Questions about lrm, validate, pentrace
(11/04/29 22:09), Frank Harrell wrote:
Yes I would select that as the final model.
Thank you for your comment. I am able to be confident about my model now. The difference you saw is caused
by different treatment of penalization of factor variables, related to the use of the sum squared differences between the estimate at one category from the average over all categories. I think that as long as you code it one way consistently and pick the penalty using that coding you are OK. But if the coefficients of the non-factor variables depend on how the binary predictor is coded, there is a bit more concern.
A lot of previous studies have demonstrated that poor outcome is more frequent in treat2 than in treat 1. So, I coded treat1 as 0, and treat2 as 1 in the first mail. Then, I came back to the original coding of treat1 and treat2 in the newer mail. According to your answer, I guess I am OK. :-) Prof Harrell, Your book (Rregression Modeling Strategies) and many kind comments helped me a lot. Thank you very much again. -- KH
Frank ???? wrote:
Thank you for you quick reply, Prof. Harrell. According to your advice, I ran pentrace using a very wide range.
> pentrace.x6factor<- pentrace(x6factor.lrm, seq(0, 100, by=0.5)) > plot(pentrace.x6factor)
I attached this figure. Then,
> pentrace.x6factor<- pentrace(x6factor.lrm, seq(0, 10, by=0.05))
It seems reasonable that the best penalty is 2.55.
> x6factor.lrm.pen<- update(x6factor.lrm, penalty=2.55) > cbind(coef(x6factor.lrm), coef(x6factor.lrm.pen),
abs(coef(x6factor.lrm)-coef(x6factor.lrm.pen)))
[,1] [,2] [,3]
Intercept -4.32434556 -3.86816460 0.456180958
stenosis -0.01496757 -0.01091755 0.004050025
T1 3.04248257 2.42443034 0.618052225
T2 -0.75335619 -0.57194342 0.181412767
procedure -1.20847252 -0.82589263 0.382579892
ClinicalScore 0.37623189 0.30524628 0.070985611
> validate(x6factor.lrm, bw=F, B=200)
index.orig training test optimism index.corrected n Dxy 0.6324 0.6849 0.5955 0.0894 0.5430 200 R2 0.3668 0.4220 0.3231 0.0989 0.2679 200 Intercept 0.0000 0.0000 -0.1924 0.1924 -0.1924 200 Slope 1.0000 1.0000 0.7796 0.2204 0.7796 200 Emax 0.0000 0.0000 0.0915 0.0915 0.0915 200 D 0.2716 0.3229 0.2339 0.0890 0.1826 200 U -0.0192 -0.0192 0.0243 -0.0436 0.0243 200 Q 0.2908 0.3422 0.2096 0.1325 0.1582 200 B 0.1272 0.1171 0.1357 -0.0186 0.1457 200 g 1.6328 1.9879 1.4940 0.4939 1.1389 200 gp 0.2367 0.2502 0.2216 0.0286 0.2080 200
> validate(x6factor.lrm.pen, bw=F, B=200)
index.orig training test optimism index.corrected n Dxy 0.6375 0.6857 0.6024 0.0833 0.5542 200 R2 0.3145 0.3488 0.3267 0.0221 0.2924 200 Intercept 0.0000 0.0000 0.0882 -0.0882 0.0882 200 Slope 1.0000 1.0000 1.0923 -0.0923 1.0923 200 Emax 0.0000 0.0000 0.0340 0.0340 0.0340 200 D 0.2612 0.2571 0.2370 0.0201 0.2411 200 U -0.0192 -0.0192 -0.0047 -0.0145 -0.0047 200 Q 0.2805 0.2763 0.2417 0.0346 0.2458 200 B 0.1292 0.1224 0.1355 -0.0132 0.1423 200 g 1.2704 1.3917 1.5019 -0.1102 1.3805 200 gp 0.2020 0.2091 0.2229 -0.0138 0.2158 200 In the penalized model (x6factor.lrm.pen), the apparent Dxy is 0.64, and bias-corrected Dxy is 0.55. The maximum absolute error is estimated to be 0.034, smaller than non-penalized model (0.0915 in x6factor.lrm) The changes in slope and intercept are substantially reduced in penalized model. I think overfitting is improved at least to some extent. Should I select this as a final model? I have one more question. The "procedure" variable was defined as 0/1 value in the previous mail. For some graphical reason, I redefined it as treat1/treat2 value. Then, the best penalty value was changed from 3.05 to 2.55. I guess change from numeric to factorial caused this reduction in penalty. Which set up should I select? I appreciate your help in advance. -- KH (11/04/26 0:21), Frank Harrell wrote:
You've done a lot of good work on this. Yes I would say you have moderate overfitting with the first model. The only thing that saved you from having severe overfitting is that there seems to be a signal present [I am assume this model is truly pre-specified and was not developed at all by looking at patterns of responses Y.] The use of backwards stepdown demonstrated much worse overfitting. This is in line with what we know about the damage of stepwise selection methods that do not incorporate shrinkage. I would throw away the stepwise regression model. You'll find that the model selected is entirely arbitrary. And you can't use the "selected" variables in any re-fit of the model, i.e., you can't use lrm pretending that the two remaining variables were pre-specified. Stepwise regression methods only seem to help. When assessed properly we see that is an illusion. You are using penalizing properly but you did not print the full table of penalties vs. effective AIC. We don't have faith that you penalized enough. I tend to run pentrace using a very wide range of possible penalties to make sure I've found the global optimum. Penalization somewhat solves the EPV problem but there is no substitute for getting more data. You can run validate specifying your final penalty as an argument. Frank ???? wrote:
According to the advice, I tried rms package. Just to make sure, I have data of 104 patients (x6.df), which consists of 5 explanatory variables and one binary outcome (poor/good) (previous model 2 strategy). The outcome consists of 25 poor results and 79 good results. Therefore, My events per variable (EPV) is only 5 (much less than the rule of thumb of 10). My questions are about validate and pentrace in rms package. I present some codes and results. I appreciate anybody's help in advance.
> x6.lrm<- lrm(outcome ~ stenosis+x1+x2+procedure+ClinicalScore,
data=x6.df, x=T, y=T)
> x6.lrm
...
Obs 104 LR chi2 29.24 R2 0.367 C 0.816
negative 79 d.f. 5 g 1.633 Dxy 0.632
positive 25 Pr(> chi2)<0.0001 gr 5.118 gamma 0.632
max |deriv| 1e-08 gp 0.237 tau-a 0.233
Brier 0.127
Coef S.E. Wald Z Pr(>|Z|)
Intercept -5.5328 2.6287 -2.10 0.0353
stenosis -0.0150 0.0284 -0.53 0.5979
x1 3.0425 0.9100 3.34 0.0008
x2 -0.7534 0.4519 -1.67 0.0955
procedure 1.2085 0.5717 2.11 0.0345
ClinicalScore 0.3762 0.2287 1.65 0.0999
It seems not too bad. Next, validation by bootstrap ...
> validate(x6.lrm, B=200, bw=F)
index.orig training test optimism index.corrected n Dxy 0.6324 0.6960 0.5870 0.1091 0.5233 200 R2 0.3668 0.4370 0.3154 0.1216 0.2453 200 Intercept 0.0000 0.0000 -0.2007 0.2007 -0.2007 200 Slope 1.0000 1.0000 0.7565 0.2435 0.7565 200 Emax 0.0000 0.0000 0.0999 0.0999 0.0999 200 D 0.2716 0.3368 0.2275 0.1093 0.1623 200 U -0.0192 -0.0192 0.0369 -0.0561 0.0369 200 Q 0.2908 0.3560 0.1906 0.1654 0.1254 200 B 0.1272 0.1155 0.1384 -0.0229 0.1501 200 g 1.6328 2.0740 1.4647 0.6093 1.0235 200 gp 0.2367 0.2529 0.2189 0.0341 0.2026 200 The apparent Dxy is 0.63, and bias-corrected Dxy is 0.52. The maximum absolute error is estimated to be 0.099. The changes in slope and intercept are also more substantial. In all, there is evidence that I am somewhat overfitting the data, right?. Furthermore, using step-down variable selection ...
> validate(x6.lrm, B=200, bw=T)
Backwards Step-down - Original Model
Deleted Chi-Sq d.f. P Residual d.f. P AIC
stenosis 0.28 1 0.5979 0.28 1 0.5979 -1.72
ClinicalScore 2.60 1 0.1068 2.88 2 0.2370 -1.12
x2 2.86 1 0.0910 5.74 3 0.1252 -0.26
Approximate Estimates after Deleting Factors
Coef S.E. Wald Z P
Intercept -5.865 1.4136 -4.149 3.336e-05
x1 2.915 0.8685 3.357 7.889e-04
procedure 1.072 0.5590 1.918 5.508e-02
Factors in Final Model
[1] x1 procedure
index.orig training test optimism index.corrected n
Dxy 0.5661 0.6755 0.5559 0.1196 0.4464 200
R2 0.2876 0.4085 0.2784 0.1301 0.1575 200
Intercept 0.0000 0.0000 -0.2459 0.2459 -0.2459 200
Slope 1.0000 1.0000 0.7300 0.2700 0.7300 200
Emax 0.0000 0.0000 0.1173 0.1173 0.1173 200
D 0.2038 0.3130 0.1970 0.1160 0.0877 200
U -0.0192 -0.0192 0.0382 -0.0574 0.0382 200
Q 0.2230 0.3323 0.1589 0.1734 0.0496 200
B 0.1441 0.1192 0.1452 -0.0261 0.1702 200
g 1.2628 1.9524 1.3222 0.6302 0.6326 199
gp 0.2041 0.2430 0.2043 0.0387 0.1654 199
If I select only two variables (x1 and procedure), bias-corrected Dxy
goes down to 0.45.
[Question 1]
I have EPV problem. Even so, should I keep the full model (5-variable
model)? or can I use the 2-variable (x1 and procedure) model which the
validate() with step-down provides?
[Question 2]
If I use 2-variable model, should I do
x2.lrm<- lrm(postopDWI_HI ~ T1+procedure2, data=x6.df, x=T, y=T)?
or keep the value showed above by validate function?
Next, shrinkage ...
> pentrace(x6.lrm, seq(0, 5.0, by=0.05))
Best penalty:
penalty df
3.05 4.015378
The best penalty is 3.05. So, I update it with this penalty to obtain
the corresponding penalized model:
> x6.lrm.pen<- update(x6.lrm, penalty=3.05, x=T, y=T) > x6.lrm.pen
.....
Penalty factors
simple nonlinear interaction nonlinear.interaction
3.05 3.05 3.05 3.05
Final penalty on -2 log L
[,1]
[1,] 3.8
Obs 104 LR chi2 28.18 R2 0.313 C 0.818
negative 79 d.f. 4.015 g 1.264 Dxy 0.635
positive 25 Pr(> chi2)<0.0001 gr 3.538 gamma 0.637
max |deriv| 3e-05 gp 0.201 tau-a 0.234
Brier 0.129
Coef S.E. Wald Z Pr(>|Z|) Penalty Scale
Intercept -4.7246 2.2429 -2.11 0.0352 0.0000
stenosis -0.0105 0.0240 -0.44 0.6621 17.8021
x1 2.3605 0.7254 3.25 0.0011 0.6054
x2 -0.5385 0.3653 -1.47 0.1404 1.2851
procedure 0.9247 0.4844 1.91 0.0563 0.8576
ClinicalScore 0.3046 0.1874 1.63 0.1041 2.4779
Arrange the coefficients of the two models side by side, and also list
the difference between the two:
> cbind(coef(x6.lrm), coef(x6.lrm.pen),
abs(coef(x6.lrm)-coef(x6.lrm.pen)))
[,1] [,2] [,3]
Intercept -5.53281808 -4.72464766 0.808170417
stenosis -0.01496757 -0.01050797 0.004459599
x1 3.04248257 2.36051833 0.681964238
x2 -0.75335619 -0.53854750 0.214808685
procedure 1.20847252 0.92474708 0.283725441
ClinicalScore 0.37623189 0.30457557 0.071656322
[Question 3]
Is this penalized model the one I should present for my colleagues?
I still have EPV problem. Or is EPV problem O.K. if I use penalization?
I am still wondering about what I can do to avoid EPV problem.
Collecting new data would be a long-time and huge work...
(11/04/22 1:46), khosoda at med.kobe-u.ac.jp wrote:
Thank you for your comment. I forgot to mention that varclus and pvclust showed similar results for my data. BTW, I did not realize rms is a replacement for the Design package. I appreciate your suggestion. -- KH (11/04/21 8:00), Frank Harrell wrote:
I think it's OK. You can also use the Hmisc package's varclus function. Frank ???? wrote:
Dear Prof. Harrel, Thank you very much for your quick advice. I will try rms package. Regarding model reduction, is my model 2 method (clustering and recoding that are blinded to the outcome) permissible? Sincerely, -- KH (11/04/20 22:01), Frank Harrell wrote:
Deleting variables is a bad idea unless you make that a formal part of the BMA so that the attempt to delete variables is penalized for. Instead of BMA I recommend simple penalized maximum likelihood estimation (see the lrm function in the rms package) or pre-modeling data reduction that is blinded to the outcome variable. Frank ???? wrote:
Hi everybody, I apologize for long mail in advance. I have data of 104 patients, which consists of 15 explanatory variables and one binary outcome (poor/good). The outcome consists of 25 poor results and 79 good results. I tried to analyze the data with logistic regression. However, the 15 variables and 25 events means events per variable (EPV) is much less than 10 (rule of thumb). Therefore, I used R package, "BMA" to perform logistic regression with BMA to avoid this problem. model 1 (full model): x1, x2, x3, x4 are continuous variables and others are binary data.
x16.bic.glm<- bic.glm(outcome ~ ., data=x16.df,
glm.family="binomial", OR20, strict=FALSE)
summary(x16.bic.glm)
(The output below has been cut off at the right edge to save space) 62 models were selected Best 5 models (cumulative posterior probability = 0.3606 ): p!=0 EV SD model 1 model2 Intercept 100 -5.1348545 1.652424 -4.4688 -5.15 -5.1536 age 3.3 0.0001634 0.007258 . sex 4.0 .M -0.0243145 0.220314 . side 10.8 .R 0.0811227 0.301233 . procedure 46.9 -0.5356894 0.685148 . -1.163 symptom 3.8 -0.0099438 0.129690 . . stenosis 3.4 -0.0003343 0.005254 . x1 3.7 -0.0061451 0.144084 . x2 100.0 3.1707661 0.892034 3.2221 3.11 x3 51.3 -0.4577885 0.551466 -0.9154 . HT 4.6 .positive 0.0199299 0.161769 . . DM 3.3 .positive -0.0019986 0.105910 . . IHD 3.5 .positive 0.0077626 0.122593 . . smoking 9.1 .positive 0.0611779 0.258402 . . hyperlipidemia 16.0 .positive 0.1784293 0.512058 . . x4 8.2 0.0607398 0.267501 . . nVar 2 2 1 3 3 BIC -376.9082 -376.5588 -376.3094 -375.8468 -374.5582 post prob 0.104 0.087 0.077 0.061 0.032 [Question 1] Is it O.K to calculate odds ratio and its 95% confidence interval from "EV" (posterior distribution mean) and?SD?(posterior distribution standard deviation)? For example, 95%CI of EV of x2 can be calculated as;
exp(3.1707661)
[1] 23.82573 -----> odds ratio
exp(3.1707661+1.96*0.892034)
[1] 136.8866
exp(3.1707661-1.96*0.892034)
[1] 4.146976 ------------------> 95%CI (4.1 to 136.9) Is this O.K.? [Question 2] Is it permissible to delete variables with small value of "p!=0" and "EV", such as age (3.3% and 0.0001634) to reduce the number of explanatory variables and reconstruct new model without those variables for new session of BMA? model 2 (reduced model): I used R package, "pvclust", to reduce the model. The result suggested x1, x2 and x4 belonged to the same cluster, so I picked up only x2. Based on the subject knowledge, I made a simple unweighted sum, by counting the number of clinical features. For 9 features (sex, side, HT2, hyperlipidemia, DM, IHD, smoking, symptom, age), the sum ranges from 0 to 9. This score was defined as ClinicalScore. Consequently, I made up new data set (x6.df), which consists of 5 variables (stenosis, x2, x3, procedure, and ClinicalScore) and one binary outcome (poor/good). Then, for alternative BMA session...
BMAx6.glm<- bic.glm(postopDWI_HI ~ ., data=x6.df,
glm.family="binomial", OR=20, strict=FALSE)
summary(BMAx6.glm)
(The output below has been cut off at the right edge to save space) Call: bic.glm.formula(f = postopDWI_HI ~ ., data = x6.df, glm.family = "binomial", strict = FALSE, OR = 20) 13 models were selected Best 5 models (cumulative posterior probability = 0.7626 ): p!=0 EV SD model 1 model 2 Intercept 100 -5.6918362 1.81220 -4.4688 -6.3166 stenosis 8.1 -0.0008417 0.00815 . . x2 100.0 3.0606165 0.87765 3.2221 3.1154 x3 46.5 -0.3998864 0.52688 -0.9154 . procedure 49.3 0.5747013 0.70164 . 1.1631 ClinicalScore 27.1 0.0966633 0.19645 . . nVar 2 2 1 3 3 BIC -376.9082 -376.5588 -376.3094 -375.8468 -375.5025 post prob 0.208 0.175 0.154 0.122 0.103 [Question 3] Am I doing it correctly or not? I mean this kind of model reduction is permissible for BMA? [Question 4] I still have 5 variables, which violates the rule of thumb, "EPV> 10". Is it permissible to delete "stenosis" variable because of small value of "EV"? Or is it O.K. because this is BMA? Sorry for long post. I appreciate your help very much in advance. -- KH
______________________________________________ R-help at 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/BMA-logistic-regression-odds-ratio-model-reduction-etc-tp3462416p3462919.html Sent from the R help mailing list archive at Nabble.com.
______________________________________________ R-help at 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/BMA-logistic-regression-odds-ratio-model-reduction-etc-tp3462416p3473354.html Sent from the R help mailing list archive at Nabble.com.
______________________________________________ R-help at 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.
--
*************************************************
????????????? ????????
??? ??
?
??650-0017?????????7??5-1
Phone: 078-382-5966
Fax : 078-382-5979
E-mail address
Office: khosoda at med.kobe-u.ac.jp
Home : khosoda at venus.dti.ne.jp
*************************************************
______________________________________________ R-help at 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/BMA-logistic-regression-odds-ratio-model-reduction-etc-tp3462416p3483634.html Sent from the R help mailing list archive at Nabble.com.
______________________________________________ R-help at 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.
*************************************************
????????????? ????????
??? ??
?
??650-0017?????????7??5-1
Phone: 078-382-5966
Fax : 078-382-5979
E-mail address
Office: khosoda at med.kobe-u.ac.jp
Home : khosoda at venus.dti.ne.jp