similar to: recommendation on B for validate.lrm () ?

Displaying 20 results from an estimated 4000 matches similar to: "recommendation on B for validate.lrm () ?"

2011 May 08
1
Hosmer-Lemeshow 'goodness of fit'
I'm trying to do a Hosmer-Lemeshow 'goodness of fit' test on my logistic regression model. I found some code here: http://sas-and-r.blogspot.com/2010/09/example-87-hosmer-and-lemeshow-goodness.html The R code is above is a little complicated for me but I'm having trouble with my answer: Hosmer-Lemeshow: p=0.6163585 le Cessie and Houwelingen test (Design library): p=0.2843620
2011 Nov 12
2
Odds ratios from lrm plot
The code library(Design) f <- lrm(y~x1+x2+x1*x2, data=data) plot(f) produces a plot of log odds vs x2 with 0.95 confidence intervals. How do I get a plot of odds ratios vs x2 instead? Thanks -- View this message in context: http://r.789695.n4.nabble.com/Odds-ratios-from-lrm-plot-tp4033340p4033340.html Sent from the R help mailing list archive at Nabble.com.
2012 Sep 20
1
validate.lrm - confidence interval for boostrap-corrected AUC ?
Hi Does anyone know whether the rms package provides a confidence interval for the bootstrap-corrected Dxy or c-index? I have fitted a logistic model, and would like to obtain the 95% confidence interval of the bootstrap-corrected area under the ROC curve estimate. Thanks. [[alternative HTML version deleted]]
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 <-
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')
2011 Jan 18
2
Baseline terms for lrm
Dear R-help and Prof. Harrell: My question concerns the baseline state for continuous variable in lrm() within the RMS package. I have a model which can be reduced to: lrm(FT ~ rcs(V1, c(0, 1,5)) The model makes perfect sense if the baseline state is where V1>=5 but the model makes no sense if the baseline category is 0 (which I had expected). Can someone point me to a reference, or
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 ~
2010 Apr 26
1
logical(0) response from lrm
What causes the error report: logical(0) to arise in the rms function lrm? Here's my data: But both the dependent and the independent variable seem fine... > str(AABB) 'data.frame': 1176425 obs. of 9 variables: $ sex : int 1 1 0 1 1 0 0 0 0 0 ... $ faint : int 0 0 0 0 0 0 0 0 0 0 ... Here's the simplified model and error AABB$model1 < lrm (faint ~ sex)
2010 Aug 11
3
extracting the standard error in lrm
Hi, I would like to extract the coefficients of a logistic regression (estimates and standard error as well) in lrm as in glm with summary(fit.glm)$coef Thanks David
2009 Sep 26
1
Summary/Bootstrap for Design library's lrm function
Can anyone tell me what I might be doing incorrectly for an ordinal logistic regression for lrm? I cannot get R(2.9.1)to run either summary nor will it let me bootstrp to validate. ### Y is a 5 value measure with a range from 1-5, the independent variables are the same. N=75 but when we knock out the NAs it comes down to 51#### > lrm(formula = Y ~ permemp + rev + gconec + scorpstat, data =
2010 Dec 25
2
predict.lrm vs. predict.glm (with newdata)
Hi all I have run into a case where I don't understand why predict.lrm and predict.glm don't yield the same results. My data look like this: set.seed(1) library(Design); ilogit <- function(x) { 1/(1+exp(-x)) } ORDER <- factor(sample(c("mc-sc", "sc-mc"), 403, TRUE)) CONJ <- factor(sample(c("als", "bevor", "nachdem",
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
2011 Apr 28
5
fisher exact for > 2x2 table
I'm using fisher.exact on a 4x2 table and it seems to work. Does anyone know exactly what is going on? I thought fisher.exact is only for 2x2 tables. Note: I can't use chi-squared because I have a couple of cells with 0 and < 5 observations. -- View this message in context: http://r.789695.n4.nabble.com/fisher-exact-for-2x2-table-tp3481979p3481979.html Sent from the R help
2010 Mar 06
1
Robust SE for lrm object
I'm trying to obtain the robust standard errors for a multinomial ordered logit model: mod6 <- lrm(wdlshea ~ initdesch + concap + capasst + qualrat + terrain,data=full2) The model is fine but when I try to get the RSE I get an error. coeftest(mod6, vcov = vcovHAC(mod6)) Error in match.arg(type) : 'arg' should be one of “ordinary”, “score”, “score.binary”, “pearson”,
2010 Oct 01
6
Interpreting the example given by Frank Harrell in the predict.lrm {Design} help
Dear list, I am relatively new to ordinal models and have been working through the example given by Frank Harrell in the predict.lrm {Design} help All of this makes sense to me, except for the responses, i,e how do i interpret them? i would be extremely grateful if someone could explain the results? First i establish the date and model - > y <- factor(sample(1:3, 400, TRUE), 1:3,
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
2009 Aug 29
3
lrm in Design
Hello everybody, I am trying to do a logistic regression model with lrm() from the design package. I am comparing to groups with different medical outcome which can either be "good" or "bad". In the help file it says that lrm codes al responses to 0,1,2,3, etc. internally and does so in alphabetical order. I would guess this means bad=0 and good=1. My question: I am trying to
2008 Oct 09
2
Singular information matrix in lrm.fit
Hi R helpers, I'm fitting large number of single factor logistic regression models as a way to immediatly discard factor which are insignificant. Everything works fine expect that for some factors I get error message "Singular information matrix in lrm.fit" which breaks whole execution loop... how to make LRM not to throw this error and simply skip factors with singularity