similar to: validate.lrm - confidence interval for boostrap-corrected AUC ?

Dear List, I am trying to assess the prediction accuracy of an ordinal model fit with LRM in the Design package. I used predict.lrm to predict on an independent dataset and am now attempting to assess the accuracy of these predictions. >From what I have read, the AUC is good for this because it is threshold independent. I obtained the AUC for the fit model output from the c score (c =

Dear Sir or Madam? I am a doctor of urology,and I am engaged in developing a nomogram of bladder cancer. May I ask for your help on below issue? I set up a dataset which include 317 cases. I got the Binary Logistic Regression model by SPSS.And then I try to reconstruct the model ?lrm(RECU~Complication+T.Num+T.Grade+Year+TS)? by R-Project,and try to internal validate the model through

Hi, I'm developing an experiment with logistic regression. I've come across the lrm function in the Design library. While I understand and can use the basic functionality, there are a ton of options that go beyond my knowledge. I've carefully read the help page for lrm, but don't understand many of the arguments and optional return values. (penalty, penalty.matrix,

Dear R-philes, I am plotting ROC curves for several cross-validation runs of a classifier (using the function below). In addition to the average AUC, I am interested in obtaining a confidence interval for the average AUC. Is there a straightforward way to do this via the ROCR package? plot_roc_curve <- function(roc.dat, plt.title) { #print(str(vowel.ROC)) pred <-

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 =

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

Hi, I have a dataset (see attached) with 2 variables "Y" is binary, "x" is a continuous variable. I want to calculate area under the curve (AUC) for the ROC curve, but I got different AUC values using ROC() from Epi package vs. rcorr.cens() from rms package: test<-read.table("test.txt",sep='\t',header=T,row.names=NULL) y<-test$y x<-test$x

Hi all, I have another question about the lrm function (from the rms library) that I cannot find the answer to. I get an error message when I try to fit a model, and I don't know what to make of it. Please forgive me for not having a toy example, but it appears the size and complexity of my data is somehow causing the error. The best I can do is show you what I type and what errors I get.

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')

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 <-

I am using logistic regression model (lrm) of package Design. Can some one please tell me how to calculate the average Area Under Curve (AUC) for n-fold cross-validation The help for lrm function says to do cross validation like this f <- lrm( cy ~ x1 + x2, x=TRUE, y=TRUE) val <- validate.lrm(f, method="cross", B=5) Now I dont know what to do with variable "val" to

I've noticed that glm and lrm give extremely different results if you attempt to fit a saturated model to a dataset with zero cells. Consider, for instance the data from, Agresti's Death Penalty example [0]. The crosstab table is: , , PENALTY = NO VIC DEF BLACK WHITE BLACK 97 52 WHITE 9 132 , , PENALTY = YES VIC DEF BLACK WHITE BLACK 6 11

> 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, 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

Greetings: I'm running R-1.9.1 on Fedora Core 2 Linux. I tested a proportional odds logistic regression with MASS's polr and Design's lrm. Parameter estimates between the 2 are consistent, but the standard errors are quite different, and the conclusions from the t and Wald tests are dramatically different. I cranked the "abstol" argument up quite a bit in the polr

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

Dear List, I have two questions about how to do predictions using lrm, specifically how to predict the ordinal response for each observation *individually*. I'm very new to cumulative odds models, so my apologies if my questions are too basic. I have a dataset with 4000 observations. Each observation consists of an ordinal outcome y (i.e., rating of a stimulus with four possible

Hello All, Despite my preference for reporting confidence intervals, I need to obtain a p-value for a hypothesis test in the context of regression using bootstrapping. I have read John Fox's chapter on bootstrapping regression models and have consulted Efron & Tibshirani's "An Introduction to the Bootstrap" but I just wanted to ask the experts here for some feedback to make

I know how to compute the ROC curve and the empirical AUC from the logistic regression after fitting the model. But here is my question, how can I compute the standard error for the AUC estimator resulting form logistic regression? The variance should be more complicated than AUC based on known test results. Does anybody know a reference on this problem? [[alternative HTML version deleted]]

Does anyone know how to get the C-index from a logistic model - not using the dataset that was used to train the model, but instead using a fresh dataset on the same model? I have a dataset of 400 points that I've split into two halves, one for training the logistic model, and the other for evaluating it. The structure is as follows: column headers are "got a loan" (dichotomous),