similar to: Obtaining p-values for coefficients from LRM function (package Design)

Sent this mail in rich text format before. Excuse me for this. ------------------------ Dear all, I'm using the lrm function from the package "Design", and I want to extract the p-values from the results of that function. Given an lrm object constructed as follows : fit <- lrm(Y~(X1+X2+X3+X4+X5+X6+X7)^2, data=dataset) I need the p-values for the coefficients printed by calling

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

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

Greetings: I'm running R-1.9.1 on Fedora Core 2 Linux. I tested a proportional odds logistic regression with MASS's polr and Design's lrm. Parameter estimates between the 2 are consistent, but the standard errors are quite different, and the conclusions from the t and Wald tests are dramatically different. I cranked the "abstol" argument up quite a bit in the polr

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,

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

inputfille snpid indid genotype gvariable probeid gene geneexpression rs1040480 CHB_NA18524 C/T 2 GI_19743926-I PTPRT 5.850586 rs1040480 CHB_NA18526 C/C 1 GI_19743926-I PTPRT 6.028641 rs1040480 CHB_NA18529 C/C 3 GI_19743926-I PTPRT 5.944392 rs1040481 CHB_NA18532 C/C 1 GI_19743926-I PTPRT 5.938578 rs1040481 CHB_NA18537 C/C 2 GI_19743926-I PTPRT 5.874439 rs1040481 CHB_NA18540 C/C 3 GI_19743926-I

I am trying to obtain the AICc after performing logistic regression using the Design package. For simplicity, I'll talk about the AIC. I tried building a model with lrm, and then calculating the AIC as follows: likelihood.ratio <- unname(lrm(succeeded~var1+var2,data=scenario,x=T,y=T)$stats["Model L.R."]) #Model likelihood ratio??? model.params <- 2 #Num params in my model AIC

Hi, I am planning to externally validate a logistic prediction model in a new cohort. Outcome is mortality. The betacoefficients were derived from a previous published article. It seems not possible in R to assign fixed beta coefficients to predictors like lrm (death ~ intercept+beta1*var1+beta2*var2...). How do i solve this problem? Thank you in advance. Joey L -- View this message in context:

Hello again R users, I have a devilishly hard problem, which should be very simple. I hope someone out there will have the answer to this on the tip of their tongue. Please consider the following toy example: x <- read.table(textConnection("y x1 x2 indv.1 bagels 4 6 indv.2 donuts 5 1 indv.3 donuts 1 10 indv.4 donuts 10 9 indv.5 bagels 0 2 indv.6 bagels 2 9 indv.7 bagels 8 5 indv.8

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

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

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 ,

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

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 =

Dear List, I have two questions about how to do predictions using lrm, specifically how to predict the ordinal response for each observation *individually*. I'm very new to cumulative odds models, so my apologies if my questions are too basic. I have a dataset with 4000 observations. Each observation consists of an ordinal outcome y (i.e., rating of a stimulus with four possible

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

Hi, there, I am having trouble using 'lrm' function in package 'Design'. Basically, the ' . ' after ' ~ ' wouldn't work. Here are some sample codes: > temp <- data.frame(a=c(rep(0,3),rep(1,3)),b=rnorm(6),c=c('a','b','c','a','b','c')) > lrm(a~.,data=temp) Error in terms.formula(formula, specials =

Hi List, I don't understand why 'lrm' doesn't recognize the '~.' formula. I'm pretty sure it was working before. Please see below: I'm using R2.3.0, WinXP, Design 2.0-12 thanks, ...Tao > dat <- data.frame(y=factor(rep(1:2,each=50)), x1=rnorm(100), x2=rnorm(100), x3=rnorm(100)) > lrm(y~., data=dat, x=T, y=T) Error in terms.formula(formula, specials =

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 +