similar to: modeling logit(y/n) using lrm

I am trying to use the sas.get function to bring a SAS dataset into R. After calling the sas.get function I got an error message stating that 'sas' is not a recognized internal or external command, etc. I am guessing that I need to specify the internal command to invoke SAS in the sasprog option for the sas.get function, but I don't know how to determine what that command might be.

I have read some very good reviews comparing R (or Splus) to SAS. Does anyone know if there are any reviews comparing R (or Splus) to Stata? I am trying to get others to try R in my department, and I have never used Stata. Regards, -Cody Cody Hamilton, Ph.D Institute for Health Care Research and Improvement Baylor Health Care System (214) 265-3618 This e-mail, facsimile, or letter

Dear all, I am trying to find a suitable R-function for parametric proportional hazard regressions. The package survival contains the coxph() function which performs a Cox regression which leaves the base hazard unspecified, i.e. it is a semi-parametric method. The package Design contains the function pphsm() which is good for parametric proportional hazard regressions when the underlying base

Hi, I am trying to construct a logistic regression model from my data (104 patients and 25 events). I build a full model consisting of five predictors with the use of penalization by rms package (lrm, pentrace etc) because of events per variable issue. Then, I tried to approximate the full model by step-down technique predicting L from all of the componet variables using ordinary least squares

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I am cleaning up the rms package to not export functions not to be called directly by users. rms uses generic functions defined in other packages. For example there is a latex method in the Hmisc package, and rms has a latex method for objects of class "anova.rms" so there are anova.rms and latex.anova.rms functions in rms. I use:

Does anyone know of a function that computes the necessary sample size to reject a null median survival in favor of a given alternative median survival in a one arm trial? Cody Hamilton, Ph.D Institute for Health Care Research and Improvement Baylor Health Care System (214) 265-3618 This e-mail, facsimile, or letter and any files or attachments transmitted with it contains information

I have a spatial data set with ordinal response variable containing four levels. I would like to know if and how spatial autocorrelation can be taken into account when ordinal logistic regression is used (e.g. the function lrm from the Design package). Thanks for your help! Christof

I'd like to estimate an ordinal logistic regression with a random effect for a grouping variable. I do not find a pre-packaged algorithm for this. I've found methods glmmML (package: glmmML) and lmer (package: lme4) both work fine with dichotomous dependent variables. I'd like a model similar to polr (package: MASS) or lrm (package: Design) that allows random effects. I was

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

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

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 ~

> 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

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

Dear useR's, I was comparing results for a logistic regression model between different library's. themodel formula is arranged as follows: response ~ (intercept) + value + group OR: glm( response ~ (intercept) + value + group , family=binomial(link='logit')) lrm( response ~ (intercept) + value + group ) ROC( from = response ~ (intercept) + value + group ,

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 <-

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",

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

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

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,