similar to: controling the size of vectors in a matrix

hie l would like to create a 6th column "actual surv time" from the following data the condition being if censoringTime>survivaltime then actual survtime =survival time else actual survtime =censoring time the code l used to create the data is s=2 while(s!=0){ n=20 m<-matrix(nrow=n,ncol=4)

l would like to create the following matrice treatmentgrp strata 1 1 1 1 1 1 1 2 1 2 1 2 2 1 2 1 2 1 2 2 2 2 2 2 l should be able to choose the size of the treatment grps and stratas the method l used intially creates the

hie how do you compute the logrank test using R what commands do you use my data looks something like just an example treatmentgrp strata censoringTime survivalTime censoring act.surv.time [1,] 2 2 42.89005 1847.3358 1 42.89005 [2,] 1 1 74.40379 440.3467 1 74.40379 [3,] 2 2

i would like to add a variable to an existing matrix by manipulating 2 previous variables eg for the data m treat strata censti survTime [1,] 1 2 284.684074 690.4961005 [2,] 1 1 172.764515 32.3990335 [3,] 1 1 2393.195400 24.6145279 [4,] 2 1 30.364771 8.0272267 [5,] 1 1 523.182282 554.7659501 l

l used the following code to generate a sample and then calculated then did a log rank test.can l get the normal version of the logrank eg sqrt of the chisqr(1) will give you the N~(0,1). from my sample can i use the above expression to get the normal dist from the result of the log rank test. thank s=1 while(s!=0){ n=100 m<-matrix(nrow=n,ncol=4)

l have the following dataset and would like to calculate the actual survival time by if censoring time > survival time then actual survival time =survival time else its= censoring time. treatmentgrp strata censoringTime survivalTime censoring actualsurvivaltim [1,] 1 1 1.012159 1137.80922 0 [2,] 2 2

hie when i use plot.survfit to plot more than one graph why I only see the last graph how do i see the other graphs.for example n=20 n1=n/2 n2=n/4 a11=4;a12=4 ;a21=4 ;a22=4 t1<-array(1,c(n1)) t2<-array(2,c(n1)) treatgrp=matrix(c(t1,t2))

hie l would like to create about ten strata l have tried using sample(c(1,2),1,replace=TRUE) but all l get is a vector of 1 and 2 if i change it to sample(c(1,10),1,replace=TRUE) it gives me a vector of 1 and 10 how do l go about it --------------------------------- [[alternative HTML version deleted]]

hie i'm tryimg to generate two survival data using the following code (I know its ugly ) but it seems to repeat two of the variables can any one tell me whats the porblem. n=20 n1=n/2 n2=n/4 a11=1 ;a12=1.4 ;a21=16 ;a22=a12 * a21 t1<-array(1,c(n1)) t2<-array(2,c(n1)) treatgrp=matrix(c(t1,t2))

l want to create a column of 1 and 2 randomly what command should l use eg treatment strata 1 1 2 0 1 1 2 1 2 0 2 1 2 0 1 0 these should be created randomly secondly if l have

Hi, I'm going crazy trying to plot a quite simple graph. i need to plot estimated hazard rate from a cox model. supposing the model i like this: coxPhMod=coxph(Surv(TIME, EV) ~ AGE+A+B+strata(C) data=data) with 4 level for C. how can i obtain a graph with 4 estimated (better smoothed) hazard curve (base-line hazard + 3 proportional) to highlight the effect of C. thanks!! laudan [[alternative

hie 1...i'm trying to carryout a relibility testusing cronbach's alpha what fuctin do i use. 2.. this is more of a statistical question.if the alpha value for all the variables is negative what does it mean. and if the alpha value is negative for all tyha variables but is greater than 0.7 for some sections of the variables what does that mean thanks in advance

hie how do I specify the number of decuimal places for my calulations thanks __________________________________________________ [[alternative HTML version deleted]]

Hello R user, I have following data frame: df=data.frame(id=c(1:10),strata=rep(c(1,2),c(5,5)),y=c( 10,12,10,NA,15,70,NA,NA,55,100),x=c(3,4,5,7,4,10,12,8,3,15)) and I would like to replace NA's with: instead of first NA tapply(na.exclude(df)$y,na.exclude(df)$strata,sum)[1]* *7 */tapply(na.exclude(df)$x,na.exclude(df)$strata,sum)[1] where 7 is the value of x (id=4) in strata 1 where y=NA

Buenas. Abajo pongo la salida de un modelo de cox , dónde he estratificado por una variable de país (Countryb) y por otra (Q6). Además hay interacción entre la variable mobilityPDurG2 (es una variable 0,1, y 0 es la categoría de referencia) país. La categoría de referencia para país es "united kingdom". Mi duda surge si quiero calcular el hazard ratio para los que tienen un 1

Dear useRs, What is the syntax to obtain survival curves for single strata on many subjects? I have a model based on Surv(time,response) object, so there is a single row per subject and no start,stop and no switching of strata. The newdata has many subjects and each subject has a strata and the survival based on the subject risk and the subject strata is needed. If I do newpred <-

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 have been experimenting with the new Survey package. Specifically, I was trying to use some of the functions on the public-use survey data from NHIS (2000 Sample Adult file). Error 1): The first error I get is when I try to specify the complex survey design. nhis.design<-svydesign(ids=~psu, probs=~probs, strata=~strata, data=nhis.df, check.strata=TRUE) Error in svydesign(ids =

There is a code from SPSS Syntax do if sub(ATVK,2,2)="01". comp strata=1. else if sub(ATVK,2,2)>="05" and sub(ATVK,2,2)<="27". comp strata=3. else if sub(ATVK,4,2)>"20" or sub(ATVK,6,2)>"20". comp strata=4. else if sub(ATVK,4,2)>"00". comp strata=2. end if. value labels strata 1 "R 2 "Li" 3 "M" 4

how do l programme the logrank test. l am trying to compare 2 survival curves --------------------------------- [[alternative HTML version deleted]]