similar to: Readjusting the OUTPUT csv file

Hello, I would like to use the parfm package: https://cran.r-project.org/web/packages/parfm/parfm.pdfhttps://cran.r-project.org/web/packages/parfm/parfm.pdf in my work. This package fits parametric frailty models to survival data. To ensure I was using it properly, I started by running some small simulations to generate some survival data (without any random effects), and analyse the data using

Dear R helpers, I have a data.frame as given below - dat1 = data.frame(date = as.Date(c("3/30/12","3/29/12","3/28/12","3/27/12","3/26/12", "3/23/12","3/22/12","3/21/12","3/20/12", "3/30/12","3/29/12","3/28/12","3/27/12",

hi there I have a dataframe abc 123 345 abc 345 456 lmn 567 345 hkl 568 535 lmn 096 456 lmn 768 094 i want the uniques of column 1 and there corresponsing column 2 and 3 output abc 123 345 lmn 567 345 hkl 568 535 cbind(DF1[,1],DF1[which(unique(DF1[,1]),c(2,3)]) but didnt work kindly let me know how to go abt it ramya -- View this message in context:

Dear R helpers Suppose I have a vector as vect1 = as.character(c("ABC", "XYZ", "LMN", "DEF")) > vect1 [1] "ABC" "XYZ" "LMN" "DEF" I want to reverse the order of this vector as vect2 = c("DEF", "LMN", "XYZ", "ABC") Kindly guide Regards Vincy [[alternative HTML

Hello, I have a question concerning fitting a cox model with a random intercept, also known as a frailty model. I am using both the coxme package, and the frailty statement in coxph. Often 'shared' frailty models are implemented in practice, to group people who are from a cluster to account for homogeneity in outcomes for people from the same cluster. I am more interested in the classic

I have a function that fits polynomial models for the orders in n: lmn <- function(d,n){ models=list() for(i in n){ models[[i]]=lm(y~poly(x,i),data=d) } return(models) } My data is: > d=data.frame(x=1:10,y=runif(10)) So first just do it for a cubic: > mmn = lmn(d,3) > predict(mmn[[3]]) 1 2 3 4 5 6 7 8

Hi there, I have a dataframe length.unique.info > length.unique.info abc 12 345 def 16 550 lmn 6 600 I want those names that fall under the condition (length.unique.info[,2][i] <=5 && length.unique.info[,3][i] >=500) abcder<-length.unique.info[which(length.unique.info[,2][i] <=5 && length.unique.info[,3][i] >= 500),1] will "&&" look for

hi there, I have a list of list objects i need to remove the top layer [[1]] [1].0 "ABC" "DEF""LMN" [1].1 "WER" "ERT" "TRY" [[2]] [2].0 "ASD","wer""qwe" [2].1 "wdv""ghj""ggj" I wanna avoid the top layer...that is [[1]] [[2]] shouldnt be there just a simple list is wat i need.

Hi there I have two dataframes Dataframe_1 column_1 colum_2 121 12345 145 1675 167 2765 Dataframe_2 column_1 column2 121 abc 345 lmn 167 efg I want a resulting dataframe 121 12345 abc 167 2765 efg how do i go abt it Ramya -- View this message in context: http://n4.nabble.com/Dataframe-help-tp947934p947934.html Sent from the R

Hi everybody, I'm fairly new to survival analysis with R and have some questions how to apply and interpret the coxph and related functions: I have time-dependent covariates with several measurements per subject with constant delta t. The covariates change in each time step. I fitted the following model: fit <- coxph(Surv(start, stop, event) ~ ratePO + rateC + BLamp + BLP80 +

Dear R Forum, I have a data.frame as df = data.frame(date = c("2013-04-15", "2013-04-14", "2013-04-13", "2013-04-12", "2013-04-11"), ABC_f = c(62.80739769,81.04525895,84.65712455,12.78237251,57.61345256), LMN_d = c(21.16794336,54.6580401,63.8923307,87.59880367,87.07693716), XYZ_p = c(55.8885464,94.1358684,84.0089114,98.99746696,64.71083712),

Dear R-help, I think I have kinda isolated the problem I had to the following: i <- 0 while( {i <- i + 1} < 5) { if(i < 3) next print(i) } This seems to go into an infinite loop. After I break the execution, i has the value 1. At the R prompt, if I start from i <- 0 and keep typing {i <- i + 1} < 5, it eventually evaluate to TRUE. So why does the while loop not work?

Hi all, How can I add or modify a route in the kernel? It seems ip_rt_ioctl() does the job. I read the code but got one question. A route is in a route table, so we have to find the table number (from 0 to 255) first. ip_rt_ioctl() called fib_convert_rtentry() to fill the fields of the req.rtm, such as the rtm_protocol, rtm_scope, rtm_type, etc. But where is rtm_table is assigned value? Or is

Prof. Gentleman (and R-help), Here's an example of what didn't work. I still don't understand why. Function: trycon <- function(file, n) { f.con <- file(file, open="rt") on.exit(close(f.con)) i <- 0 while( length(readln <- readLines(f.con, 1)) > 0 ) { x <- unlist(strsplit(readln, " ")) if(length(x) <= 6 && x[3] ==

Prof. Tierney, Thanks very much for the info. Why does the loop work if I move the assignment out of the condition? E.g., the following works: i <- 0 while(i < 5) { i <- i + 1 if(i < 3) next print(i) } Regards, Andy > -----Original Message----- > From: Luke Tierney [mailto:luke at nokomis.stat.umn.edu] > Sent: Tuesday, October 02, 2001 3:36 PM > To: Liaw, Andy

Hello ! I am having troubles setting up Busy Light Fields (BLF) in asterisk 1.4.18 The things work up to 80%, I can transfer the call by BLF button and I can see the green (free) status and red (busy) status. What I cannot do is to accept the call when someone rings a remote extension. The BLF button starts to blink in red telling me that the call is ringing on remote extenson, but if I press it,

Hello list. I'm trying to rebuild the 2.6.9.67.0.20.EL kernel, but it fails even without modifications. How did I try it? Created a (non-root) build environment (not a mock ) Installed the kernel.scr.rpm and did a rpmbuild -ba --target=`uname -m` kernel-2.6.spec 2> prep-err.log | tee prep-out.log The build failed at the end: Processing files: kernel-xenU-devel-2.6.9-67.0.20.EL Checking

Signed-off-by: Scott Seago <sseago at redhat.com> --- AUTHORS | 17 ++++++ README | 10 +++ conf/ovirt-agent | 12 ++++ conf/ovirt-db-omatic | 12 ++++ conf/ovirt-host-browser | 12 ++++