similar to: Survival analysis and comparing survival curves

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. I get .

Hi, I ran the example on pp. 799-800 from Machael Crawley's "The R Book" using package survival v. 2.36-5, R 2.13.0 and RStudio 0.94.83. The model is a Cox's Proportional Hazards model. The result was quite different compared to the R Book. I have compared my code to the code in the book but can not find any differences in the function call. My results are attached as well as a

Hi, I have a data frame: > class(B27.vec) [1] "data.frame" > head(B27.vec) AGE Gend B27 AgeOn DD uveitis psoriasis IBD CD UC InI BASDAI BASFI Smok UV 1 57 1 1 19 38 2 1 1 1 1 1 5.40 8.08 NA 1 2 35 1 1 33 2 2 1 1 1 1 1 1.69 2.28 NA 1 3 49 2 1 40 9 1 1 1 1 1 1 8.30 9.40 NA

I have a dataset which for the sake of simplicity has two endpoints. We would like to test if two different end-points have the same eventual meaning. To try and take an example that people might understand better: Lets assume we had a group of subjects who all received a treatment. The could stop treatment for any reason (side effects, treatment stops working etc). Getting that data is very

I read through Harrington and Fleming (1982) but it is beyond my statistical comprehension. I have survival data for insects that have a very finite expiration date. I'm trying to test for differences in survival distributions between different groups. I understand that the medical field is most often dealing with censored data and that survival analysis, at least in the package survival,

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) ~

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

> 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

> On Feb 14, 2018, at 5:26 PM, David Winsemius <dwinsemius at comcast.net> wrote: > >> >> 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. >>

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 of freedom,

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.html), Fleming-Harrington weighted log-rank test should have 2 parameters

R Folk: Please forgive what I'm sure is a fairly na?ve question; I hope it's clear. A colleague and I have been doing a really simple one-off survival 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

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, 25.7833333333333, 72, 72, 72, 72, 72,

Dear all: I have the following codes: Xdata<-c(2,3,8,9,10) Ydata<-1:5 gap.plot(Xdata, Ydata,gap=c(5,6),gap.axis="x",type="o") However, the type='o' seems only work on the first part of gap plot, the second half of the plot always just points, you can not add lines on that part, any help will be highly appreciated. I would like to have these two parts of

The number of recent questions from umn.edu 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) ~

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.

Estimados miembros de la lista, Estoy haciendo una análisis de supervivencia con R. Adjunto mis datos. Quiero analizar la supervivencia de 5 grupos diferentes y compararla. Para ello estoy utilizando el paquete survival. > s = Surv(c$tiempo, c$estado) > f = survfit(s ~ tratamiento, data = c) > d = survdiff(s ~ tratamiento, data = c) > d Call: survdiff(formula = s ~ tratamiento,

Good morning, Sorry to trouble the list. I have a problem I hope to seek your advice on. Essentially, I am trying to 'validate' a multivariate Cox proportional hazards model built in a training set, by testing it on an external test set. I have performed a survfit using the Cox model to predict survival for the test set, and obtained individual predictions for survival time, with

Hi I have a problem with this error, I have searched the archives and found previous discussion about this, can I cannot understand how the explanations apply to what I am trying to do. I am trying to do Log_rank Survival analysis, I have included tables and str command, is it a factor/integer problem? If so how do I correct this, as all my attempt to recode the data have failed. >

A user reported a problem with the survdiff function and the use of variables that contain a space.? Here is a simple example.? The same issue occurs in survfit for the same reason. lung2 <- lung names(lung2)[1] <- "in st"?? # old name is inst survdiff(Surv(time, status) ~ `in st`, data=lung2) Error in `[.data.frame`(m, ll) : undefined columns selected In the body of the code