search for: ovarian

Hi I am trying to understand how to get the validate() function in Design to work with the subset option. I tried this: ovarian.cph=cph(Surv(futime, fustat) ~ age+factor(ecog.ps)+strat(rx), time.inc=1000, x=T, y=T, data=ovarian) validate(ovarian.cph) #fine when no subset is used, but the following two don't work: > validate(ovarian.cph, subset=ovarian$ecog.ps==2) Error in order(c(1L, 1L, 1L, 2L, 1L, 1L, 2L, 2L,...

...a 26 LC:A549 -2.66221 0.71215 Lung 61 RE:A498 -2.89402 0.93287 Renal 9 BR:HS578T -2.94118 1.1217 Breast 34 LC:NCI_H522 -2.94381 0.3859 Lung 66 RE:TK_10 -2.95281 1.26245 Renal 52 OV:NCI_ADR_RES -3.04456 0.17046 Ovarian 57 OV:SK_OV_3 -3.04477 2.15405 Ovarian 53 OV:OVCAR_3 -3.0705 -0.31743 Ovarian 14 CNS:SF_295 -3.09348 -1.00095 CNS 54 OV:OVCAR_4 -3.13137 -0.47497 Ovarian 36 LE:HL_60 -3.16745 -3.16745 Leukemia 38 LE:MOLT_4 -3.20055 -1.72841 Le...

...below works with a standard data set, but not with my data, but I cannot locate the problem. Note that I am using an older package of survival to avoid a problem with the newly renamed function in survival meeting Design. Dieter # First, check standard example to make sure library(Design) data(ovarian) dd = datadist(ovarian) options(datadist="dd") ovarian$rx = as.factor(ovarian$rx) cp = cph(Surv(futime,fustat)~rx,data=ovarian,surv=TRUE,x=TRUE,y=TRUE) summary(cp) # works ok survest(cp,levels(ovarian$rx),times=500) # Small data set, 223 rows, 3650 bytes cc = read.table("http://www....

...3 4.073 11.569 30 Melanoma 1 5.829 11.183 31 Melanoma 1 2.533 10.812 32 Melanoma 1 3.253 11.278 33 Melanoma 1 2.427 10.954 34 Melanoma 1 2.522 10.585 35 Melanoma 1 6.019 9.384 36 Melanoma 1 2.711 9.801 37 Melanoma 1 6.570 10.049 38 Melanoma 3 7.838 11.364 39 Ovarian 1 9.067 11.060 40 Ovarian 1 8.645 11.849 41 Ovarian 3 7.079 10.937 42 Ovarian 3 9.626 11.911 43 Ovarian 3 8.478 10.954 44 Ovarian 3 8.890 12.076 45 Prostate 3 8.356 12.486 46 Prostate 3 9.074 11.841 47 Renal 3 9.117 12.324 48 Renal 3 9.522 12.362 4...

[code]>require("survival") > coxph(Surv(futime,fustat)~age + strata(rx),ovarian) Call: coxph(formula = Surv(futime, fustat) ~ age + strata(rx), data = ovarian) coef exp(coef) se(coef) z p age 0.137 1.15 0.0474 2.9 0.0038 Likelihood ratio test=12.7 on 1 df, p=0.000368 n= 26, number of events= 12 > coxph(Surv(futime,fustat)~age, ovarian, subset=rx==1)...

Dear all, I am using R 3.4.3 on Windows 10. I am preparing some teaching materials and I'm having trouble matching the by-hand version with the R code. I have fitted a Cox model - let's use the ovarian data as an example: library(survival) data(ovarian) ova_mod <- coxph(Surv(futime,fustat)~age+rx,data=ovarian) If I want to make predict survival for a new set of individuals at 100 days then that is trivial using predict.coxph e.g.: newdata <- data.frame(futime=rep(100,5),fustat=rep(1,5),age...

Hi all, Using: R version 2.8.1 Patched (2009-03-07 r48068) on OSX (10.5.6) with survival version: Version: 2.35-3 Date: 2009-02-10 I get the following using the first example in ?summary.survfit: > summary( survfit( Surv(futime, fustat)~1, data=ovarian)) Call: survfit(formula = Surv(futime, fustat) ~ 1, data = ovarian) time n.risk n.event survival std.err lower 95% CI upper 95% CI 59 26 1 0.962 0.0377 0.890 1.000 115 25 1 0.923 0.0523 0.826 1.000 156 24 1 0.885 0.0...

...rayprofile at yahoo.com> wrote: > > > > Jim, > > > > I tried a few things, I found that clip() works if I just do some regular graphing tasks. But as long as I run lines(fit) with "fit" object is a survfit object, this would reset to default plot region. See the ovarian example below: > > > > library(survival) > > ovarian1<-ovarian > > ovarian1$fustat[ovarian$futime>450]<-0 > > ovarian1$futime[ovarian$futime>450]<-450 > > ovarian2<-subset(ovarian,futime>450) > > > > fit1 <- survfit(Surv(futime...

...ions in "muhaz" package to estimate and plot hazard funciton. However function "muhaz" always gives error message "Error in Surv(times, delta) : object "times" not found". I could not even run their sample codes in the user's manual as follows: data(ovarian) attach(ovarian) fit1 <- muhaz(futime, fustat) it gave the same error message. By the way, the "survival" package is installed and functions well on my computer. Does anyone have the same problem? Thanks! Deming

...e output of survreg function with standard error, z, p.value etc? Does it mean that intercept was fitted with the covariates? Does Value column represent coefficients or some thing else? Regards, ------------------------------------------------- tmp = survreg(Surv(futime, fustat) ~ ecog.ps + rx, ovarian, dist='weibull',scale=1) > summary(tmp) Call: survreg(formula = Surv(futime, fustat) ~ ecog.ps + rx, data = ovarian, dist = "weibull", scale = 1) Value Std. Error z p (Intercept) 6.962 1.322 5.267 1.39e-07 ecog.ps -0.433 0.587 -0.7...

Dear All, Sorry to bother you again. I have a model: coxfita=coxph(Surv(rem.Remtime/365,rem.Rcens)~all.sex,data=nearma) and I'm trying to do a plot of Schoenfeld residuals using the code: plot(cox.zph(coxfita)) abline(h=0,lty=3) The error message I get is: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In sqrt(x$var[i, i] * seval) : NaNs

Hola Estoy tratando de correr un survival analysis usando un Cox regression model. Tengo una duda respecto a la organizacion del script. Tengo una variable que es -tamano del individuo- y quiero ver si hay diferencia en sobrevivencia respecto a tamano. Como diseno de campo los tamanos fueron ubicados de forma aleatoria en bloques al azar. Cuado planteo el script tengo algo como:

Dear all, I am doing library(survival) fit <- coxph(Surv(futime,fustat) ~ rx, ovarian) plot(survfit(fit,newdata=ovarian),col=c(1,2)) legend("bottomleft", legend=c("rx = 0", "rx = 1"), lty=c(1,2),col=c(1,2)) Is this correct to compare these two groups? Is the 0.31 the p-value that the median f two groups are equal Why lty does not work here? Man...

Dear All: I have some questions about the output of coxph. Below is the input and output: ---------------------------------------- > coxph(formula = Surv(futime, fustat) ~ age + rx + ecog.ps, data = + ovarian, x = TRUE) Call: coxph(formula = Surv(futime, fustat) ~ age + rx + ecog.ps, data = ovarian, x = TRUE) coef exp(coef) se(coef) z p age 0.147 1.158 0.0463 3.17 0.0015 rx -0.815 0.443 0.6342 -1.28 0.2000 ecog.ps 0.103 1.109 0.6064 0.17 0.8600...

...h lower bounds of zero, despite the fact that plnorm(0)==0 and pnorm(-Inf)==0 are well defined. Below is a short example to reproduce the problem. Does anyone know why survreg() must behave that way? Is there an alternate solution to this problem? Sincerely, Jerome Asselin library(survival) data(ovarian) newovarian <- ovarian newovarian$lower59 <- newovarian$futime-59 newovarian$time59 <- Surv(newovarian$lower59,newovarian$futime, event=rep(3,nrow(newovarian)),type="interval") survreg(time59~ecog.ps+rx,data=newovarian,dist="lognormal") #THIS DOES NOT WORK BECAUSE...

...in the package survAUC. The relevant function is UnoC. The question has to do with what happens when I specify a time point t for the upper limit of the time range under consideration (we want to avoid using the right-end tail of the KM curve). Copying from the example in the help file: TR <- ovarian[1:16,] TE <- ovarian[17:26,] train.fit <- coxph(Surv(futime, fustat) ~ age,x=TRUE, y=TRUE, method="breslow", data=TR) lpnew <- predict(train.fit, newdata=TE) Surv.rsp <- Surv(TR$futime, TR$fustat) Surv.rsp.new <- Surv(TE$futime, TE$fustat) If time is left NU...

Hi, I'm using natural cubic splines from splines::ns() in survival regression (regressing inter-arrival times of patients to a queue on queue size). The queue size fluctuates between 3600 and 3900. I would like to be able to run predict.survreg() for sizes <3600 and >3900 by assuming that the rate for <3600 is the same as for 3600 and that for >4000 it's the same as for

...iables as additional arguments is a matter of personal preference. f.coxph.zph<-function(x, timeVar, statusVar) { cox.fit <- coxph(Surv(timeVar, statusVar) ~ x, na.action = na.exclude, method = "breslow", x=TRUE) fit.zph<-cox.zph(cox.fit) fit.zph$table[,3] } time.cox <- ovarian$futime status.cox <- ovarian$fustat apply(ovarian[,-(1:2)],2, f.coxph.zph, timeVar = time.cox, statusVar = status.cox) --Matt -----Original Message----- From: s-news-owner at lists.biostat.wustl.edu [mailto:s-news-owner at lists.biostat.wustl.edu]On Behalf Of array chip Sent: Tuesday, Novem...

I am wondering how to estimate the survival curve for a particular case(s) given a coxph model using this example code: #fit a cox proportional hazards model and plot the #predicted survival curve fit <- coxph( Surv(futime,fustat)~resid.ds+strata(rx)+ecog.ps+age,data=ovarian[1:23,]) z <- survfit(fit,newdata=ovarian[24:26,],individual=F) zs <- z$surv zt <- z$time I get the following output: 24 25 26 [1,] 0.9740399 0.91737529 0.9873785 [2,] 0.9431988 0.82552974 0.9721557 [3,] 0.9023088 0.71387936 0.9515702...

How can I get the Log - Rank p value to be output? The chi square value can be output, so I was thinking if I can also have the degrees of freedom output I could generate the p value, but can't see how to find df either. > (survtest <- survdiff(Surv(time, cens) ~ group, data = surv,rho=0)) Call: survdiff(formula = Surv(time, cens) ~ group, data = surv, rho = 0) N Observed