similar to: simple save question

Hello, I create a plot from a coxph object called fit.ads4: plot(survfit(fit.ads4)) plot is located at: https://www.dropbox.com/s/9jswrzid7mp1u62/survfit%20plot.png I also create the following survfit statistics: > print(survfit(fit.ads4),print.rmean=T) Call: survfit(formula = fit.ads4) records n.max n.start events *rmean *se(rmean) median 0.95LCL 0.95UCL 203.0

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

Hi, I'm not sure if this is the proper way to ask questions, sorry if not. But here's my problem: I'm trying to do a bootstrap estimate of the mean for some survival data. Is there a way to specifically call upon the rmean value, in order to store it in an object? I've used print(...,print.rmean=T) to print the summary of survfit, but I'm not sure how to access only rmean

Hello, I need some results from the survival analysis of my data that I do not know whether exist in Survival Package or how to obtain if they do: 1. The Mean survival time 2. The standard error of the mean 3. Point and 95% Lower & Upper Confidence Intervals estimates Any help will be greatly appreciated. Cem [[alternative HTML version

Hi, I'm going crazy trying to plot a quite simple graph. i need to plot estimated hazard rate from a cox model. supposing the model i like this: coxPhMod=coxph(Surv(TIME, EV) ~ AGE+A+B+strata(C) data=data) with 4 level for C. how can i obtain a graph with 4 estimated (better smoothed) hazard curve (base-line hazard + 3 proportional) to highlight the effect of C. thanks!! laudan [[alternative

Dear all, I am confused with the behaviour of survfit with newdata option. I am using the latest version R-2-15-0. In the simple example below I am building a coxph model on 90 patients and trying to predict 10 patients. Unfortunately the survival curve at the end is for 90 patients. Could somebody please from the survival package confirm that this behaviour is as expected or not - because I

Hi, when I generated a survfit() object, I can get number of patients at risk at various time points by using summary(): fit<-survfit(Surv(time,status)~class,data=mtdata) summary(fit) class=1 time n.risk n.event survival std.err lower 95% CI upper 95% CI 9.9 78 1 0.987 0.0127 0.963 1 41.5 77 1 0.974 0.0179 0.940 1 54.0 76

Using R 2.10 on Windows: I have a filtered database of 650k event observations in a data frame with 20+ variables. I'd like to be able to quickly generate estimate and plot survival curves. However the survfit and cph() functions are extremely slow. As an example: I tried results.cox<-coxph(Surv(duration, success) ~ start_time + factor1+ factor2+ variable3, data=filteredData) #(took a

Dear Sirs, I want to estimate the survival mean of a few specific teams. I'm trying to calculate it through a Kaplan Meier estimator. For doing so, I load the "survival" package and run the following instructions: "options(survfit.print.mean=TRUE)" allows showing the mean and mean standard error "KM=survfit(Surv(Dias,Censura))"

Hello, I have estimated a survival model with six strata: >model.b <- survfit(Surv(time=start.tijd,time2=eind.tijd2,event=va)~strata(product.code) , data=wu.wide) I would like to save the output of >print(model.b,print.n="records",show.rmean=FALSE) in a dataframe so that I can export it later. How do I do this? Note that summary(model.b) gives an error: Error in

I would like to be able to construct hazard rates (or unconditional death prob) for many subjects from a given survfit. This will involve adjusting the ( n.event/n.risk) with (coxph object )$linear.predictors I must be having another silly day as I cannot reproduce the linear predictor: fit <- coxph(Surv(futime, fustat) ~ age, data = ovarian) fit$linear.predictors[1] [1] 2.612756

Since the fact that gfortran performance has improved over the major releases, I decided to benchmark the current releases on a MacPro with the Polyhedron 2005 benchmarks using -ffast-math -funroll-loops -msse3 -O3. The results are... gcc release gcc 4.2.4 gcc 4.3.3 gcc 4.4-pre gcc 4.3.3/ gcc 4.4-pre/

Hello all, Here goes one of my first functions. I want to make a nonparametric multiple sample comparison with unequal sample sizes (see Zar?s Biostatistical Analysis, 3rd. Ed., pg. 201 Example 10.11, pg. 288 Example 11.10). In the real world, I want to compare samples of fish length captured with different fishing gears. After using the Kruskal-Wallis test I want to check the differences

Hi all, Using: R version 2.8.1 Patched (2009-03-07 r48068) on OSX (10.5.6) with survival version: Version: 2.35-3 Date: 2009-02-10 I get the following using the first example in ?summary.survfit: > summary( survfit( Surv(futime, fustat)~1, data=ovarian)) Call: survfit(formula = Surv(futime, fustat) ~ 1, data = ovarian) time n.risk n.event survival

Hi everyone. There's been several threads on baseline hazard in Cox model but I think they were all on cumulative baseline hazard, for instance http://tolstoy.newcastle.edu.au/R/help/01a/0464.html http://tolstoy.newcastle.edu.au/R/help/01a/0436.html "basehaz" in package survival seems to do a cumulative hazard. extract from the basehaz function: sfit <- survfit(fit) H

Dear List, Does anybody no how to compare mean survival times for two (more) groups in R? What test statistics should I use? Thank you very much! Joe [[alternative HTML version deleted]]

Dear r-help, I have a matrix, suppose, 10x10, and I need the matrix 5x5, having in each cell a mean value of the cells from the initial matrix. Please, point me to a function in R, which can help me doing that. Digging the documentation and mail archives didn't give me a result. Thank you. --- Best regards, Vladimir mailto:wl at eimb.ru

Dear Thomas, The head of my dataset > head(wsuv) parcel sp time censo treatment species 1 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 2 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 3 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 4 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1

Hi there, This problem has been dogging me for a bit, and I'm trying to figure out why. When running the the subsets function in the leaps library, R is giving me the following error message > lvodsub <- subsets(pred, resp$LVOD) Warning message: XHAUST returned error code -999 in: leaps.exhaustive(a, really.big = really.big) but this still happens if I add the really.big option:

Hi, I want to fit multiple cox models using the coxph() function. To do this, I use a for-loop and save the relevant results in a separate matrix. In the example below, only two models are fitted (my actual matrix has many more columns), one gives a warning message, while the other does not. Right now, I see all the warning message(s) after the for-loop is completed but have no idea which model