similar to: the Surv function

Hi Everyone, I think I found a problem with the ifelse function: If the condition argument is NA, it treats it as true. Anyone agree or disagree with this? Jen -- Jennifer Dillon Doctoral Student Harvard Biostatistics Room 414B, Building 1 [[alternative HTML version deleted]]

Hello, I am almost a total novice, and I am sure there must be an easy (and basic) way to turn a variable of 1's and 2's into a variable of zeros and ones. This is in a data frame, but if I could do it with vectors, that's all I need. Can someone tell me how? Thanks so much, Jen

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

One one of my new vmware guests... puppetd -v spits out a bunch of err & warnings about "State group failed"... what does this mean? <snip> info: Caching configuration at /var/lib/puppet/localconfig.yaml err: Could not create root: Could not find a default provider for group warning: file=/etc/yum.repos.d/: State group failed: Could not find a default provider for group

Dear R users, I use the excelent Anova function of the library car because the easy way to get sphericity correction. Unless I use the scan function. I have not been able to access the values ​​of sum squares and degrees of freedom for each effect in the univariate summary table. Example of the car library for Anova function: library(car) phase <- factor(rep(c("pretest",

hello: my data looks like: time1  time2   event  catagoria 2004    2006        1            C 2004    2005        0            C 2005    2010        1            E 2007    2009        1            C 2006    2007        0            E 2008    2010        0            C 2008    2010        1            E ... and the census interval is 1 year I have tried  this

Dear list, I would like to run an ANCOVA using car::Anova with repeated measures factors, but I can't figure out how to do it. My (between-subjects) covariate always interacts with my within-subject factors. As far as I understand ANCOVA, covariates usually do not interact with the effects of interest but are simply additive (or am I wrong here?). More specifically, I can add a covariate as

Hello dear R members. I have been learning the Anova syntax in order to perform an SS type III Anova with repeated measures designs (thank you Prof. John Fox!) And another question came up: where/what are the (between/within) residuals for my model? ############ Play code: phase <- factor(rep(c("pretest", "posttest", "followup"), c(5, 5, 5)),

I'm trying to fit a Cox proportional hazards model to some hospital admission data. About 25% of the patients have had at least one admission, and of these, 40% have had two admissions within the 12 month period of the study. Each patients has had one of 4 treatments, and one of the treatment groups has had no admissions for the period. I used:

Hi r users, I have two data sets (X1, X2). For example, time1<-c( 0, 8, 15, 22, 43, 64, 85, 106, 127, 148, 169, 190 ,211 ) outpue1<-c(171 ,164 ,150 ,141 ,109 , 73 , 47 ,26 ,15 ,12 ,6 ,2 ,1 ) X1<-cbind(time1,outpue1) time2<-c( 0 ,8 ,15 , 22 ,43 , 64 ,85 ,106 ,148) output2<-c( 5 ,5 ,4 ,5 ,5 ,4 ,1 ,2 , 1 ) X2<-cbind(time2,output2) I want to

Greetings listeRs - Given a data frame such as times time1 time2 time3 time4 1 70.408543 48.92378 7.399605 95.93050 2 17.231940 27.48530 82.962916 10.20619 3 20.279220 10.33575 66.209290 30.71846 4 NA 53.31993 12.398237 35.65782 5 9.295965 NA 48.929201 NA 6 63.966518 42.16304 1.777342 NA one can use "transform" to

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]]

Hi, I would like to merge two data frames. It is just that I want the merging to be done with some kind of a fuzzy criterion. Let me explain. My first data frame looks like this : ID1 time1 dt 1 2008-01-02 13:11 10 2 2008-01-02 14:20 20 3

Hello all, I've 2 strings that representing the start and end values of a date and time. For example, time1 <- c("21/04/2005","23/05/2005","11/04/2005") time2 <- c("15/07/2009", "03/06/2008", "15/10/2005") as.difftime(time1,time2) Time differences in secs [1] NA NA NA attr(,"tzone") [1] "" How can i

To Whomever is Interested, I have spent several days searching the web, help files, the R wiki and the archives of this mailing list for a solution to this problem, but nonetheless I apologize in advance if I have missed something obvious. The problem is this; I have a 5-column data frame with about 4.2 million rows, and want to create a new (and hopefully much smaller) data frame that

Hello, Below is the code for a basic bar graph. I was seeking advice regarding the following: (a) For each time period there are values from 16 people. How I can change the colour value so that each person has a different colour, which recurs across each of the three graphs/tie epriods? (b) I have seen much more sophisticated examples using lattice (e.g each person has a separate

Hi all I am running a 2-way repeated measure anova with 1 between-subjects factor (Group=treatment, control), and 1 within-subject factor (Time of measurement: time1, time2). I extract the results of the anova with: summary(aov(effect ~ Group*Time + Error=Subj/Time, data=mydata)) Now, this must be clearly a dumb question, but how can I quickly extract in R all the post-hoc t-tests for the

Hi. I am using Windows version of R 1.8.1. Being somewhat new to survival analysis, I am trying to compare cph (Design) with coxph (survival) for use with a survival data set. I was wondering why cph and coxph provide me with different confidence intervals for the hazard ratios for one of the variables. I was wondering if I am doing something wrong? Or if the two functions are calculating hazard

Further pursuing my curiosity to measure the efficiency of R/C++ interface, I conducted a simple matrix-vector multiplication test using .C and .Call functions in R. In each case, I measured the execution time in R, as well as inside the C++ function. Subtracting the two, I came up with a measure of overhead associated with each call. I assume that this overhead would be non-existent of the entire

Dear All,   #I have the following code   Dose<-1000 Tinf <-0.5 INTERVAL <-8 TIME8 <-matrix(c((0*INTERVAL):(1*INTERVAL))) TIME7 <-matrix(c((0*INTERVAL):(2*INTERVAL))) TIME6 <-matrix(c((0*INTERVAL):(3*INTERVAL))) TIME5 <-matrix(c((0*INTERVAL):(4*INTERVAL))) TIME4 <-matrix(c((0*INTERVAL):(5*INTERVAL))) TIME3 <-matrix(c((0*INTERVAL):(6*INTERVAL))) TIME2