similar to: p-value using 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. I get .

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) ~

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

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,

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) ~

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

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,

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

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.

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

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 everyone, I'm using the very good imaptest [0] tool to test my little imap server implementation. I've tried to use the dovecot-crlf [1] file, but it looks like there are some major issues : $ grep -n "In-Reply-To.*;" tests/data/dovecot-crlf 479:In-Reply-To: <20020806175441.GA7148 at linux.taugt.net>; from rueckert at informatik.uni-rostock.de on Tue, Aug 06, 2002

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

Hi, I'm doing kolmogorv-smirnov test but I don't know what conclusions to take, I want to know if my data has a normal distribution: > ks.test(dados1,"pnorm") One-sample Kolmogorov-Smirnov test data: dados1 D = 0.972, p-value < 2.2e-16 alternative hypothesis: two-sided Warning message: cannot compute correct p-values with ties in: ks.test(dados1,

Hi all, I'm having this error, since I'm working with a data matrix I don't understand what's happening; I've tried several ways to solve this, even working with sparse matrix, but nothing seems to solve it, I've also tried svm (with a simple matrix 3*3 and still got the same error. > dados<-read.table("b.txt",sep="",nrows=30000) >

my program given below can some one make it presentable. I trying to simulate survival data and calculate the power. I think i could have done better. s=10 number=0 count1=0;count2=0;count3=0;count4=0;count5=0;count6=0;count7=0;count8=0; count9=0; count11=0;count22=0;count33=0;count44=0;count55=0;count66=0;count77=0; count88=0;count99=0; while(s!=0){ n=100 n1=n/2 n2=n/4

hie Im trying to calculate the power of the logrank test for different values of rho .I was just wandering whether the following programme would do it. any suggestions are welcome s=50 number=1 count1=0;count2=0;count3=0;count4=0;count5=0;count6=0;count7=0;count7=0; count8=0;count9=0 while(s!=0){ n=20 n1=n/2

> 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

Dear helpers I constructed a data frame with this structure > str(dados1) `data.frame': 485 obs. of 16 variables: $ Emissor : int 1 1 1 1 1 1 1 1 1 1 ... $ Marisca.Rio : int 1 1 1 1 1 1 1 1 1 1 ... $ Per?odo : int 1 1 1 1 1 1 1 1 1 1 ... $ Reproducao : int 3 3 3 3 3 3 3 3 3 3 ... $ Estacao : int 2 2 2 2 2 2 2 2 2 2 ... $ X30cm : int

> 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. >>