similar to: Extended summary

I'm writing an publication in two column format and need to shrink some plots. After increasing the axis labels it does not look nice at all. The y-axis label and tick labels almost touch each other and the x-axis tick labels expand into the plot instead of away from it. Is there a better way than "cex" to control the: 1) font size of axis and tick labels 2) font thickness 3)

Hi all! I've got a datstructure like this: subject week value 1 1 4 2 1 8 3 1 3 4 1 5 1 2 5 2 2 6 3 2 2 4 2 6 1 3 3 2 3 7 3 3 3 4 3 7 I'd like to plot the mean of 'value' against week. Is there a direct way of doing this or do I have to make a new structure with the calculated values and then plot it? All the best! -- ########################################### Kare

I'd like to avoid looping through an array in order to change values in the array as it takes too long. I red from an earlier post it can be done by "do.call" but never got it to work. The Idea is to change the value of "y" according to values in "x". Wherever "x" holds the value 3, the corresponding value in "y" should be set to 1. So I

Dear R-gurus! Is it possible within boxplot to break the y-scale into two (or anything)? I'd like to have a normal linear y-range from 0-10 and the next tick mark starting at, say 60 and continue to 90. The reason is for better visualising the outliers. All the best, Kare [[alternative HTML version deleted]]

R-gurus... I've got a 5 column dataframe where I'd like to plot each ID's "b" against "c" with "b" in ascending order (within the same ID). How do I sort "b" so that the other variables are altered equally? ID a b c d 101 1 240 26.7 21.85 101 2 335 21.8 21.85 101 3 1387 26.6 21.85 101 4 877 24.1 21.85

Hi r-helpers... Why do I get this strange huge jump of 36524 days when changing "origin" from 1969-01-01 to 1968-12-31. It should still be close to zero! This really messes up my calculations of follow-up times in my analyses. > julian(strptime("010169", format = "%d%m%y"),origin = as.Date("1969-01-01")) > Time difference of -0.04166667 days >

Hi all. If I run the simple regression when x is a categorical variable ( x <- factor(x) ): > MyFit <-coxph( Surv(start, stop, event) ~ x ) How can I get the overall p-value on x other than for each dummy variable? > anova(MyFit) does NOT provide that information as previously suggested on the list. All the best, Kare [[alternative HTML version deleted]]

Hello r-help! I have data with two kind of ratings on status of 100 occupations. The first kind of rating is on the percieved "objective" status that these occupations have in society at large, and the second kind or rating is on the status that the respondents think that these occuption *should* have. The ratings were originally integer values in the rage 1-9, but in the current data,

Dear R-helpers, I'm for sure not familiar with R, but it seem like a nice sofware tool, so I've decided to try using it. Here is my problem I just can't figure out: I'd like to do least square fit of a straight horizontal (a = 0) line y = ax + b through some data points x = (3,4,5,6,7,8) y = (0.62, 0.99, 0.83, 0.69, 0.76, 0.82) How would i find b????? All the best, Ked

I am unable to get the basehaz function to apply a proportional hazards model to a new data frame. I replicated my specific situation with the example for coxph in the help, where I changed the x value of the first record from 0 to 1. Is there something incorrect in the syntax that I am using? Thanks in advance! test1 <- list(time= c(4, 3,1,1,2,2,3), status=c(1,NA,1,0,1,1,0),

Hello, When I try to to obtain the expected risk for a new dataset using coxph in the survival package I get an error. Using the example from ?coxph: > test1 <- list(time= c(4, 3,1,1,2,2,3),+ status=c(1,NA,1,0,1,1,0),+ x= c(0, 2,1,1,1,0,0),+ sex= c(0, 0,0,0,1,1,1))> cox<-coxph( Surv(time, status) ~ x + strata(sex), test1)

I stumbled onto this working on an update to coxph. The last 6 lines below are the question, the rest create a test data set. tmt585% R R version 2.12.2 (2011-02-25) Copyright (C) 2011 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-unknown-linux-gnu (64-bit) # Lines of code from survival/tests/singtest.R > library(survival) Loading required package: splines

Dear all, I'm having difficulty getting access to data generated by survfit and print.survfit when they are using with a Cox model (survfit.coxph). I would like to programmatically access the median survival time for each strata together with the 95% confidence interval. I can get it on screen, but can't get to it algorithmically. I found myself examining the source of print.survfit to

I'm testing out some changes to survreg and got the following output, the likes of which I've never seen before: ---------------------------------------------------------------------- R version 2.7.1 (2008-06-23) Copyright (C) 2008 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it

Hi, I'm trying to print the p-values from the output of a CPH test onto a Kaplan Meier plot. Can this be done? I only really want the p-values from the CPH test to appear but if this can't be done I am willing to have the entire CPH output. This is what I am currently trying: (it doesn't print the CPH output) plot_KM <- function(field) { library(survival) y =

Hi Gang, I've got the latest SVN branch of 1.4 downloaded onto SUSE 11.0. Everything is happy EXCEPT, I can't get fax to be recognized by make menuselect. I tried copying app_rxfax.c and app_txfax.c to the apps directory and starting again from ./configure, but no joy. Any suggestions? Danny Nicholas -------------- next part -------------- An HTML attachment was

Hi there, After I fitted a cox model, I used vcov to obtain the variance of the parameter estimate. It worked correctly in R 1.9.1. But it failed in R 2.0.0 and the error message is Error in vcov(cox.1) : no applicable method for "vcov" I don't know if it is a bug or there is some update on this function. Thanks! Lei Liu Assistant Professor Division of Biostatistics and

I'm trying to add 'gplots' from install.packages("gplots", lib = "/usr/lib/R/library", dependencies = TRUE) on a linux host, but R does not seem to figure out there's a new package installed (yes, I've restarted the R-session). I've tried the default, and different library folders, without success. Installation does not report any warnings or bad exit

I'm trying to devide x into tertiles, but ends up with integer limits even x holds one decimal. The analysis is extremely sensitive to the limits and I like to keep them right. How can that be done? quartiles <- quantcut( x[x >= 0], q=seq(0,1, by=(1/3)) > table(quartiles) quartiles [180,344] (344,448] (448,644] 16467 16476 16452 [[alternative HTML version deleted]]

I hope one and all will allow a stats question: When running a cox proportional hazards model ,there are two ways to deal with age, including age as a covariate, or to include age as part of the follow-up time, viz, Age as a covariate: tetest1 <- list(time= c(4, 3,1,1,2,2,3), status=c(1,NA,1,0,1,1,0), age= c(0, 2,1,1,1,0,0),