similar to: Reporting Kaplan-Meier / Cox-Proportional Hazard Standard Error, km.coxph.plot, survfit.object

Hi All, Please pardon me if I am missing something obvious here. How do I get the Kaplan-Meier estimate function that is created by survfit and plotted by the code. fit <- survfit(Surv(time, status) , data=aml) plot(fit) That is, I need a function that will give me the survival estimate at a given time: \hat{S}(t). Thanks in advance. Ritwik Sinha ritwik.sinha at gmail.com | +12033042111 |

When I try to the code from library(survival) of library(ISwR), the following code survfit(Surv(days,status==1)) that could produce Kaplan-Meier estimates shows the following error "Error in survfit(Surv(days, status == 1)) : Survfit requires a formula or a coxph fit as the first argument" How it can be done in R.2.10 -- View this message in context:

I am attempting to model data with the following variables: timepoint - n=48, monthly over 4 years hospital - n=3 opsn1 - no of outcomes total.patients skillmixpc - skill mix percentage nurse.hours.per.day Aims To determine if skillmix affects rate (i.e. no.of.outcomes/total.patients). To determine if nurse.hours.per.day affects rate. To determine if rates vary between

Hi, I'm having difficulty importing my textfile that looks something like this: #begin text file Timepoint 1 ObjectNumber Volume SurfaceArea 1 5.3 9.7 2 4.9 8.3 3 5.0 9.1 4 3.5 7.8 Timepoint 2 ObjectNumber Volume SurfaceArea 1 5.1

Hi Folks, I have repeated measures for data on association time (under 2 acoustic condtions) in male and female frogs as they grow to adulthood (6 timepoints). Thus, two within-subject variables (Acoustic Condition: 2 levels, Timepoint: 6 levels) and one between-subject variable (Sex:male or female). I am pretty sure my distributions depart from normality but I would first like to simply run a

Hello everyone, I am a new user of R. Does anybody know how hazard ratios are extracted for each factor level in a stratified Cox proportional hazard regression model? I have a cancer data set where the variable ?differentiation? is a factor with three levels: poor, intermediate and good. I would like to extract the hazard ratio for each grade level and relate it to another prognostic factor.

He, I used the "coxph" function, with four covariates. Let's say something like that > model.1 <- coxph(Surv(Time,Event)~X1+X2+X3+X4,data=DATA) So I obtain the 4 coefficients B1,B2,B3,B4 such that h(t) = h0(t) exp(B1*X1+ B2*X2 + B3*X3 + B4*X4). When I use the function on the same data > predict.coxph(model.1,type="lp") how it works in making the prediction?

I am using a syntax like coxph(Surv(start, stop, event) ~ X, data) to estimate the effect of X, which may change at each measurement (every 6 months). Is there anyone who knows a way to test the proportional hazard assumption in that case? Thank you in advance Jean-François Boudreau Sherbrooke University [[alternative HTML version deleted]]

Hi, Here is the summary-output of the Coxph-model I used (the output is based on the best final model i.e. all significant explanatory variables and their interactions are included): coxph(formula = Y ~ LT + Food + Temp2 + LT:Food + LT:Temp2 + Food:Temp2 + LT:Food:Temp2) n= 555 coef exp(coef) se(coef) z

Hi all, I wish to do build discrete choice model to analyse habitat selection of wolverines. This can be done with a 'tricked' stratified Cox proportional hazard model. For each individual animal each selected position, and possible alternative non-used available positions are combined into strata. This means that one stratum contains a set of 1 used position and several positions which

Hi all, Is there an equivalent to predict(...,type="linear") of a Proportional hazard model for a Cox model instead? For example, the Figure 13.12 in MASS (p384) is produced by: (aids.ps <- survreg(Surv(survtime + 0.9, status) ~ state + T.categ + pspline(age, df=6), data = Aidsp)) zz <- predict(aids.ps, data.frame(state = factor(rep("NSW", 83), levels =

Hi, I'm new to R and using it for Cox survival analysis. Thanks to this great forum I learned how to compute the HR with its confidence interval. My question would be: Is there any way to get the p-value for a hazard ratio in addition to the confidence interval? Thanks, Thierry -- Thierry Panje Visiting Student Researcher Department of Psychology Stanford

I am executing a Repeated Measures Analysis of Variance with 1 DV (LOCOMOTOR RESPONSE), 2 Within-Subjects Factors (AGE, ACOUSTIC CONDITION), and 1 Between-Subjects Factor (SEX). Does anyone know whether the between-subjects factor (SEX) belongs in the Error Term of the aov or not? And if it does belong, where in the Error Term does it go? The 3 possible scenarios are listed below: e.g., 1.

Dear friends, I have used R for some time now and have a tricky question about the coxph-function: To sum it up, I am not sure whether I can use coxph in conjunction with missing covariate data in a model with time-variant covariates. The point is: I know how "old" every piece that I oberserve is, but do not have fully historical information about the corresponding covariates. Maybe you

dear list, i am using a multifactorial design with two treatments (factor A: drugs, three levels; factor B: theraphy, two levels) and a time factor (three levels, different timepoint). hypothetically, i measured the same subjects for all treatements and timepoints, so its a repeated measurement design. now i ran an anova in R and also some Tukey post-hoc tests using glht. but what i am actually

Hello, I have a number of files output1.dat, output2.dat, ... , output20.dat, each of which monitors several variables over a fixed number of timepoints. From this I want to create a data frame which contains the mean value between all files, for each timepoint and each variable. The code below works, but it seems like I should be able to do the second part without a for loop. I played

dear list, i am running a within-design ANOVA with 4 factors (4,4,2 and 2 levels each). the last one is a time factor comprising two different treatment timepoints. i fit a mixed-effects model using lme and apply the anova function to the outcome. according to this analysis, there are highly significant main effect on the first and the time factor. i then checked the boxplots for the two 4-level

library(ggplot2) a<- read.table("data", header=T) b = na.omit(a) ggplot(data=b) + geom_line(aes(x=timepoint, y=value,group=sample, colour= factor(sample))) +? geom_point(aes(x=timepoint, y=value, group=s ample)) + facet_wrap(~bio, scales = "free",ncol = 5) +theme_bw() + opts(legend.direction = "horizontal",??? legend.position = "top",????

Hi useRs, Many studies of the link between red meat and colorectal cancer use Cox proportional hazards with (among other things) a gender covariate. If it is true that men eat more red meat, drink more alcohol and smoke more than women, and if it is also true that alcohol and tobacco are known risk factors then why does it make sense to "adjust" for gender? I would think that in this

Hello, I need to implement variable selection procedures (such as stepwise and backward selection) in Cox proportional hazards model, but can't seem to find an R or S-plus command for these procedures. I am aware that these can be done in SAS. I would appreciate help from anyone who knows how to implement these procedures in Cox models using S-plus or (preferably) R. Thanks! Regards,