similar to: Design package lrm summary and factors

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 <-

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) +

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')

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

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

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 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 +

> 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

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

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

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

Dear All, using the example from the help of summary.rms library(rms) 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 <- factor(sample(c('female','male'), n,TRUE)) label(age) <- 'Age'

Hello,   I am using mod1 <- lrm(y~x1+x2,na.action=na.pass,method="lrm.fit") summary(mod1) and I've got the following error: Error in summary.Design(mod1) : adjustment values not defined here or with datadist for x1 x2   Many thank, Amor [[alternative HTML version deleted]]

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

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

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 =

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

Hi, just making some experiments with design library i get an error if i want plot(fit) - show below from onlineHelp !? ..perhaps is here another mask problem?, but label from xtable which was my first problem is now off ! Thanks for advance & regards, Christian $ n <- 1000 # define sample size $ set.seed(17) # so can reproduce the results $ age <- rnorm(n, 50, 10)

dear£º in the example£¨nomogram£©£¬I don't understand the meanings of the program which have been marked by red line.And how to compile the program(L <- .4*(sex=='male') + .045*(age-50) + (log(cholesterol - 10)-5.2)*(-2*(sex=='female') + 2*(sex=='male'))). n <- 1000 # define sample size set.seed(17) # so can reproduce the results age <- rnorm(n, 50, 10)