similar to: logistic regression lrm() output

Displaying 20 results from an estimated 600 matches similar to: "logistic regression lrm() output"

2017 Sep 14
3
Help understanding why glm and lrm.fit runs with my data, but lrm does not
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 <-
2017 Sep 14
0
Help understanding why glm and lrm.fit runs with my data, but lrm does not
> 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
2017 Sep 14
1
Help understanding why glm and lrm.fit runs with my data, but lrm does not
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
2008 May 29
2
Troubles plotting lrm output in Design Library
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 <-
2008 Apr 03
1
Design package lrm summary and factors
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
2010 Dec 09
1
error in lrm( )
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
2005 Aug 22
1
How to add values on the axes of the 3D bi-variable lrm fit?
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
2007 Jun 26
1
Marginal Effects of continuous variable in lrm model (Design package)
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
2010 Jun 18
1
Fitting a polynomial using lrm from the Design library
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 +
2004 Jan 29
2
Calculating/understanding variance-covariance matrix of logistic regression (lrm $var)
Hallo! I want to understand / recalculate what is done to get the CI of the logistic regression evaluated with lrm. As far as I came back, my problem is the variance-covariance matrix fit$var of the fit (fit<-lrm(...), fit$var). Here what I found and where I stucked: ----------------- library(Design) # data D<-c(rep("a", 20), rep("b", 20)) V<-0.25*(1:40) V[1]<-25
2011 Aug 06
1
help with predict for cr model using rms package
Dear list, I'm currently trying to use the rms package to get predicted ordinal responses from a conditional ratio model. As you will see below, my model seems to fit well to the data, however, I'm having trouble getting predicted mean (or fitted) ordinal response values using the predict function. I have a feeling I'm missing something simple, however I haven't been able to
2004 Feb 16
1
Binary logistic model using lrm function
Hello all, Could someone tell me what I am doing wrong here? I am trying to fit a binary logistic model using the lrm function in Design. The dataset I am using has a dichotomous response variable, 'covered' (1-yes, 0-no) with explanatory variables, 'nepall', 'title', 'abstract', 'series', and 'author1.' I am running the following script and
2011 May 05
7
Draw a nomogram after glm
Hi all R users I did a logistic regression with my binary variable Y (0/1) and 2 explanatory variables. Now I try to draw my nomogram with predictive value. I visited the help of R but I have problem to understand well the example. When I use glm fonction, I have a problem, thus I use lrm. My code is: modele<-lrm(Y~L+P,data=donnee) fun<- function(x) plogis(x-modele$coef[1]+modele$coef[2])
2001 Jul 28
2
Re: [S] Labels wrong with lrm
Dear Jan, Thank you very much for your excellent description of the problem and the self-contained test code. This is a problem that I've been meaning to either document better or solve for some time. The root of the problem is with the builtin S-Plus terms.inner function: > attr(terms.inner(asthma ~ pol(age,kx) + smok),'variables') expression(age, kx, smok) You can see that
2008 Feb 26
0
lrm error message
I'm trying to learn how to use the lrm() function by simulating data using an old dataset but it's giving me an error I don't understand: (nst$regular already exists) nst$regular<-as.ordered(nst$regular) nst$age<-rnorm(n=942,mean=43.20488,sd=17.03) nst$age<-round(age,digits=0) regform<-regular~age reglrm<-lrm(regform,nst) summary(reglrm) Error in
2011 Feb 08
1
Error in example Glm rms package
Hi all! I've got this error while running example(Glm) library("rms") > example(Glm) Glm> ## Dobson (1990) Page 93: Randomized Controlled Trial : Glm> counts <- c(18,17,15,20,10,20,25,13,12) Glm> outcome <- gl(3,1,9) Glm> treatment <- gl(3,3) Glm> f <- glm(counts ~ outcome + treatment, family=poisson()) Glm> f Call: glm(formula = counts ~
2001 Dec 09
1
Help for Power analysis
Dear colleague, I not sure this R code is correctly ? I would to show the number of Sample Size at Sample Size Axis that line draw from Power Axis (80%) from R code. How I show this and select the most appropriate of this power (.79955687 - 80983575). Thank for your help and answer. Best Regards, Nikom Thanomsieng, Email: nikom at kku.ac.th .... #Power analysis: Sample size for
2005 Sep 12
0
Multiple comparisons like a Chi2 or Fisher's exact test
Dear all I have an experiment where plots either have or have not regrown (in 40 plots) after receiving 12 different herbicide treatments and a control (no treatment). The data are significant with a Chi2, but to later distinguish if the differences are significant between each of the 12 treatments apart I need to do multiple Chi2 s between each. Is there a way to run those multiple Chi2 or
2008 Feb 10
1
Error while using fitdistr() function or goodfit() function
Try changing your method to "ML" and try again. I tried the run the first example from the documentation and it failed with the same error. Changing the estimation method to ML worked. @List: Can anyone else verify the error I got? I literally ran the following two lines interactively from the example for goodfit: dummy <- rnbinom(200, size = 1.5, prob = 0.8) gf <- goodfit(dummy,
2011 Apr 12
1
Datadist error
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