[R] Questions about lrm, validate, pentrace

Frank Harrell f.harrell at vanderbilt.edu
Fri Apr 29 15:09:02 CEST 2011


Yes I would select that as the final model.  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.

Frank


細田弘吉 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
>>
>>
>>
>> 細田弘吉 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
>>>>>
>>>>>
>>>>> 細田弘吉 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
>>>>>>>
>>>>>>>
>>>>>>> 細田弘吉 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.
> 
> 
> -- 
> *************************************************
>  神戸大学大学院医学研究科 脳神経外科学分野
>  細田 弘吉
>  
>  〒650-0017 神戸市中央区楠町7丁目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
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