similar to: Logistic regression - confidence intervals

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, 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')

I was unsure of what suitable goodness-of-fit tests existed in R for logistic regression. After searching the R-help archive I found that using the Design models and resid, could be used to calculate this as follows: d <- datadist(mydataframe) options(datadist = 'd') fit <- lrm(response ~ predictor1 + predictor2..., data=mydataframe, x =T, y=T) resid(fit, 'gof'). I set up a

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

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 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, 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) +

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

I've been using R to do logistic regresssion, and that's working well, but there are two things I haven't figured out how to do. (1) Is there some pre-existing function that will let you compute the odds ratios and confidence intervals for them for a specific fit. I know how to do this manually or even write a function that I can call with the coefficients and se, but

dear professor: thank you for your help,witn your help i develop the nomogram successfully. after that i want to do the internal validation to the model.i ues the bootpred to do it,and then i encounter problem again,just like that.(´íÎóÓÚerror to :complete.cases(x, y, wt) : ²»ÊÇËùÓеIJÎÊý¶¼Ò»Ñù³¤(the length of the augment was different)) i hope you tell me where is the mistake,and maybe i have

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-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, I have a question about the computation of confidence intervals in the zyp package, in particular using the functions zyp.sen and confint.zyp, or zyp.yuepilon. (1) I'm a bit confused about the confidence intervals given by zyp.sen and confint.zyp. When I request a certain confidence interval in the function, the R output seems to deliver another confidence interval, e.g. when I set

dear professor: I have a problem about the nomogram.I have got the result through analysing the dataset "exp2.sav" through multinominal logistic regression by SPSS 17.0. and I want to deveop the nomogram through R-Projject,just like this : > n<-100 > set.seed(10) > T.Grade<-factor(0:3,labels=c("G0", "G1", "G2","G3")) >

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

I have fitted a logistic regression model > failed.lr2$call lrm(formula = failed ~ Age + task2 + Age:task2, data = time.long, na.action = na.omit) using the Design package functions and would like to generate a nomogram from this model. the datadist information is generated and stored in > ddist time.long$Age time.long$task2 Low:effect 45

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

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])

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)

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