similar to: Odds ratios from lrm plot

Displaying 20 results from an estimated 2000 matches similar to: "Odds ratios from lrm plot"

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",
2012 Jun 20
2
Odds Ratios in rms package
Hi, I'm using the rms package to do regression analysis using the lrm function. Retrieving odds ratios is possible using summary.rms. However, I could not find any information on how exactly the odds ratios for continuous variables are calculated. It doesn't appear to be the odds ratio at 1 unit increase, because the output of summary.rms did not match the coefficient's value. E.g.
2010 Feb 06
4
Plot of odds ratios obtained from a logistic model
Hi all! I am trying to develop a plot a figure in which I would like to show the odds ratios obtained from a logistic model. I have tried with the dotplot option but no success. Could you help me? Is there any option when modelling the logistic model in R? Thank you in advance
2003 Dec 05
3
Odds ratios for categorical variable
Dear R-users: How does one calculate in R the odds ratios for a CATEGORICAL predictor variable that has 4 levels. I see r-help inquiries regarding odds ratios for what looked like a continuous predictor variable. I was wondering how to get the pairwise odds ratios for comparisons of levels of a categorical predictor variable. I can't seem to get the correct output using: >
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,
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
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 ~
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
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 <-
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
2006 Nov 14
1
Using lrm
Hi, I have to build a logistic regression model on a data set that I have. I have three input variables (x1, x2, x3) and one output variable (y). The syntax of lrm function looks like this lrm(formula, data, subset, na.action=na.delete, method="lrm.fit", model=FALSE, x=FALSE, y=FALSE, linear.predictors=TRUE, se.fit=FALSE, penalty=0, penalty.matrix, tol=1e-7,
2009 Sep 04
2
lrm in Design package--missing value where TRUE/FALSE needed
Hi, A error message arose while I was trying to fit a ordinal model with lrm() I am using R 2.8 with Design package. Here is a small set of mydata: RC RS Sex CovA CovB CovC CovD CovE 2 1 0 1 1 0 -0.005575280 2 2 1 0 1 0 1 -0.001959580 2 3 0 0 0 1 0 -0.004725880 2 0 0 0 1 0 0 -0.005504850 2 2 1 1 0 0 0 -0.003880170 1 2 1 0 0 1 0 -0.006074230 2 2 1 0 0 1 1 -0.003963920 2 2 1 0 0 1 0
2009 Aug 21
1
Possible bug with lrm.fit in Design Library
Hi, I've come across a strange error when using the lrm.fit function and the subsequent predict function. The model is created very quickly and can be verified by printing it on the console. Everything looks good. (In fact, the performance measures are rather nice.) Then, I want to use the model to predict some values. I get the following error: "fit was not created by a Design
2008 Mar 03
1
using 'lrm' for logistic regression
Hi R, I am getting this error while trying to use 'lrm' function with nine independent variables: > res = lrm(y1994~WC08301+WC08376+WC08316+WC08311+WC01001+WC08221+WC08106+WC0810 1+WC08231,data=y) singular information matrix in lrm.fit (rank= 8 ). Offending variable(s): WC08101 WC08221 Error in j:(j + params[i] - 1) : NA/NaN argument Now, if I take choose only four
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)
2009 Aug 17
3
Help understanding lrm function of Design library
Hi, I'm developing an experiment with logistic regression. I've come across the lrm function in the Design library. While I understand and can use the basic functionality, there are a ton of options that go beyond my knowledge. I've carefully read the help page for lrm, but don't understand many of the arguments and optional return values. (penalty, penalty.matrix,