similar to: Two ways to deal with age in Cox model

Hi all, I have a relationship (no really!) class RiskMatrix < ActiveRecord::Base has_many :severities, :order => :position, :uniq => true end class RiskFactor < ActiveRecord::Base belongs_to :risk_matrix validates_presence_of :descriptor, :example validates_uniqueness_of :descriptor, :example, :scope=> :risk_matrix_id end class Severity < RiskFactor

Hello all, After some months doing ok with R, I am embarrassed that I have to make this my first posting to the help list. I am trying to run the following (actually in a loop but shortened for the post): risk.factors <- c("file$A", "file$B", "file$C", "file$D", "file$E") table(paste(risk.factors[1])) but run into problems

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 know that mvrnorm from MASS (generously provided by Profs. Venables and Ripley) can be used to generate multivariable normal data that can be used in a linear regression with certain desired characteristics (e.g. a given mean for each variable as well as a given variance-covariance pattern). Is there any similar facility that can be used to generate data for (1) a logistic regression and (2) a

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

You asked about survival curves with age scale versus follow-up scale. > fit1 <- coxph(Surv(time/365.25, status) ~ t5 + id + age, data=stanford2) > surv1<- survfit(fit1) > surv1 n events median 0.95LCL 0.95UCL 157.000 102.000 1.999 0.898 3.608 > summary(surv1, times=3) time n.risk n.event survival std.err lower 95% CI upper 95% CI 3 46 85 0.451

Dear R users, My question is more methodology related rather than specific to R usage. Using time on study as time in a cox model, eg: library(Design) stanf.cph1=cph(Surv(time, status) ~ t5+id+age, data=stanford2, surv=T) #In this case the 1000-day survival probability would be: stanf.surv1=survest(stanf.cph1, times=1000) #Age in this case is a covariate. #I now want to compare the above

I would like to plot 3-dimensional data on a two-dimensional scatter-plot. Is there a way I can automatically modify the plot symbol (e.g. changing size or color) to indicate the value of a third variable? E.g. How can I plot weight vs. age and indicate the value of muscle mass for each value weight-age pair by making the plot point proportional to the subject's muscle mass? Thanks, John

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 =

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

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 am hoping for help with a genetic analysis. I am trying to perform an analysis of the relation between genes at a given locus (rs2304795) and a phenotypic trait (zerotg). Multiple subjects are recruited from each family (and so share a part of their genome and are correlated). Family groups are identified by the variable FAMILY. For each subject multiple measurements are made of the trait of

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

I am using read.csv to read a CSV file (produced by saving an Excel file as a CSV file). The columns containing dates are being read as factors. Because of this, I can not compute follow-up time, i.e. Followup<-postDate-preDate. I would appreciate any suggestion that would help me read the dates as dates and thus allow me to calculate follow-up time. Thanks John John Sorkin M.D., Ph.D. Chief,

I want to place four plots on a page, and I would like to have all four plots share a common title. I have tried the following code, but the title is centered over the fourth graph and not centered across all four plots. Does anyone have any suggestions? R 2.1.1 windows xp oldpar<-par(mfcol =c(1,4),ask=TRUE) plot(p,varp) plot(p,SEp) plot(p,CVp) plot(p,ppval) title(paste("P and 95%CI

I am having trouble selecting rows of a dataframe that will be included in a regression. I am trying to select those rows for which the variable Meno equals PRE. I have used the code below: difffitPre<-lm(data[,"diff"]~data[,"Age"]+data[,"Race"],data=data[data[,"Meno"]=="PRE",]) summary(difffitPre) The output from the summary indicates that

R 2.1.1 Win 2k Would someone suggest a method (or methods) that can be used to determine ntile cutpoints of a vector, i.e. to determine values that can be used to divide a vector into thirds such as 0-33 centile, 34-66 centile, 67-100 centile. If for example I had a vector: 1,2,3,4,5,6,7,8,9 and wanted to divide the vector into thirds I would have cut-points of 3, and 6. Thanks, John John

An apparatus exists whereby a collection of balls is displaced to the top of a stack by suction. A top level (Level 1) each ball is shifted 1 unit to the left or 1 unit to the right at random with equal probability. The ball then drops down to level 2. At Level 2, each ball is again shifted 1 unit to the left or 1 unit to the right at random. The process continues for 15 levels and the balls are

I am trying to model data in which subjects are followed through time to determine if they fall, or do not fall. Some of the subjects fall once, some fall several times. Follow-up time varies from subject to subject. I know how to model time to the first fall (e.g. Cox Proportional Hazards, Kaplan-Meir analyses, etc.) but I am not sure how I can model the data if I include the data for those