similar to: Repost - 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

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

Hello, I am trying to get the 'rank' function to work for me, but not sure what I am doing wrong. Please help. I ran the following commands: data = read.table("test1.csv", head=T, as.is=T, na.string=".", row.nam=NULL) X1 = as.factor(data[[3]]) X2 = as.factor(data[[4]]) X3 = as.factor(data[[5]]) Y = data[[2]] model = lm(Y ~ X1*X2*X3, na.action = na.exclude) fmodel =

Hello, Fitting a piecewise smooth curve to a set of points (and a piecewise linear function in particular) seems to be a recurring question on this list. Nevertheless, I was not able to find an answer to a question that bothers me. Suppose I have the following data set, and would want to fit it with a piecewise smooth curve, In this model data, one curve is valid for up to 3 and another one for

> 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 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

Dear R helpers, >From the lrm( ) model used for binary logistic regression, we used the L.R. model value (or the G2 value, likelihood-ratio chi-squared statistic) to evaluate the goodness-of-fit of the models. The model with the lowest G2 value consequently, has the best performance and the highest accuracy. However our model includes rsc() functions to account for non-linearity. We

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 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 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 =

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)

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.

Hi, I am wondering if there is a way to specify the prevalence of events in logistic regression using lrm() from Design package? Linear Discriminant Analysis using lda() from MASS library has an argument "prior=" that we can use to specify the prevalent of events when the actual dataset being analyzed does not have a representative prevalence. How can we incorporate this information in

Dear all, just a quick question regarding weights in logistic regression. I do results <- lrm(y.js ~ h.hhsize + h.death1 + h.ill1 + h.ljob1 + h.fin1 + h.div1 + h.fail1 + h.sex + h.ch.1

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

Hello all, Could someone tell me what I am doing wrong here? I am trying to fit a binary logistic model using the lrm function in Design. The dataset I am using has a dichotomous response variable, 'covered' (1-yes, 0-no) with explanatory variables, 'nepall', 'title', 'abstract', 'series', and 'author1.' I am running the following script and

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

Dear List, I am trying to assess the prediction accuracy of an ordinal model fit with LRM in the Design package. I used predict.lrm to predict on an independent dataset and am now attempting to assess the accuracy of these predictions. >From what I have read, the AUC is good for this because it is threshold independent. I obtained the AUC for the fit model output from the c score (c =

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, 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 ~