similar to: lrm error message

I entered the following: formula<-nst~age+soc+inc+reg+imp pnstlm<-lm(formula,nst) summary(pnstlm) imp and soc are ordered categorical variables but the summary does not give an output of the overall p-values, just individual comparisons. I can't find help for this in the manual. Is there a command or option change in the summary to find out the overall p-value? thank you in

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, I have question regarding the lrm function and estimating the odds ratio between different levels of a factored variable. The following code example illustrates the problem I am having. I have a data set with an outcome variable (0,1) and an input variable (A,B,C). I would like to estimate the effect of C vs B, but when I perform the summary I only get A vs B and A vs C, even though I

> 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

Dear r-list, When I try to plot the following 3D lrm fit I obtain only arrows with labels on the three axes of the figure (without values). fit <- lrm(y ~ rcs(x1,knots)+rcs(x2,knots), tol=1e-14,X=T,Y=T) dd <- datadist(x1,x2);options(datadist='dd'); par(mfrow=c(1,1)) plot(fit,x1=NA, x2=NA, theta=50,phi=25) How can I add values to the axes of this plot? (axes with the

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

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 R-helpers, I'm having a problem in using plot.design in Design Library. Tho following example code produce the error: > n <- 1000 # define sample size > set.seed(17) # so can reproduce the results > age <- rnorm(n, 50, 10) > blood.pressure <- rnorm(n, 120, 15) > cholesterol <- rnorm(n, 200, 25) > sex <-

Dear Jan, Thank you very much for your excellent description of the problem and the self-contained test code. This is a problem that I've been meaning to either document better or solve for some time. The root of the problem is with the builtin S-Plus terms.inner function: > attr(terms.inner(asthma ~ pol(age,kx) + smok),'variables') expression(age, kx, smok) You can see that

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 <-

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

Hi, Design isn't strictly an R base package, but maybe someone can explain the following. When lrm is called within a function, it can't find the dataset dd: > library(Design) > age <- rnorm(30, 50, 10) > cholesterol <- rnorm(30, 200, 25) > ch <- cut2(cholesterol, g=5, levels.mean=TRUE) > fit <- function(ch, age) + { + d <- data.frame(ch, age) +

Dear all, I have performed a simple logistic regression using the lrm function from the Design library. Now I want to plot the summary, or make a nomogram. I keep getting a datadist error: options(datadist= m.full ) not created with datadist. I have tried to specify datadist beforhand (although I don't know why it should be done): ddist<-datadist(d) ##where d is my dataset

Hi everyone, I'm doing a logistic regression with an ordinal variable. I'd like to set the contrasts on the ordinal variable. However, when I set the contrasts, they work for ordinary linear regression (lm), but not logistic regression (lrm): ddist = datadist(bin.time, exp.loc) options(datadist='ddist') contrasts(exp.loc) = contr.treatment(3, base = 3, contrasts = TRUE) lrm.loc =

Hi all, I am looking to fit a logistic regression using the lrm function from the Design library. I am interested in this function because I would like to obtain "pseudo-R2" values (see http://tolstoy.newcastle.edu.au/R/help/02b/1011.html). Can anyone help me with the syntax? If I fit the model using the stats library, the code looks like this: model <- glm(x$trait ~ x$PC1 +

Hi I got an error message using datadist() from Design package: > library(Design,T) > dd <- datadist(beta.final) > options(datadist="dd") > lrm(Disease ~ gsct+apcct+rarct, x=TRUE, y=TRUE) Error in eval(expr, envir, enclos) : object "Disease" not found All variables inclduing response variable "Disease" are in the data frame

Hi! Does anyone know how to do the test for goodness of fit of a logistic model (in rms package) after running fit.mult.impute? I am using the rms and Hmisc packages to do a multiple imputation followed by a logistic regression model using lrm. Everything works fine until I try to run the test for goodness of fit: residuals(type=c("gof")) One needs to specify y=T and x=T in the fit. But

Dear R-users, I have used the nomogram function from the rms package for a logistic regresison model made with lrm(). Everything works perfectly (r version 2.15.1 on a mac). My question is this: if my final model is not the one created by lrm, but I internally validated the model and 'shrunk' the regression coefficients and computed a new intercept, how can I build a nomogram using that

Dear all: When I am trying to get the marginal effects: summary(result7,adv_inc_ratio=mean(m9201 $adv_inc_ratio),adv_price_ratio=mean(m9201$adv_price_ratio), ...(SOME MORE CONTINUOUS AND DISCRETE VARIABLES BUT I AM NOT LISTING)... regW=c (0,mean(m9201$regW),1), regWM=c(0,mean(m9201$regWM),1)) It gave out an error message: Error in summary.Design(result7, adv_inc_ratio = mean(m9201

Dear experts, Please forgive the puzzled title and the length of this message - I thought it would be best to be as complete as possible and to show the avenues I have explored. I'm trying to fit a linear model to data with a binary dependent variable (i.e. Target.ACC: accuracy of response) using lrm, and thought I would start from the most complex model (of which "sample1.lrm1" is