88 140 205 345 1 0 3 16378 1996 248 108 100 248 348 1 0 4 20383 1996 241 98 204 241 445 1 0...

Hello All, Learning to use the odfWeave package. I really like the package. It has good documentation, makes some very nice looking tables, and seems to have lots of options for customizing output. There are a few things I'd like to do that don't seem to be covered in the documentation though. So I'm not sure if they're possible or not. Here's a list of some things I'd

Dear R members, I am new in using frailty models in survival analyses and am getting some contrasting results when I compare the Wald and likelihood ratio tests provided by the r output. I am testing the survivorship of different sunflower interspecific crosses using cytoplasm (Cyt), Pollen and the interaction Cyt*Pollen as fixed effects, and sub-block as a random effect. I stratified the analysis by developmental stage (G_stageSM) as an ordered factor (two classes). There is a lot of tied deaths in t...

Hello all R-helpers, I've performed an experiment to test for differential effects of elevated temperatures on three different groups of corals. I'm currently performing a cox proportional hazards regression with censoring on the survivorship (days to mortality) of each individual in the experiment with two factors: Temperature Treatment (2 levels: ambient and elevated) and experimental group (3 levels: say 1,2,3). In my experiment, all three groups survived equally well in the ambient control treatment, but two of three of the gro...

dear all, I want to plot a kaplan Meier plot with the following functions, but I fail to produce the plot I want: library(survival) tim <- (1:50)/6 ind <- runif(50) ind[ind > 0.5] <- 1; ind[ind < 0.5] <- 0; MS <- runif(50) pred <- vector() pred[MS < 0.3] <- 0; pred[MS >= 0.3] <- 1 df <- as.data.frame(cbind(MS, tim, pred, ind)) names(df) <-