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...et. Frank ------------------------------ Frank E Harrell Jr Professor School of Medicine Department of *Biostatistics* *Vanderbilt University* On Thu, Sep 14, 2017 at 10:48 AM, David Winsemius <dwinsemius at comcast.net> wrote: > > > On Sep 14, 2017, at 12:30 AM, Bonnett, Laura < > L.J.Bonnett at liverpool.ac.uk> wrote: > > > > Dear all, > > > > I am using the publically available GustoW dataset. The exact version I > am using is available here: https://na01.safelinks. > protection.outlook.com/?url=https%3A%2F%2Fdrive.google....

...0.6327 0.2003 3.16 0.0016 I was therefore hoping someone would explain why the "lrm" code is producing an error message, while "lrm.fit" and "glm" do not. In particular I would welcome a solution to ensure I can produce a nomogram. Kind regards, Laura Dr Laura Bonnett NIHR Post-Doctoral Fellow Department of Biostatistics, Waterhouse Building, Block F, 1-5 Brownlow Street, University of Liverpool, Liverpool, L69 3GL 0151 795 9686 L.J.Bonnett at liverpool.ac.uk [[alternative HTML version deleted]]

> On Sep 14, 2017, at 12:30 AM, Bonnett, Laura <L.J.Bonnett at liverpool.ac.uk> wrote: > > Dear all, > > I am using the publically available GustoW dataset. The exact version I am using is available here: https://drive.google.com/open?id=0B4oZ2TQA0PAoUm85UzBFNjZ0Ulk > > I would like to produce a nomogram for 5...

...anad[,3])),varcov=covmat) while(data[i,2]>86) data[i,]<-rmnorm(n=1,mean=c(mean(sanad[,1]),mean(sanad[,2]),mean(sanad[,3])),varcov=covmat) } return(data) } Many thanks for any assistance you can offer. Kind regards, Laura Laura Bonnett Research Assistant Department of Biostatistics University of Liverpool Telephone: 0151 7944059 Email: L.J.Bonnett@liv.ac.uk [[alternative HTML version deleted]]

Dear all, I am using R version 2.9.2 in Windows. I would like to output the results of a function I have written to a .txt file. I know that I can do this by using the code write.table(boothd(10),"boothd10.txt",sep="\t",append=TRUE) etc. However, I would like to bootstrap my function 'boothd' several times and get each vector of results as a new line in my text

Hi Everyone, I have data in a long format e.g. there is one row per patient but each follow-up appointment is included in the row. So, a snippet of the data looks like this: TrialNo Drug Sex Rand Adate1 Date1 Dose1 Time1 Adate2 Date2 Dose2 Time2 B1001029 LTG M 15719 30/04/2003 15825 150 106 29/08/2003 15946 200 227 B1117003 LTG M 15734 30/04/2003 15825 200 91 03/09/2003 15951 250 217

Hi All, This sounds a relatively simple query, and I hope it is! I am looking at a continuous variable, age. I am looking at time to 12-month remission and can calculate the HR and 95% confidence interval are follows: coxfita = coxph(Surv(rem.Remtime,rem.Rcens)~nearma$all.age,data=nearma) exp(coxfita$coefficients) exp(confint(coxfita)) However, because I am looking at age as a continuous

...1021 2993.4531767 0.3445589 The unusually large "probability" arises when there is only 1 row of data for the relevant trial number. Can anyone therefore explain why there is a problem when "sdata" is only 1 row, and ideally provide a solution? Many thanks, Laura Dr Laura Bonnett NIHR Post-Doctoral Fellow Department of Biostatistics, Waterhouse Building, Block F, 1-5 Brownlow Street, University of Liverpool, Liverpool, L69 3GL 0151 795 9686 L.J.Bonnett at liverpool.ac.uk

...N = 422)) and I get following results; Model Chisquare = 12.524 Df = 1 Pr(>Chisq) = 0.00040179 Chisquare (null model) = 812.69 Df = 3 Goodness-of-fit index = 0.98083 Adjusted goodness-of-fit index = 0.885 RMSEA index = 0.16545 90% CI: (0.09231, 0.25264) Bentler-Bonnett NFI = 0.98459 Tucker-Lewis NNFI = 0.9573 Bentler CFI = 0.98577 SRMR = 0.027022 BIC = 6.4789 Parameter Estimates Estimate Std Error z value Pr(>|z|) x1-y1 -0.67833 0.033967 -19.970 0 y1 <--- x1 x2-x1 3.88384 0.293743 13.222 0 x1 <--> x2 x2-x2 12.0...

Dear Sirs, I am using both the Bioconductor adds on (Affy, AffyPLM,...) and the 'standard' R-package. I am trying to select a list of genes which all have expression values below a certain threshold. I have done this by creating a vector which has 0s where the expression is greater than the threshold and 1s where it is less than or equal to it. Multiplying this vector by the expression

Dear all, I have decided after much deliberation to use backward elimination and forward selection to produce a multivariate model. Having read about the problems with choosing selection values I have chosen to base my decisions of inclusion and exclusion on the AIC and am consequently using the stepAIC function. This post however does not relate to whether or not this is the correct decision!

Hi All, I have a data frame which has columns comprised mainly of "NA"s. I know there are functions na.pass and na.omit etc which can be used in these situations however I can't them to work in this case. I have a function which returns the data according to some rule i.e. removal of N in this code: nep <- function(data) { dummy <- rep(0,378) for(i in 1:378){

Dear all, I am running a simulation study to test variable imputation methods for Cox models using R 2.9.0 and Windows XP. The code I have written (which is rather long) works (if I set nsim = 9) with the following starting values. >bootrs(nsim=9,lendevdat=1500,lenvaldat=855,ac1=-0.19122,bc1=-0.18355,cc1=-0.51982,cc2=-0.49628,eprop1=0.98,eprop2=0.28,lda=0.003) I need to run the code 1400

Dear R-help, I am using R-3.3.2 on Windows 10. I teach on a course which has 4 computer practical sessions related to the development and validation of clinical prediction models. These are currently written for Stata and I am in the process of writing them for use in R too (as I far prefer R to Stata!) I notice that predictions made from a Cox model in Stata are based on un-centred variables,

...> FARSCH, x2x1, NA"), I DO get all the usual statistics: Model Chisquare = 1303.7 Df = 1 Pr(>Chisq) = 0 Chisquare (null model) = 8526.8 Df = 6 Goodness-of-fit index = 0.95864 Adjusted goodness-of-fit index = 0.58639 RMSEA index = 0.30029 90% CI: (NA, NA) Bentler-Bonnett NFI = 0.84711 Tucker-Lewis NNFI = 0.082726 Bentler CFI = 0.84712 BIC = 1294.1 My understanding is the you should always put in the correlation between exogenous predictors, but when I do this I don't get fit statistics. Can anyone help me understand what is happening here? Thank...

Dear R-Help, I am using the 'mfp' package. It produces three plots (as I am using the Cox model) simultaneously which can be viewed together using the following code: fit <- mfp(Surv(rem.Remtime,rem.Rcens)~fp(age)+strata(rpa),family=cox,data=nearma,select=0.05,verbose=TRUE) par(mfrow=c(2,2)) plot(fit) They can be viewed separately but the return key must be pressed after each graph

Dear All, Sorry to bother you again. I have a model: coxfita=coxph(Surv(rem.Remtime/365,rem.Rcens)~all.sex,data=nearma) and I'm trying to do a plot of Schoenfeld residuals using the code: plot(cox.zph(coxfita)) abline(h=0,lty=3) The error message I get is: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In sqrt(x$var[i, i] * seval) : NaNs

Dear all, I am using R 3.4.3 on Windows 10. I am preparing some teaching materials and I'm having trouble matching the by-hand version with the R code. I have fitted a Cox model - let's use the ovarian data as an example: library(survival) data(ovarian) ova_mod <- coxph(Surv(futime,fustat)~age+rx,data=ovarian) If I want to make predict survival for a new set of individuals at 100

...val estimate. I did not get any warning messages with the output. RESULTS: Model Chisquare = 1374 Df = 185 Pr(>Chisq) = 0 Chisquare (null model) = 12284 Df = 210 Goodness-of-fit index = 0.903 Adjusted goodness-of-fit index = 0.88 RMSEA index = 0.0711 90% CI: (NA, NA) Bentler-Bonnett NFI = 0.888 Tucker-Lewis NNFI = 0.888 Bentler CFI = 0.902 SRMR = 0.0682 BIC = 51.4 SYNTAX rm(sem.enf.rq) mdl.rq <- specify.model() enf -> law2, NA, 1 enf -> law3, lam2, 1 enf -> law4, lam3, 1...

Dear R-help, I am trying to obtain the baseline survival estimate of a fitted Cox model (S_0 (t)). I know that previous posts have said use 'basehaz' but this gives the baseline hazard function and not the baseline survival estimate. Is there a way to obtain the baseline survival estimate or do I have to use the formula which does something like S(t) = exp[- the integral from 0 to t of