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

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 "fit". fit$coef (gives a list of only the coefficients) fit$pval, fit$p,

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

I am trying to test goodness of fit for my legalistic regression using several options as shown below.  Hosmer-Lemeshow test (whose function I borrowed from a previous post), Hosmer–le Cessie omnibus lack of fit test (also borrowed from a previous post), Pearson chi-square test, and deviance test.  All the tests, except the deviance tests, produced p-values well above 0.05.  Would anyone please

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

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 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

> 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

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",

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

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 ~

Dear r users I have a question in coding continuous variables in logistic regression. When "rcs" is used in transforming variables, sometime it gives implausible associations with the outcome although the model x2 is high. So what's your tips and tricks in coding continuous variables. P.S. How to code variables as linear+square in the formula such as lrm. lrm(y~x+sqrt(x))

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, 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

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 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

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,

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

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 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 =