I think you are doing this correctly except for one thing. The validation and other inferential calculations should be done on the full model. Use the approximate model to get a simpler nomogram but not to get standard errors. With only dropping one variable you might consider just running the nomogram on the entire model. Frank
細田弘吉 wrote: > > 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 (ols in rms package) as > the followings. I would like to know whether I am doing right or not. > >> library(rms) >> plogit <- predict(full.model) >> full.ols <- ols(plogit ~ stenosis+x1+x2+ClinicalScore+procedure, sigma=1) >> fastbw(full.ols, aics=1e10) > > Deleted Chi-Sq d.f. P Residual d.f. P AIC R2 > stenosis 1.41 1 0.2354 1.41 1 0.2354 -0.59 0.991 > x2 16.78 1 0.0000 18.19 2 0.0001 14.19 0.882 > procedure 26.12 1 0.0000 44.31 3 0.0000 38.31 0.711 > ClinicalScore 25.75 1 0.0000 70.06 4 0.0000 62.06 0.544 > x1 83.42 1 0.0000 153.49 5 0.0000 143.49 0.000 > > Then, fitted an approximation to the full model using most imprtant > variable (R^2 for predictions from the reduced model against the > original Y drops below 0.95), that is, dropping "stenosis". > >> full.ols.approx <- ols(plogit ~ x1+x2+ClinicalScore+procedure) >> full.ols.approx$stats > n Model L.R. d.f. R2 g Sigma > 104.0000000 487.9006640 4.0000000 0.9908257 1.3341718 0.1192622 > > This approximate model had R^2 against the full model of 0.99. > Therefore, I updated the original full logistic model dropping > "stenosis" as predictor. > >> full.approx.lrm <- update(full.model, ~ . -stenosis) > >> validate(full.model, bw=F, B=1000) > index.orig training test optimism index.corrected n > Dxy 0.6425 0.7017 0.6131 0.0887 0.5539 1000 > R2 0.3270 0.3716 0.3335 0.0382 0.2888 1000 > Intercept 0.0000 0.0000 0.0821 -0.0821 0.0821 1000 > Slope 1.0000 1.0000 1.0548 -0.0548 1.0548 1000 > Emax 0.0000 0.0000 0.0263 0.0263 0.0263 1000 > >> validate(full.approx.lrm, bw=F, B=1000) > index.orig training test optimism index.corrected n > Dxy 0.6446 0.6891 0.6265 0.0626 0.5820 1000 > R2 0.3245 0.3592 0.3428 0.0164 0.3081 1000 > Intercept 0.0000 0.0000 0.1281 -0.1281 0.1281 1000 > Slope 1.0000 1.0000 1.1104 -0.1104 1.1104 1000 > Emax 0.0000 0.0000 0.0444 0.0444 0.0444 1000 > > Validatin revealed this approximation was not bad. > Then, I made a nomogram. > >> full.approx.lrm.nom <- nomogram(full.approx.lrm, > fun.at=c(0.05,0.1,0.2,0.4,0.6,0.8,0.9,0.95), fun=plogis) >> plot(full.approx.lrm.nom) > > Another nomogram using ols model, > >> full.ols.approx.nom <- nomogram(full.ols.approx, > fun.at=c(0.05,0.1,0.2,0.4,0.6,0.8,0.9,0.95), fun=plogis) >> plot(full.ols.approx.nom) > > These two nomograms are very similar but a little bit different. > > My questions are; > > 1. Am I doing right? > > 2. Which nomogram is correct > > I would appreciate your help in advance. > > -- > KH > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > ----- Frank Harrell Department of Biostatistics, Vanderbilt University -- View this message in context: http://r.789695.n4.nabble.com/Question-on-approximations-of-full-logistic-regression-model-tp3524294p3525372.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
