similar to: Calculating/understanding variance-covariance matrix of logistic regression (lrm $var)

Displaying 20 results from an estimated 4000 matches similar to: "Calculating/understanding variance-covariance matrix of logistic regression (lrm $var)"

2006 May 16
2
Interrater and intrarater variability (intraclass correlationcoefficients)
It sounds as thought you are interested in Hoyt's Anova which is a form of generalizability theory. This is usually estimated using by getting the variance components from ANOVA. > -----Original Message----- > From: r-help-bounces at stat.math.ethz.ch > [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Karl Knoblick > Sent: Tuesday, May 16, 2006 6:10 AM > To: r-help at
2006 May 17
1
Response to query re: calculating intraclass correlations
Karl, If you use one of the specialized packages to calculate your ICC, make sure that you know what you're getting. (I haven't checked the packages out myself, so I don't know either.) You might want to read David Futrell's article in the May 1995 issue of Quality Progress where he describes six different ways to calculate ICCs from the same data set, all with different
2003 Dec 31
2
Plot grouped data: How to change x-axis? (nlme)
Hallo! GENERAL QUESTION: I'm trying to change the tick marks of the x-axis in a grouped data plot (nlme). CONCRETE EXAMPLE: In the example (see below) I want the x-axis to have tick marks at 0, 6, 12, 18, 24. How can I do this? WHAT I TRIED I tried "normal" methods like axis(...) but this does not work with this plot. And I also tried xlim=c(0,24) but the ticks are unchanged and
2008 Nov 10
6
Variable passed to function not used in function in select=... in subset
Hello! I have the problem that in my function the passed variable is not used, but the variable name of the dataframe itself?- difficult to explain, but an easy example: TestFunc<-function(df, group) { ??? print(names(subset(df, select=group))) } df1<-data.frame(group="G1", visit="V1", value=0.9) TestFunc(df1, c("group", "visit")) Result: [1]
2006 May 16
5
Interrater and intrarater variability (intraclass correlation coefficients)
Hello! I want to calculate the intra- and interrater reliability of my study. The design is very simple, 5 raters rated a diagnostic score 3 times for 19 patients. Are there methods/funtions in R? I only found packages to calculate interrater variability and intraclass correlation coefficients for matrices of n*m (n subjects, m raters) - I have n subjects, m raters and r repetitions. Can
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
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
2011 May 18
1
logistic regression lrm() output
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')
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
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 +
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
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
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 <-
2004 Jun 29
1
nls fitting problems (singularity)
Hallo! I have a problem with fitting data with nls. The first example with y1 (data frame df1) shows an error, the second works fine. Is there a possibility to get a fit (e.g. JMP can fit also data I can not manage to fit with R). Sometimes I also got an error singularity with starting parameters. # x-values x<-c(-1,5,8,11,13,15,16,17,18,19,21,22) # y1-values (first data set)
2003 Nov 14
4
LOCF - Last Observation Carried Forward
Hi! Is there a possibilty in R to carry out LOCF (Last Observation Carried Forward) analysis or to create a new data frame (array, matrix) with LOCF? Or some helpful functions, packages? Karl --------------------------------- Gesendet von http://mail.yahoo.de Schneller als Mail - der neue Yahoo! Messenger. [[alternative HTML version deleted]]
2010 Oct 06
2
A problem --thank you
dear:teacher i have a problem which about the polr()(package "MASS"), if the response must have 3 or more levels? and how to fit the polr() to 2 levels? thank you. turly yours [[alternative HTML version deleted]]
2005 Jul 12
1
Design: predict.lrm does not recognise lrm.fit object
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
2004 Mar 22
2
Handling of NAs in functions lrm and robcov
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
2012 May 27
2
Unable to fit model using “lrm.fit”
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 ~