similar to: How to assign fixed beta coefficients in lrm for external validation

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 <-

> 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

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 all, I'm using the lrm function from the package "Design", and I want to extract the p-values from the results of that function. Given an lrm object constructed as follows : fit <- lrm(Y~(X1+X2+X3+X4+X5+X6+X7)^2, data=dataset) I need the p-values for the coefficients printed by calling "fit". fit$coef (gives a list of only the coefficients) fit$pval, fit$p,

Dear list, I'm currently trying to use the rms package to get predicted ordinal responses from a conditional ratio model. As you will see below, my model seems to fit well to the data, however, I'm having trouble getting predicted mean (or fitted) ordinal response values using the predict function. I have a feeling I'm missing something simple, however I haven't been able to

Sent this mail in rich text format before. Excuse me for this. ------------------------ Dear all, I'm using the lrm function from the package "Design", and I want to extract the p-values from the results of that function. Given an lrm object constructed as follows : fit <- lrm(Y~(X1+X2+X3+X4+X5+X6+X7)^2, data=dataset) I need the p-values for the coefficients printed by calling

Dear All, reposting, because I did not find a solution, maybe someone could check the example below. It's taken from the help page of survdiff. Executing it, gives the error "Error in floor(temp) : Non-numeric argument to mathematical function" best regards, Heinz library(survival) ## Example from help page of survdiff ## Expected survival for heart transplant patients based

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

Dear R helpers, >From the lrm( ) model used for binary logistic regression, we used the L.R. model value (or the G2 value, likelihood-ratio chi-squared statistic) to evaluate the goodness-of-fit of the models. The model with the lowest G2 value consequently, has the best performance and the highest accuracy. However our model includes rsc() functions to account for non-linearity. We

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",

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 ~

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'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 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

Dear Robert, The differences have to do with diffent scaling defaults. lrm by default standardizes the covariates to unit sd before applying penalization. penalized by default does not do any standardization, but if asked standardizes on unit second central moment. In your example: x = c(-2, -2, -2, -2, -1, -1, -1, 2, 2, 2, 3, 3, 3, 3) z = c(0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1) You

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 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.

Dear all, this might be not only an R-question but also a statistical. When I do a logistic regression analysis (species distribution modeling) with function lrm (Design package) I get the follwoing error message: > tadl1<-lrm(triad~fd+dista+fd2+dista2+fd:dista+dista:geo2, x=T, y=T) Error in if (!length(fname) || !any(fname == zname)) { : missing value where TRUE/FALSE needed The