similar to: summary of Surv object crashes R 2.2.0 (PR#8203)

Dear All, there seems to be some strange influence of the Design package on data.frame. If I build a data.frame containing a Surv object without loading the package Design, the data frame is usable to coxph. If instead I just load Design and build a data.frame afterwards, the naming of the Surv object is different and it does not work with coxph. (In my real application I loaded Design to use the

Does Surv in psm handle interval2 data? The argument list seems to indicate it does but I get an error. Thanks, Chris # code library('survival') left <- c(1, 3, 5, NA) right <-c(2, 3, NA, 4) Surv(left, right, type='interval2') survreg(Surv(left, right, type='interval2') ~ 1) library('rms') Surv(left, right, type='interval2') # error args(Surv)

1. A Surv object is a matrix with some extra attributes. The as.matrix.Surv function removes the extras but otherwise leaves it as is. 2. The last several versions of the survival library were accidentally missing the S3method('as.matrix', 'Surv') line from their NAMESPACE file. (Instead it's position is held by a duplicate of the line just above it in the NAMESPACE file,

Dear All, since years I am struggling with Surv objects in data.frames. The following seems to have to do with it. See below the modified example from the help page of survSplit. The original works, as expected. If, however, a Surv object is added to the data.frame, each record gets doubled. Is there some solution other than avoiding Surv objects in data.frames? Thanks, Heinz

Dear everyone, My question involves the use of the survival object. We can have Surv(time,time2,event, type=, origin = 0) (1) As detailed on p.65 of: http://cran.r-project.org/web/packages/survival/survival.pdf My data (used in my study) is 'right censored' i.e. my variable corresponding to 'event' indicates whether a person is alive (0) or dead (1) at date last seen

Dear All, a Surv object I put in a data frame behaves somehow unexpected (see example). If I do a Cox regression on the original Surv object it works. If I put it in a data.frame and do the regression on the data frame it does not work. Seemingly it has to do with the class attribute, because if I change the class attribute to let "Surv" appeare first, again it works. Is this known?

 Dear R-users,   I  have two questions regarding validation and calibration of Survival regression models.   1.  I am trying to calibrate and validate a cox model using val.surv. here is my code:  f.1<-cph(Surv(time,event)~age, x=T, y=T, data=train)  test1<-test[,"age"]  val.surv(f.1, newdata=data.frame(test1), u=10)    but I get an error message:    Error in val.surv(f.1, newdata

Hi, I am trying to write a function to automate multiple graph generation. My data looks like: Table of numeric values with the following headers: timeM1 statusM1 xM1 timeM2 statusM2 xM2 timeM3 statusM3 xM3 1 2 3 4 5 6 Where M1,M2, M3 hve no similarity except they have a max string length of 7. Examples are mcw0045, adl0003, lei0101. Now, what I want to do is Function(M1, M2,

Dear R People: I used the "Surv" function to produce the following object: >a <- Surv(1:4,2:5,c(0,1,1,0)) a [1] (1,2+] (2,3 ] (3,4 ] (4,5+] >str(a) Error in "[.Surv"(object, 1:ile) : subscript out of bounds > Why does str(a) give an error, please? Or did I do something wrong? Thanks in advance. R Version 1.9.1 Windows Sincerely, Laura Holt mailto: lauraholt_983

Hi, I'm trying to do a simple survival analysis on some data, and I'm having the following problem (here's my code and the error message): out <- Surv(fup,event=status) Error in Surv(fup, event = status) : argument "time2" is missing, with no default >From reading the documentation, it seems that I should be able to simply write: Surv(time1, event) if my data is

Can someone help me. I am very new to R. I am fitting a Cox model using Frank Harrell's cph as I want to produce a Nomogram. This is what I have done: Srv<- Surv(time,cens) f.cox<- cph(Srv~ v1+v2+v3+v4, x=T, y=T, surv=T) As soon as I press enter, Windows XP crashes. If I remove surv=T, then it works. I have R version 2.9.0. Is there a way of displaying the parameter estimates (ie beta

Dear R users: I tried to use val.surv to give an internal validation of survival prediction model. I used the sample sources. # Generate failure times from an exponential distribution set.seed(123) # so can reproduce results n <- 1000 age <- 50 + 12*rnorm(n) sex <- factor(sample(c('Male','Female'), n, rep=TRUE, prob=c(.6, .4))) cens <- 15*runif(n) h

Dear, I want to analyze two-level survival data using a shared frailty model, for which I want to use the R package 'Frailtypack", proposed by Rondeau et al. The dataset was built using SAS software. I also tried to change the format using SPSS and Excell. My (reduced) dataset has following column names: ID entry time status family var1 I used following command: >

Can survreg() handle interval-censored data like the documentation says? I ask because the command: survreg(Surv(start, stop, event) ~ 1, data = heart) fails with the error message Invalid survival type yet the documentation for Surv() states: "Presently, the only methods allowing interval censored data are the parametric models computed by 'survreg'"

Dear All, trying to write a data.frame, containing Surv objects to a csv-file I get "Error in dimnames(X) <- list(dn[[1L]], unlist(collabs, use.names = FALSE)) : length of 'dimnames' [2] not equal to array extent". See example below. May be, I overlooked something, but I expected that also data.frames containing Surv objects may be written to csv files. Is there a

Hello, I have srv <- Surv(sample(1:10), sample(0:1, 10, replace = TRUE)) srv [1] 1 10 2+ 8 6+ 7+ 3 5+ 4+ 9+ srv.char <- as.character(srv) srv.char [1] " 1 " "10 " " 2+" " 8 " " 6+" " 7+" " 3 " " 5+" " 4+" " 9+" Is there an inverse to as.character(srv). That is, I would

Can anybody give me a neat example of interval censored data analysis codes in R? Given that suvreg(Surv(c(1,1,NA,3),c(2,NA,2,3),type="interval2")~1) works why does survreg(Surv(data[,1],data[,2],type="interval2")~1) not work where data is : T.1 T.2 Status 1 0.0000000 0.62873036 1 2 0.0000000 2.07039068 1 3 0.0000000

Dear R users: I want to externally validate a model with val.surv. Can I use only calculated survival (at 1 year) and actual survival? Or I needed the survival function and actual survival. Thanks *Yao Zhu* *Department of Urology Fudan University Shanghai Cancer Center Shanghai, China* [[alternative HTML version deleted]]

Dear Prof. Therneau I was impressed to discover that the 'survConcordance' now handles Surv() objects in counting format (example below to clarify what I mean). This is not documented in the help page for the function. I am very curious to see how a c-index is estimated in this case, using just the linear predictors. It was my impression that with left truncation the ordering of

OS X R 4.3.3 Colleagues I have created objects using the Surv function in the survival package: > FIT.1 Call: survfit(formula = FORMULA1) n events median 0.95LCL 0.95UCL SUBDATA$ARM=1, SUBDATA[, EXP.STRAT]=0 18 13 345 156 NA SUBDATA$ARM=2, SUBDATA[, EXP.STRAT]=1 13 5 NA 186 NA SUBDATA$ARM=2, SUBDATA[, EXP.STRAT]=2 5