similar to: prevalence in logistic regression lrm()

Hi R, I am getting this error while trying to use 'lrm' function with nine independent variables: > res = lrm(y1994~WC08301+WC08376+WC08316+WC08311+WC01001+WC08221+WC08106+WC0810 1+WC08231,data=y) singular information matrix in lrm.fit (rank= 8 ). Offending variable(s): WC08101 WC08221 Error in j:(j + params[i] - 1) : NA/NaN argument Now, if I take choose only four

Hi, I am trying to run a simple logistic regression using lrm() to calculate a odds ratio. I found a confusing output when I use summary() on the fit object which gave some OR that is totally different from simply taking exp(coefficient), see below: > dat<-read.table("dat.txt",sep='\t',header=T,row.names=NULL) > d<-datadist(dat) > options(datadist='d')

Hello all, Could someone tell me what I am doing wrong here? I am trying to fit a binary logistic model using the lrm function in Design. The dataset I am using has a dichotomous response variable, 'covered' (1-yes, 0-no) with explanatory variables, 'nepall', 'title', 'abstract', 'series', and 'author1.' I am running the following script and

Hallo! I want to understand / recalculate what is done to get the CI of the logistic regression evaluated with lrm. As far as I came back, my problem is the variance-covariance matrix fit$var of the fit (fit<-lrm(...), fit$var). Here what I found and where I stucked: ----------------- library(Design) # data D<-c(rep("a", 20), rep("b", 20)) V<-0.25*(1:40) V[1]<-25

Hello, I'm an MD working in an eye clinic. I'm learning by myself to use R for use in my research works and for implementation in a software project. There are some authors who recomends the use of Cox regression as a substitute for Logistic regression (<a href="http://www.biomedcentral.com/1471-2288/3/21.pdf"> Barros AJD, Hirakata VN. BMCMedical Research Methodology, 2003;

I am interested in a model diagnostic for logistic regression which is normally distributed (much like the residuals in linear regression with are ~ N(0,variance unknown). My understanding is that most (all?) of the residuals returned by residuals.lrm {design} either don't have a well defined distribution or are distributed as Chi-Square. Have I overlooked a residual measure or would it be

Hello I'm using logistic regression from the Design library (lrm), then fastbw to undertake a backward selection and create a reduced model, before trying to make predictions against an independent set of data using predict.lrm with the reduced model. I wouldn't normally use this method, but I'm contrasting the results with an AIC/MMI approach. The script contains: # Determine full

Hi all! I'm new to this forum so please excuse me if I don't conform perfectly to the protocols on this board! I'm trying to get an estimate of true prevalence based upon results from an imperfect test. I have various estimates of se/sp which could inform my priors (at least upper and lower limits even if with a uniform distribution) and found the following code on this website..

> 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 covariates - AGE, HYP, KILLIP, HRT and ANT. I have

Hi all I have run into a case where I don't understand why predict.lrm and predict.glm don't yield the same results. My data look like this: set.seed(1) library(Design); ilogit <- function(x) { 1/(1+exp(-x)) } ORDER <- factor(sample(c("mc-sc", "sc-mc"), 403, TRUE)) CONJ <- factor(sample(c("als", "bevor", "nachdem",

Fixed 'maxiter' in the help file. Thanks. Please give the original source of that dataset. That dataset is a tiny sample of GUSTO-I and not large enough to fit this model very reliably. A nomogram using the full dataset (not publicly available to my knowledge) is already available in http://biostat.mc.vanderbilt.edu/tmp/bbr.pdf Use lrm, not lrm.fit for this. Adding maxit=20 will

Hi R-helpers I have a dataframe DF (lets say with the variables, y, x1, x2, x3, ..., clust) containing relatively many NAs. When I fit an ordinal regression model with the function lrm from the Design library: model.lrm <- lrm(y ~ x1 + x2, data=DF, x=TRUE, y=TRUE) it will by default delete missing values in the variables y, x1, x2. Based on model.lrm, I want to apply the robust covariance

Hi, I am running a logistic regression model using lrm library and I get the following error when I run the command: mod1 <- lrm(death ~ factor(score), x=T, y=T, data = env1) Unable to fit model using ?lrm.fit? where score is a numeric variable from 0 to 6. LRM executes fine for the following commands: mod1 <- lrm(death ~ score, x=T, y=T, data = env1) mod1<- lrm(death ~

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 covariates - AGE, HYP, KILLIP, HRT and ANT. I have successfully fitted a logistic regression model using the "glm" function as shown below. library(rms) gusto <-

Hi, I've come across a strange error when using the lrm.fit function and the subsequent predict function. The model is created very quickly and can be verified by printing it on the console. Everything looks good. (In fact, the performance measures are rather nice.) Then, I want to use the model to predict some values. I get the following error: "fit was not created by a Design

I am trying to obtain the AICc after performing logistic regression using the Design package. For simplicity, I'll talk about the AIC. I tried building a model with lrm, and then calculating the AIC as follows: likelihood.ratio <- unname(lrm(succeeded~var1+var2,data=scenario,x=T,y=T)$stats["Model L.R."]) #Model likelihood ratio??? model.params <- 2 #Num params in my model AIC

Hi, I have to build a logistic regression model on a data set that I have. I have three input variables (x1, x2, x3) and one output variable (y). The syntax of lrm function looks like this lrm(formula, data, subset, na.action=na.delete, method="lrm.fit", model=FALSE, x=FALSE, y=FALSE, linear.predictors=TRUE, se.fit=FALSE, penalty=0, penalty.matrix, tol=1e-7,

Hi, A error message arose while I was trying to fit a ordinal model with lrm() I am using R 2.8 with Design package. Here is a small set of mydata: RC RS Sex CovA CovB CovC CovD CovE 2 1 0 1 1 0 -0.005575280 2 2 1 0 1 0 1 -0.001959580 2 3 0 0 0 1 0 -0.004725880 2 0 0 0 1 0 0 -0.005504850 2 2 1 1 0 0 0 -0.003880170 1 2 1 0 0 1 0 -0.006074230 2 2 1 0 0 1 1 -0.003963920 2 2 1 0 0 1 0

Hello everybody, I am trying to do a logistic regression model with lrm() from the design package. I am comparing to groups with different medical outcome which can either be "good" or "bad". In the help file it says that lrm codes al responses to 0,1,2,3, etc. internally and does so in alphabetical order. I would guess this means bad=0 and good=1. My question: I am trying to

Hi List, I don't understand why 'lrm' doesn't recognize the '~.' formula. I'm pretty sure it was working before. Please see below: I'm using R2.3.0, WinXP, Design 2.0-12 thanks, ...Tao > dat <- data.frame(y=factor(rep(1:2,each=50)), x1=rnorm(100), x2=rnorm(100), x3=rnorm(100)) > lrm(y~., data=dat, x=T, y=T) Error in terms.formula(formula, specials =