search for: survdiff

ExpeRts, I'm trying to compare three survival curves using the function "survdiff" in the survival package. Following is my code and corresponding error message. > survdiff(Surv(st_months, status) ~ strata(BOR), data=mydata) Error in survdiff(Surv(st_months, status) ~ strata(BOR), data = mydata) : No groups to test When I check the "strata" of the variable....

...,time,censorTime) colnames(m)=c(“treat”,”strata”,”time”,”censorTime”) eventTime<-pmin(m[,"time"],m[,"censorTime"]) m<-cbind(m,eventTime) m<-cbind(m,0) m[m[,3]<m[,4],6]<-1 colnames(m)[6]<-"censoring" b=data.frame(m) c1= survdiff(Surv(eventTime,censoring)~treatgrp ,data=b,rho=-.15) c11= survdiff(Surv(eventTime)~treatgrp ,data=b,rho=-1.5) c2= survdiff(Surv(eventTime,censoring)~treatgrp ,data=b,rho=-1) c22= survdiff(Surv(eventTime)~treatgrp ,data=b,rho=-1) c3= survdiff(Surv(eventTime,censoring)~treatgrp ,data...

Hi everyone!! I have dataset composed of a numbers of survival analyses. ( for batch survival analyses by using for-loop) . Here are code !! ####### dim(svsv) Num_t<-dim(svsv) Num<-Num_t[2] # These are predictors !! names=colnames(svsv) for (i in 1:Num ) { name_tt=names[i] survdiff(Surv(survival.m, survival) ~ names[i], data=svsv) fit.Group<-survfit(Surv(survival.m, survival) ~ names[i] , data=svsv) plot(fit.Group, col=2:1, xlab="Survival", ylab="Prob") } ##### names[i] is not working in the survdiff. According to help R , the predictor must be singl...

Hi list, I want to use the p-value from the survdiff function (package survival) to reuse within a function in a Kaplan-Meier plot. The p-value is somehow not a component of the value list ?! Thanks in advance -- A. Goralczyk G?ttingen, Ger.

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 Expected (O-E)^2/E (O-E)^2/V group=1 20 16 11.0 2.23 4.64 group=2 20 12 17.0 1.45 4.64 Chisq= 4.6 on 1 degrees of fr...

Dear all, Does anyone knows how I could extract the p-value in: > survdiff(Surv(tempo,status) ~ grupo,data=dados1,rho=1) Call: survdiff(formula = Surv(tempo, status) ~ grupo, data = dados1, rho = 1) N Observed Expected (O-E)^2/E (O-E)^2/V grupo=1 21 5.12 12.00 3.94 14.5 grupo=2 21 14.55 7.68 6.16 14.5 Chisq= 14.5 on 1 degrees...

Hello, I am trying to set up a loop that can run the survdiff function with the ultimate goal to generate a csv file with the p-values reported. However, whenever I try a loop I get an error such as "invalid type (list) for variable 'survival_data_variables[i]". This is a subset of my data: structure(list(time = c(1.51666666666667, 72, 72, 25...

...rvival analysis, but this is an area with which we are not very familiar, we just happen to have gathered some data that needs this type of analysis. We've done quite a bit of reading, but answers escape us, even though the question below seems simple. Considering the following example from ?survdiff: > survdiff(Surv(time, status) ~ pat.karno, data=lung) Call: survdiff(formula = Surv(time, status) ~ pat.karno, data = lung) n=225, 3 observations deleted due to missingness. N Observed Expected (O-E)^2/E (O-E)^2/V pat.karno=30 2 1 0.658 0.1774 0.179 pat.karn...

...classical biological perspective. For example, for lifetable analysis of insects there is often no need to estimate survival using a Kaplan-Meir estimate because it is relatively easy to follow a cohort of individuals through the entire course of life. Thus I question the appropriateness of using survdiff in my analysis; I have exact data yet I would be testing on the Kaplan-Meir estimate of these data in survdiff. Thanks for any help.

...makes me wonder if there's homework involved.... Simpler for your example is to use get and subset. dat <- structure(..... as found below var.to.test <- names(dat)[4:6] #variables of interest nvar <- length(var.to.test) chisq <- double(nvar) for (i in 1:nvar) { tfit <- survdiff(Surv(time, completion==2) ~ get(var.to.test[i]), data=dat, subset=(group==3)) chisq[i] <- tfit$chisq } write.csv(data.frame(var.to.test, chisq)) On 10/19/2012 05:00 AM, r-help-request at r-project.org wrote: > Hello, > > I am trying to set up a loop that can run the survdiff...

...ames(m)=c(?treat?,?strata?,?time?,?censorTime?) act.surv.time<-pmin(m[,"time"],m[,"censorTime"]) m<-cbind(m,act.surv.time) m<-cbind(m,0) m[m[,3]>m[,4],6]<-1 colnames(m)[6]<-"censoring" c1= survdiff(Surv(act.surv.time,censoring)~treatgrp ,data=b,rho=-2) c2= survdiff(Surv(act.surv.time,censoring)~treatgrp ,data=b,rho=-.15) c3= survdiff(Surv(act.surv.time,censoring)~treatgrp ,data=b,rho=-1) c4= survdiff(Surv(act.surv.time,censoring)~treatgrp ,data=b,rho=-0.5)...

Hi all,? The survdiff() from survival package has an argument "rho" that implements Fleming-Harrington weighted long rank test.? But according to several sources including "survminer" package (https://cran.r-project.org/web/packages/survminer/vignettes/Specifiying_weights_in_log-rank_comparisons.htm...

dear all, I would like to know whether survdiff and survReg function in the survival package work for left-truncated and right-censored data. If not, what other functions can i use to make comparison between two survival curves with LTRC data. thanks for any help given sing yee

Hello, As I am quitte an ignorant user of R, excuse me for any wrongfull usage of all the terms. My question relates to the statistics behind the survdiff function in the package survival. My textbook knowledge of the logrank test tells me that if I want to compare two survival curves, I have to take the sum of the factors: (O-E)^2/E of both groups, which will give me the Chisq. If I calculate this by hand, I get a different value than the one R is g...

Hei, I have a one simple question which does not seem to be that simple as I cannot find any solution/answer: Is it possible to compare multiple survival curves in R with survdiff-function when there is interaction term involved in predictor variables (and this interaction is significant)? Example: survdiff(Surv(death,status)~treatment*gapsize) R is making "problems" with it ie.e. it does not want to perform the test. And all the examples I have found so far, ma...

...7 1 1 1 1 1 1 6.63 6.52 NA 0 6 27 1 1 23 4 1 2 1 1 1 1 7.19 6.51 NA 0 I am trying to perform survival analysis but continually get errors when extracting from this data.frame: attempt 1: > X <- Surv(B27.vec$AgeOn,B27.vec$UV) > survdiff(X,rho=0,data=uvf) Error in x$terms : $ operator is invalid for atomic vectors attempt 2: > X <- Surv(B27.vec[,4],B27.vec[,15]) > survdiff(X,rho=0,data=uvf) Error in x$terms : $ operator is invalid for atomic vector attempt 3: > AO <- B27.vec[["AgeOn", exact = TRUE]] >...

> On Feb 13, 2018, at 4:02 PM, array chip via R-help <r-help at r-project.org> wrote: > > Hi all, > > The survdiff() from survival package has an argument "rho" that implements Fleming-Harrington weighted long rank test. > > But according to several sources including "survminer" package (https://cran.r-project.org/web/packages/survminer/vignettes/Specifiying_weights_in_log-rank_compa...

Hi Does anyone know if there is a function like survdiff which can also handle left-truncated and right-censored data? When I use it on left-truncated and right-censored data I get an error message saying Right censored data only. Many thanks Rajen [[alternative HTML version deleted]]

Hello all- I am doing a survival analysis for two species of invasive plants I outplanted to edges and interiors of island and mainland sites in a local reservoir. I am using the KM estimate and had no problem doing survdiff for my data using the following code: S4<-Surv(outplant$SurvTime, outplant$StatusD6) diff4=survdiff(S4 ~ outplant$Species+outplant$SiteType+outplant$EdgInt) diff4 Species = ALBIJU or LONIJA SiteType = Island or Mainland EdgInt = Edge or Interior The overall test for difference among the 8 cur...

...nTx = B, TimeToFailure = C, NotFailed = D) We can do a survival analysis on those individually: OnTxFit = survfit (Surv ( TimeOnTx, StillOnTx==0 ) ~ 1 , data = myData) FailedFit = survfit (Surv ( TimeToFailure , NotFailed==0 ) ~ 1 , data = myData) plot(OnTxFit) lines(OnTxFit) But how can I do a survdiff type of comparison between the two? Do I have to restructure the data so that Time's are all in one column, Event in another and then a Group to indicate what type of event it is? Seems a complex way to do it (especially as the dataset is of course more complex than I've just shown)... so...