similar to: bootcov and cph error

Hello! I have created a bplot-figure using this code: *file <- "2dcali_red.ttt" ux<-as.matrix(read.table(file, dec = ",")) mode(ux)<-'numeric' vel<-ux[,1] ang<-ux[,2] x<-ux[,3] y<-ux[,4] dat<- data.frame(ang=ang, x=x,y=y) require(rms) ddist2 <- datadist(dat) options(datadist="ddist2") fitn <- lrm(ang ~ rcs(x,4) +

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

Dear r-help list, I'm trying to use multiple imputation for my MSc thesis. Having good exemples using the Hmisc package, I tried the aregImpute function. But with my own dataset, I have the following error : Erreur dans matxv(X, xcof) : columns in a (51) must be <= length of b (50) De plus : Warning message: In f$xcoef[, 1] * f$xcenter :   la taille d'un objet plus long n'est pas

I have a quick question, and I apologize in advance if, in asking, I expose my woeful ignorance of R and its packages. I am trying to use the bootcov function to estimate the standard errors for some regression quantiles using a cluster bootstrap. However, it seems that bootcov passes arguments that rq.fit doesn't like, preventing the command from executing. Here is an example:

Hello, I am using R.12.2.0. I am trying to generate bootstrap confidence intervals using bootcov from the rms package. I am able to impute the missing data using aregImpute and to perform a linear regression on the imputed datasets using fit.mult.impute, but I am unable to use bootcov to generate the confidence intervals for the R-squared. Here is a small example that should duplicate the

Dear users, I'm trying to implement the nonparametric "two-stage" bootstrap (Davison and Hinkley 1997, pag 100-102) in R. As far as I understood, 'bootcov' is the most appropriate method to implement NONPARAMETRIC bootstrap in R when you have clustered data (?). I read the 'bootcov' manual but I still have a few questions: 1 - When the variable 'cluster' is

Dear list, I made a logistic regression model (MyModel) using lrm and penalization by pentrace for data of 104 patients, which consists of 5 explanatory variables and one binary outcome (poor/good). Then, I found bootcov and robcov function in rms package for calculation of confidence range of coefficients and odds ratio by bootstrap covariance matrix and Huber-White sandwich method,

Hi, I am trying to figure out exactly what the bootcov() function in the Design package is doing within the context of clustered data. From reading the documentation/source code it appears that using bootcov() with the cluster argument constructs standard errors by resampling whole clusters of observations with replacement rather than resampling individual observations. Is that right, and is

SSweibull() :  problems with step factor and singular gradient Hello I am working with growth data of ~4000 tree seedlings and trying to fit non-linear Weibull growth curves through the data of each plant. Since they differ a lot in their shape, initial parameters cannot be set for all plants. That’s why I use the self-starting function SSweibull(). However, I often got two error messages:

John, You can make a local version of bootcov which either: deletes these arguments from the call to fitter, or modify the switch statement to include rq.fit, the latter would need to also modify rq() to return a fitFunction component, so the first option is simpler. One of these days I'll incorporate clustered se's into summary.rq, but meanwhile this seems to be a good alternative.

Hi, I am having trouble with the survival library, particualrily the coxph function. the following works coxph(jtree9$cph.call,z,rep(1,dim(z)[1])) Call: coxph(formula = jtree9$cph.call, data = z, weights = rep(1, dim(z)[1])) coef exp(coef) se(coef) z p SM 0.2574 1.294 0.0786 3.274 1.1e-03 Sex -0.1283 0.880 0.1809 -0.709

Package: logcheck Version: 1.2.51 Severity: normal Logcheck has an implicit assumption that the default locale should be used by grep when processing log files. However, that's not always the case. For example, I use the locale "en_US.UTF-8", and consequently grep assumes that its inputs are encoded as UTF-8. But the log files appear to be encoded as ISO 8859-1, which means that

Hi folks, Using R Version 1.6.2 (2003-01-10) on SuSE Linux 7.2, I just installed Hmisc_1.5-3.tar.gz and Design_1.1-5.tar.gz These were taken from http://hesweb1.med.virginia.edu/biostat/s/library/r Checked the dependencies: Hmisc: grid, lattice, mva, acepack -- all already installed Design: Hmisc, survival -- survival already installed, so installed Hmisc first All seems to go

Dear list members, I am looking for a goodness of test that will tell me if a sample is likely to have come from a standard normal distribution. I can find plenty of omnibus tests for normality in the nor.test package, but none of them appear to allow me to test against the specific alternative that the data are not standard normal. My back up option is to use a Kolmogorov-Smirnov test, but my

Hi all, I want to create a simple plot with 2 type='s' lines on it: plot(a, b, type='s') lines(x, y, type='s') I wish to then fill the area between the curves with a colour to accentuate the differences eg col=gray(0.95). I cant seem to come up with a simple method for this. Any pointers in the right direction much appreciated. Cheers Scott

I am trying to build a model to aid a clinical decision. Certain patients have a blood marker measured at each visit - a rise of this may indicate recurrence of the cancer after treatment (endpoint is "clinical recurrence", censored). In a proportion (up to 30%), this rise is a false positive - hence I wish to correlate factors at the time of the rising test to clinical recurrence,

Hi, I have fitted a cox model with time-varying covariates (counting process style) using the cph function of the Design package. Now I want to know the survival probabilities at each time point given the history of a single individual. I know the survest function, but I am not sure how to interpretet its output when using time-varying covariates. Does it just give the probabilities as if it

Hi, I am trying to run a conditional logistic model on a nested case-control study using cph() and then estimate survival based on the model. The data came from Prof Bryan Langholz website where he also has the SAS code to this, so I am trying to replicate the SAS results. The data attached. Basically, the variables are: rstime: risk set age rsentry: fake entry time, just before rstime setno:

Sorry everybody, temporary dyslexia. Sent from my BlackBerry wireless smartphone

You wrote: "It may be best to either write to the package maintainer (me, as you did) or post to the group but not both." This is just a note that I disagree wrt my own packages: I go on vacation or trips, or have other projects so won't always answer Other folks on the list often have good ideas that I'd miss My preferred standard is "ask the list, with a copy