similar to: KM estimation for interval censoring?

in R when carring out the log rank test is the censored variable denoted by 1 or 0 or its of no consequence. thanks --------------------------------- always stay connected to friends. [[alternative HTML version deleted]]

Usually the y-axis is shown on the left-hand-side of a graph, is it possible to artifically creat one more y-axis on the right-hand-side in R? What is the main reference? Thank you in advance. [[alternative HTML version deleted]]

I have the following data ID x y time 1 10 20 0 1 10 30 1 1 10 40 2 2 12 23 0 2 12 25 1 2 12 28 2 2 12 38 3 3 5 10 0 3 5 15 2 ..... x is time invariant, ID is the subject id number, y is changing over time. I want to find out the difference between the first and last observed y value for each subject and get a table like ID x y 1 10 20 2 12 15 3 5 5 ...... Is there any easy way to generate

So, I want to build the greatest online radio ever! (for my, currently crappy, faderwave.net radio station. Right now, I''m running Icecast straight up. The only problem is that managing different DJs and keeping them off the air when they aren''t supposed to be there is a pain. There also is not one single management package that does everything. My current system is a hacked

I have the complete data like id time censor 1 10 0 1 20 0 1 30 0 2 10 0 2 20 1 2 30 0 2 40 0 3 10 0 3 20 0 3 30 1 .... for id 1, i want to select the last row since all censor indicator is 0; for id 2, i want to select the row where censor ==1; for id 3, i also want to select the row where censor==1. So if there is a 1 for censor, then I want to select such a row, otherwise I want to select the

Dear all, I am looking for a good reference on "Survival analysis". I am looking for a booking containing both applications and Maths. Explaining different methods in survival analysis .... Many thanks Bernard --------------------------------- [[alternative HTML version deleted]]

find the position of the first value who equals certain number in a vector: Say a=c(0,0,0,0,0.2, 0.2, 0.4,0.4,0.5) i wish to return the index value in a for which the value in the vector is equal to 0.4 for the first time. in this case, it is 7. [[alternative HTML version deleted]]

Dear r-helpers, I have a very simple question. Suppose my data is like id=c(rep(1,2),rep(2,2)) b=c(2,3,4,5) m=cbind(id,b) > m id b [1,] 1 2 [2,] 1 3 [3,] 2 4 [4,] 2 5 I wish to select the first observation for each id. That is, I want to quickly select two rows: id b 1 2 2 4 only. how should i do this? [[alternative HTML version deleted]]

Has anybody written an actuarial (life) survival procedure, this does not appear to be an option in the survival package? This approach is common in orthopaedic surgery to demonstrate the survival of prostheses. I need to apply the "modified" lower conf.int because of the censoring over time. I want a life table which I can then easily plot. Many thanks

say, I am plotting x=seq(0,5,len=100) y=-(x-5)^2 plot(x,y) how can I put some color or verticle lines below the plotted curve? [[alternative HTML version deleted]]

i have a data set with repeated measures on same people, structure like below: id x1 x2 ... 001 10 20 ... 001 8 45 ... 001 4 2 ... 002 .... 002 ... 002 .... 002 .... 003 .... ....... what is the easist way to show how many observations for each subject id? [[alternative HTML version deleted]]

I am trying to plot a graph but the points on the graph should be different symbols and colors. It should represent what is in the legend. I tried using the points command but this does not work. Is there another command in R that would allow me to use different symbols and colors for the points? Thank you kindly. data(mtcars) plot(mtcars$wt,mtcars$mpg,xlab= "Weight(lbs/1000)",

i have converted my data into date format like below: > day=as.Date(originaldate,"%m/%d/%Y") > day[1:5] [1] "2008-04-12" "2011-07-02" "2011-09-02" "2008-04-12" "2008-04-12" I wish to select only those observations from 2007 to 2009, how can I select from this list? [[alternative HTML version deleted]]

I have two vectors for values collected from a group of subjects, say a=c(100,200,150,120,140,180) b=c(200,300,420,130) I also have two vectors which indicate the corresponding subjects for a and b, say for a, the subjects are suba=c(1,2,3,4,5,6) for b, the subjects are subb=c(1,3,5,6) Then, I want to find out the paired values from a and b, such as (100, 200) (from subject 1), (150,300)

I have the following list of observations of calendar time: [1] 03-Nov-1997 09-Oct-1991 27-Aug-1992 01-Jul-1994 19-Jan-1990 12-Nov-1993 [7] 08-Oct-1993 10-Nov-1982 08-Dec-1986 23-Dec-1987 02-Aug-1995 20-Oct-1998 [13] 29-Apr-1991 16-Mar-1994 20-May-1991 28-Dec-1987 14-Jul-1999 27-Nov-1998 [19] 09-Sep-1999 26-Aug-1999 20-Jun-1997 05-May-1995 26-Mar-1998 15-Aug-1994 [25] 24-Jun-1996 02-Oct-1996

Hello all, I was just wondering if anybody knew whether mechanize is supposed to be thread-safe or not? I didn''t really find any information about it anywhere. I''ve been getting a strange error in protocol.rb when I run a script that uses mechanize in a multi threaded fashion, but not with a single thread. I''m trying to write a spider that does multiple gets in

Currently, I am doing it this way. x <- mtcars$mpg h<-hist(x, breaks=10, col="red", xlab="Miles Per Gallon", main="Histogram with Normal Curve") xfit<-seq(min(x),max(x),length=40) yfit<-dnorm(xfit,mean=mean(x),sd=sd(x)) yfit <- yfit*diff(h$mids[1:2])*length(x) lines(xfit, yfit, col="blue", lwd=2) But since, ggplot2 has more appealing

Dear all, This is a rather lengthy message, but I don't know what I made wrong in my real example since the simple code works. I have two variables a, b and a function f for which I would like to calculate all possible combinations of the values of a and b. If f is multiplication, I would simply do: a <- 1:5 b <- 1:5 outer(a,b) ## A bit more complicated is this: f <-

Dear all, This is a rather lengthy message, but I don't know what I made wrong in my real example since the simple code works. I have two variables a, b and a function f for which I would like to calculate all possible combinations of the values of a and b. If f is multiplication, I would simply do: a <- 1:5 b <- 1:5 outer(a,b) ## A bit more complicated is this: f <-

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