similar to: difference between lrm's "Model L.R." and anova's "Chi-Square"

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

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

On Sat, 2011-10-29 at 15:16 -0500, Hal Finkel wrote: > On Sat, 2011-10-29 at 14:02 -0500, Hal Finkel wrote: > > On Sat, 2011-10-29 at 12:30 -0500, Hal Finkel wrote: > > > Ralf, et al., > > > > > > Attached is the latest version of my autovectorization patch. llvmdev > > > has been CC'd (as had been suggested to me); this e-mail contains > >

On Sat, 2011-10-29 at 14:02 -0500, Hal Finkel wrote: > On Sat, 2011-10-29 at 12:30 -0500, Hal Finkel wrote: > > Ralf, et al., > > > > Attached is the latest version of my autovectorization patch. llvmdev > > has been CC'd (as had been suggested to me); this e-mail contains > > additional benchmark results. > > > > First, these are preliminary

Dear R list, From the lrm() binary logistic model we derived the G2 value or the likelihood-ratio chi-squared statistic given as L.R. model, in the output of the lrm(). How can this value be penalized for non-linearity (we used splines in the lrm function)? lrm.iRVI <- lrm(arson ~ rcs(iRVI,5), penalty=list(simple=10,nonlinear=100,nonlinear.interaction=4)) This didn’t work

Pete Cooper and I have been looking at memory profiles of running llc on verify-uselistorder.lto.opt.bc (ld -save-temps dump just before CodeGen of building verify-uselistorder with -flto -g). I've attached leak-backend.patch, which we're using to make Intrustruments more accurate (instead of effectively leaking things onto BumpPtrAllocators, really leak them with malloc()). (I've

Dear all, sorry to bother you but I was looking for the "scale Inverse Chi square distribution" and I could not find it! If I remember well , when X~Scale-Inv-Chi-Square (a,b) then X~Inv-Gamma(a/2,a*b/2) but with the shape, scale and rate parameters I always get confused! Any suggestions? Thanks a lot for your prompt reply. Best regards, Giorgio Di Gessa

On Sat, 2011-10-29 at 12:30 -0500, Hal Finkel wrote: > Ralf, et al., > > Attached is the latest version of my autovectorization patch. llvmdev > has been CC'd (as had been suggested to me); this e-mail contains > additional benchmark results. > > First, these are preliminary results because I did not do the things > necessary to make them real (explicitly quiet the

Ralf, et al., Attached is the latest version of my autovectorization patch. llvmdev has been CC'd (as had been suggested to me); this e-mail contains additional benchmark results. First, these are preliminary results because I did not do the things necessary to make them real (explicitly quiet the machine, bind the processes to one cpu, etc.). But they should be good enough for discussion.

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 2016-05-25 19:13, Kelly Lesperance wrote: > Hdparm didn?t get far: > > [root at r1k1 ~] # hdparm -tT /dev/sda > > /dev/sda: > Timing cached reads: Alarm clock > [root at r1k1 ~] # Hi Kelly, Try running 'iostat -xdmc 1'. Look for a single drive that has substantially greater await than ~10msec. If all the drives except one are taking 6-8msec, but one is very

> 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

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, 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 ~

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, 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

Hej, i need the which() funktion to find the positions of an entry in a matrix. the entries i'm looking for are : seq(begin,end,0.01) and there are no empty spaces i'm searching in the right range. so i was looking for the results R can find and i recieved this answer. for (l in

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