A natural technique to select variables in the context of generalized linear models is to use a stepŵise procedure. It is natural, but contreversial, as discussed by Frank Harrell in a great post, clearly worth reading. Frank mentioned about 10 points against a stepwise procedure.
- It yields R-squared values that are badly biased to be high.
- The F and chi-squared tests quoted next to each variable on the printout do not have the claimed distribution.
- The method yields confidence intervals for effects and predicted values that are falsely narrow (see Altman and Andersen (1989)).
- It yields p-values that do not have the proper meaning, and the proper correction for them is a difficult problem.
- It gives biased regression coefficients that need shrinkage (the coefficients for remaining variables are too large (see Tibshirani (1996)).
- It has severe problems in the presence of collinearity.
- It is based on methods (e.g., F tests for nested models) that were intended to be used to test prespecified hypotheses.
- Increasing the sample size does not help very much (see Derksen and Keselman (1992)).
- It allows us to not think about the problem.
- It uses a lot of paper.
In order to illustrate that issue of variable selection, consider a dataset I’ve been using many times on the blog,
MYOCARDE=read.table( "http://freakonometrics.free.fr/saporta.csv", head=TRUE,sep=";")
where we have observations from people entering E.R., because of a (potential) infarctus, and we want to understand who did survive, and to build a predictive model.
What if we use a forward stepwise logistic regression here? I want to use a forward construction since it usually yields to models with less explanatory variables. We can use Akaike Information Criterion
> reg0=glm(PRONO~1,data=MYOCARDE,family=binomial) > reg1=glm(PRONO~.,data=MYOCARDE,family=binomial) > step(reg0,scope=formula(reg1), + direction="forward",k=2) # AIC Start: AIC=98.03 PRONO ~ 1 Df Deviance AIC + REPUL 1 46.884 50.884 + INSYS 1 51.865 55.865 + INCAR 1 53.313 57.313 + PRDIA 1 78.503 82.503 + PAPUL 1 82.862 86.862 + PVENT 1 87.093 91.093 <none> 96.033 98.033 + FRCAR 1 94.861 98.861 Step: AIC=50.88 PRONO ~ REPUL Df Deviance AIC + INCAR 1 44.530 50.530 + PVENT 1 44.703 50.703 + INSYS 1 44.857 50.857 <none> 46.884 50.884 + PAPUL 1 45.274 51.274 + PRDIA 1 46.322 52.322 + FRCAR 1 46.540 52.540 Step: AIC=50.53 PRONO ~ REPUL + INCAR Df Deviance AIC <none> 44.530 50.530 + PVENT 1 43.134 51.134 + PRDIA 1 43.772 51.772 + INSYS 1 44.305 52.305 + PAPUL 1 44.341 52.341 + FRCAR 1 44.521 52.521 Call: glm(formula = PRONO ~ REPUL + INCAR, family = binomial, data = MYOCARDE) Coefficients: (Intercept) REPUL 1.633668 -0.003564 INCAR 1.618479 Degrees of Freedom: 70 Total (i.e. Null); 68 Residual Null Deviance: 96.03 Residual Deviance: 44.53 AIC: 50.53
or Schwarz Bayesian Information Criterion,
> step(reg0,scope=formula(reg1), + direction="forward",k=log(n)) # BIC Start: AIC=98.11 PRONO ~ 1 Df Deviance AIC + REPUL 1 46.884 51.043 + INSYS 1 51.865 56.024 + INCAR 1 53.313 57.472 + PRDIA 1 78.503 82.662 + PAPUL 1 82.862 87.021 + PVENT 1 87.093 91.252 <none> 96.033 98.113 + FRCAR 1 94.861 99.020 Step: AIC=51.04 PRONO ~ REPUL Df Deviance AIC + INCAR 1 44.530 50.768 + PVENT 1 44.703 50.942 <none> 46.884 51.043 + INSYS 1 44.857 51.095 + PAPUL 1 45.274 51.512 + PRDIA 1 46.322 52.561 + FRCAR 1 46.540 52.778 Step: AIC=50.77 PRONO ~ REPUL + INCAR Df Deviance AIC <none> 44.530 50.768 + PVENT 1 43.134 51.452 + PRDIA 1 43.772 52.089 + INSYS 1 44.305 52.623 + PAPUL 1 44.341 52.659 + FRCAR 1 44.521 52.838 Call: glm(formula = PRONO ~ REPUL + INCAR, family = binomial, data = MYOCARDE) Coefficients: (Intercept) REPUL 1.633668 -0.003564 INCAR 1.618479 Degrees of Freedom: 70 Total (i.e. Null); 68 Residual Null Deviance: 96.03 Residual Deviance: 44.53 AIC: 50.53
With those two approaches, we have the same story: the most important variable (or say with the highest predictive value) is REPUL. And we can improve the model by adding INCAR. And that’s it. We can get a good model with those two covariates.
Now, what about using cross-validation here? We should keep in ming that AIC is asymptotically equivalent to One-Leave-Out Cross Validation (see Stone (1977)), while BIC is equivalent to -fold Cross Validation (see Shao (1997)), where
- Using Leave-One-Out Cross Validation
In order to select the first variable, consider 7 logistic regression, each on a single different variable. Each time, we estimate the model on observations and get a prediction on the remaining one,
on. Set . The function to get those values is
> name_var=names(MYOCARDE) > pred_i=function(i,k){ + fml = paste(name_var[8],"~",name_var[k],sep="") + reg=glm(fml,data=MYOCARDE[-i,],family=binomial) + predict(reg,newdata=MYOCARDE[i,], + type="response") + }
then for each variable , we get the ROC curve using and,
> library(AUC) > ROC=function(k){ + Y=MYOCARDE[,8]=="Survival" + S=Vectorize(function(i) pred_i(i,k)) + (1:length(Y)) + R=roc(S,as.factor(Y)) + return(list(roc=cbind(R$fpr,R$tpr), + auc=AUC::auc(R))) + }
Here, for each variable, we compute the area under the curve (AUC criterion)
> AUC=rep(NA,7) > for(k in 1:7){ + AUC[k]=ROC(k)$auc + cat("Variable ",k,"(",name_var[k],") :", + AUC[k],"\n") } Variable 1 ( FRCAR ) : 0.4934319 Variable 2 ( INCAR ) : 0.8965517 Variable 3 ( INSYS ) : 0.909688 Variable 4 ( PRDIA ) : 0.7487685 Variable 5 ( PAPUL ) : 0.7134647 Variable 6 ( PVENT ) : 0.6584565 Variable 7 ( REPUL ) : 0.9154351
But we can also visualize those curves,
> plot(0:1,0:1,col="white",xlab="",ylab="") > for(k in 1:7) + lines(ROC(k)$roc,type="s",col=CL[k]) > legend(.8,.45,name_var,col=CL,lty=1,cex=.8)
(there is no PRONO here, there is a typo in the Legend)
where here colors were obtained using
> library(RColorBrewer) > CL=brewer.pal(8, "Set1")[-7]
Here ROC curves were obtained using a Leave-one-Out strategy. And the best variable (if we should keep one, and one only) is
> k0=which.max(AUC) > name_var[k0] [1] "REPUL"
Now, consider a stepwise procedure: we keep that ‘best’ variable in our model, and we try to add another one.
> pred_i=function(i,k){ + vk=c(k0,k) + fml = paste(name_var[8],"~",paste(name_var[vk], + collapse="+"),sep="") + reg=glm(fml,data=MYOCARDE[-i,],family=binomial) + predict(reg,newdata=MYOCARDE[i,], + type="response") + } > library(AUC) > ROC=function(k){ + Y=MYOCARDE[,8]=="Survival" + S=Vectorize(function(i) pred_i(i,k)) + (1:length(Y)) + R=roc(S,as.factor(Y)) + return(list(roc=cbind(R$fpr,R$tpr), + auc=AUC::auc(R))) + } > plot(0:1,0:1,col="white",xlab="",ylab="") > for(k in (1:7)[-k0]) lines(ROC(k)$roc,type="s",col=CL[k]) > segments(0,0,1,1,lty=2,col="grey") > legend(.8,.45, + name_var[-k0], + col=CL[-k0],lty=1,cex=.8)
We were already quite good. And we might expect to find another variable that will increase the predictive power of our classifier.
> AUC=rep(NA,7) > for(k in (1:7)[-k0]){ + AUC[k]=ROC(k)$auc + cat("Variable ",k,"(",name_var[k],") :", + AUC[k],"\n") + } Variable 1 ( FRCAR ) : 0.9064039 Variable 2 ( INCAR ) : 0.9195402 Variable 3 ( INSYS ) : 0.9187192 Variable 4 ( PRDIA ) : 0.9137931 Variable 5 ( PAPUL ) : 0.9187192 Variable 6 ( PVENT ) : 0.9137931
And, of course, we can move foreward, add another variable, etc,
> k0=c(k0,which.max(AUC)) > pred_i=function(i,k){ + vk=c(k0,k) + fml = paste(name_var[8],"~",paste( + name_var[vk],collapse="+"),sep="") + reg=glm(fml,data=MYOCARDE[-i,],family=binomial) + predict(reg,newdata=MYOCARDE[i,], + type="response") + } > library(AUC) > ROC=function(k){ + Y=MYOCARDE[,8]=="Survival" + S=Vectorize(function(i) pred_i(i,k)) + (1:length(Y)) + R=roc(S,as.factor(Y)) + return(list(roc=cbind(R$fpr,R$tpr), + auc=AUC::auc(R))) + } > > plot(0:1,0:1,col="white",xlab="",ylab="") > for(k in (1:7)[-k0]) lines(ROC(k)$roc,type="s",col=CL[k]) > segments(0,0,1,1,lty=2,col="grey") > legend(.8,.45,name_var[-k0], + col=CL[-k0],lty=1,cex=.8)
But here, the gain is rather small (if any)
> AUC=rep(NA,7) > for(k in (1:7)[-k0]){ + AUC[k]=ROC(k)$auc + cat("Variable ",k,"(",name_var[k],") :", + AUC[k],"\n") + } Variable 1 ( FRCAR ) : 0.9121511 Variable 3 ( INSYS ) : 0.9170772 Variable 4 ( PRDIA ) : 0.910509 Variable 5 ( PAPUL ) : 0.907225 Variable 6 ( PVENT ) : 0.909688
With that stepwise algorithm, the best strategy is to keep, first, REPUL, and then to add INCAR. Which is consistent with the stepwise procedure using Akaike Information Criterion.
An alternative could be to select the best pair among all possible pairs. It will be time consuming, but it can be used to avoid the stepwise drawback.
> pred_i=function(i,k){ + fml = paste(name_var[8],"~",paste(name_var[ + as.integer(k)],collapse="+"),sep="") + reg=glm(fml,data=MYOCARDE[-i,],family=binomial) + predict(reg,newdata=MYOCARDE[i,], + type="response") + } > library(AUC) > ROC=function(k){ + Y=MYOCARDE[,8]=="Survival" + L=list() + n=length(Y) + nk=trunc(n/trunc(n/10)) + for(i in 1:(nk-1)) L[[i]]=((i-1)* + trunc(n/10)+1:(n/10)) + L[[nk]]=((nk-1)*trunc(n/10)+1):n + S=unlist(Vectorize(function(i) + pred_i(L[[i]],k))(1:nk)) + R=roc(S,as.factor(Y)) + return(AUC::auc(R)) + } > v=data.frame(k1=rep(1:7,each=7),k2=rep(1:7,7)) > v=v[v$k1<v$k2,] > v$auc=NA > for(i in 1:nrow(v)) v$auc[i]=ROC(v[i,1:2]) > v k1 k2 auc 2 1 2 0.9047619 3 1 3 0.9047619 4 1 4 0.6990969 5 1 5 0.6395731 6 1 6 0.6334154 7 1 7 0.8817734 10 2 3 0.9072250 11 2 4 0.9088670 12 2 5 0.8940887 13 2 6 0.8801314 14 2 7 0.8899836 18 3 4 0.8916256 19 3 5 0.8817734 20 3 6 0.9014778 21 3 7 0.8768473 26 4 5 0.6925287 27 4 6 0.7138752 28 4 7 0.8825944 34 5 6 0.6912972 35 5 7 0.8834154 42 6 7 0.8834154
Here the best pair is
> v[which.max(v$auc),] k1 k2 auc 11 2 4 0.908867 > name_var[as.integer(v[which.max(v$auc),1:2])] [1] "INCAR" "PRDIA"
which is different, compared with the one we got above. What is odd here is that we get a smaller AUC than the ones we got at step 2 in the stepwise procedure.
Nevertheless, even with a few observations (our dataset is rather small here), it is time consuming to look at all ROC curves, for all pairs. An alternative might be to use –Fold Cross Validation.
- Using -Fold Cross Validation
Here we consider a partition of indices, , and we define
based on observations. For all, set . Then, we can use the stepwise method described above
> pred_i=function(i,k){ + fml = paste(name_var[8],"~",name_var[k],sep="") + reg=glm(fml,data=MYOCARDE[-i,],family=binomial) + predict(reg,newdata=MYOCARDE[i,], + type="response") + } > library(AUC) > ROC=function(k){ + Y=MYOCARDE[,8]=="Survival" + L=list() + n=length(Y) + nk=trunc(n/trunc(n/10)) + for(i in 1:(nk-1)) L[[i]]=((i-1)* + trunc(n/10)+1:(n/10)) + L[[nk]]=((nk-1)*trunc(n/10)+1):n + S=unlist(Vectorize(function(i) + pred_i(L[[i]],k))(1:nk)) + R=roc(S,as.factor(Y)) + return(list(roc=cbind(R$fpr,R$tpr), + auc=AUC::auc(R))) + } > plot(0:1,0:1,col="white",xlab="",ylab="") > for(k in (1:7)) lines(ROC(k)$roc,col=CL[k]) > segments(0,0,1,1,lty=2,col="grey") > legend(.8,.45,name_var,col=CL,lty=1,cex=.8)
with
> AUC=rep(NA,7) > for(k in 1:7){ + AUC[k]=ROC(k)$auc + cat("Variable ",k,"(",name_var[k],") :", + AUC[k],"\n") + } Variable 1 ( FRCAR ) : 0.3932677 Variable 2 ( INCAR ) : 0.8940887 Variable 3 ( INSYS ) : 0.908046 Variable 4 ( PRDIA ) : 0.7278325 Variable 5 ( PAPUL ) : 0.6756979 Variable 6 ( PVENT ) : 0.63711 Variable 7 ( REPUL ) : 0.8834154
So, this time, INSYS is probably the best covariate to use. Now, if we keep that variable, and move forward,
> k0=which.max(AUC) > pred_i=function(i,k){ + vk=c(k0,k) + fml = paste(name_var[8],"~",paste(name_var[vk], + collapse="+"),sep="") + reg=glm(fml,data=MYOCARDE[-i,],family=binomial) + predict(reg,newdata=MYOCARDE[i,], + type="response") + } > library(AUC) > ROC=function(k){ + Y=MYOCARDE[,8]=="Survival" + L=list() + n=length(Y) + nk=trunc(n/trunc(n/10)) + for(i in 1:(nk-1)) L[[i]]=((i-1)* + trunc(n/10)+1:(n/10)) + L[[nk]]=((nk-1)*trunc(n/10)+1):n + S=unlist(Vectorize(function(i) + pred_i(L[[i]],k))(1:nk)) + R=roc(S,as.factor(Y)) + return(list(roc=cbind(R$fpr,R$tpr), + auc=AUC::auc(R))) + } > plot(0:1,0:1,col="white",xlab="",ylab="") > for(k in (1:7)[-k0]) + lines(ROC(k)$roc,col=CL[k]) > segments(0,0,1,1,lty=2,col="grey") > legend(.8,.45,name_var[-k0], + col=CL[-k0],lty=1,cex=.8)
and our best choice for the second variable would be INCAR
> AUC=rep(NA,7) > for(k in (1:7)[-k0]){ + AUC[k]=ROC(k)$auc + cat("Variable ",k,"(",name_var[k],") :", + AUC[k],"\n") + } Variable 1 ( FRCAR ) : 0.9047619 Variable 2 ( INCAR ) : 0.907225 Variable 4 ( PRDIA ) : 0.8916256 Variable 5 ( PAPUL ) : 0.8817734 Variable 6 ( PVENT ) : 0.9014778 Variable 7 ( REPUL ) : 0.8768473 > which.max(AUC) [1] 2
Here again, it is possible to look at all pairs
> pred_i=function(i,k){ + fml = paste(name_var[8],"~",paste(name_var[ + as.integer(k)],collapse="+"),sep="") + reg=glm(fml,data=MYOCARDE[-i,],family=binomial) + predict(reg,newdata=MYOCARDE[i,], + type="response") + } > library(AUC) > ROC=function(k){ + Y=MYOCARDE[,8]=="Survival" + L=list() + n=length(Y) + nk=trunc(n/trunc(n/10)) + for(i in 1:(nk-1)) L[[i]]=((i-1)* + trunc(n/10)+1:(n/10)) + L[[nk]]=((nk-1)*trunc(n/10)+1):n + S=unlist(Vectorize(function(i) + pred_i(L[[i]],k))(1:nk)) + R=roc(S,as.factor(Y)) + return(AUC::auc(R)) + } > v=data.frame(k1=rep(1:7,each=7),k2=rep(1:7,7)) > v=v[v$k1<v$k2,] > v$auc=NA > for(i in 1:nrow(v)) v$auc[i]=ROC(v[i,1:2]) > v[which.max(v$auc),] k1 k2 auc 11 2 4 0.908867 > name_var[as.integer(v[which.max(v$auc),1:2])] [1] "INCAR" "PRDIA"
which is the same as what we got using an One-Leave-Out strategy: we have again the same two covariates. And again, the AUC is lower than the one we got using a stepwise procedure (I still don’t understand how this is possible). An alternative for the code would be to store all the regression models in a list,
> L=list() > n=nrow(MYOCARDE) > nk=trunc(n/trunc(n/10)) > for(i in 1:(nk-1)) L[[i]]=((i-1)*trunc(n/10)+ + 1:(n/10)) > L[[nk]]=((nk-1)*trunc(n/10)+1):n > REG=list() > for(k in 1:7){ + REG[[k]]=list() + fml = paste(name_var[8],"~", + paste(name_var[as.integer(k)],collapse="+"), + sep="") + for(i in 1:nk) REG[[k]][[i]]=reg=glm(fml, + data=MYOCARDE[-L[[i]],],family=binomial) + }
and then to call them, properly
> pred_i=function(i,k){ + I=which(sapply(1:10,function(j) i%in%L[[j]])) + predict(REG[[k]][[I]],newdata=MYOCARDE[i,], + type="response") + }
One has to check about the efficiency, especially with a large dataset.
- Using Trees and Random Forests
Another quick, but popular (from what I’ve seen), technique is to use trees. Important variables should appear in the output,
> library(rpart) > tree=rpart(PRONO~.,data=MYOCARDE) > library(rpart.plot) > rpart.plot(tree)
Here, the first variable that appears in the tree construction is INSYS, and the second on is REPUL. Which is rather different with what we got above. But using one tree is maybe not sufficient. One can use the variable importance function (described in a previous post) obtained using random forests.
> library(randomForest) > rf=randomForest(PRONO~.,data=MYOCARDE, + importance=TRUE) > rf$importance[,4] FRCAR INCAR INSYS 1.042006 7.363255 8.954898 PRDIA PAPUL PVENT 3.149235 2.571267 3.152619 REPUL 7.510110
Here, we have the same story as the one we got with a simple tree: the ‘most important’ variable is INSYS while the second one is REPUL. But here, we consider tree based predictors. And not a logistic regression.
- Using Dedicated R functions
It is possible to use some dedicated R functions for variable selection. For instance, since we consider a logistic regression, use
> library(bestglm) > Xy=as.data.frame(MYOCARDE) > Xy[,8]=(Xy[,8]=="Death")*1 > names(Xy)=names(MYOCARDE) > B=bestglm(Xy) > B$Subsets[,2:8] FRCAR INCAR INSYS PRDIA PAPUL PVENT REPUL 0 FALSE FALSE FALSE FALSE FALSE FALSE FALSE 1 FALSE FALSE FALSE FALSE FALSE FALSE TRUE 2* FALSE TRUE FALSE FALSE FALSE FALSE TRUE 3 FALSE TRUE FALSE FALSE FALSE TRUE TRUE 4 TRUE TRUE FALSE FALSE FALSE TRUE TRUE 5 TRUE TRUE TRUE FALSE FALSE TRUE TRUE 6 TRUE TRUE FALSE TRUE TRUE TRUE TRUE 7 TRUE TRUE TRUE TRUE TRUE TRUE TRUE
With only one variable, we should consider REPUL (row 1 of the matrix), while with two variables, we should consider REPUL and INCAR (and that is the best model, based on some Bayesian Information Criterion). Here, cross validation techniques can be used also,
> B=bestglm(Xy, IC = "CV", CVArgs = + list(Method = "HTF", K = 10, REP = 1)) > cverrs = B$Subsets[, "CV"] > sdCV = B$Subsets[, "sdCV"] > CVLo = cverrs - sdCV > CVHi = cverrs + sdCV > ymax = max(CVHi) > ymin = min(CVLo) > k = 0:(length(cverrs) - 1) > plot(k, cverrs, ylim = c(ymin, + ymax), type = "n", yaxt = "n") > points(k,cverrs,cex = 2,col="red",pch=16) > lines(k, cverrs, col = "red", lwd = 2) > axis(2, yaxp = c(0.6, 1.8, 6)) > segments(k, CVLo, k, CVHi,col="blue", lwd = 2) > eps = 0.15 > segments(k-eps, CVLo, k+eps, CVLo, + col = "blue", lwd = 2) > segments(k-eps, CVHi, k+eps, CVHi, + col = "blue", lwd = 2) > indMin = which.min(cverrs) > fmin = sdCV[indMin] > cutOff = fmin + cverrs[indMin] > abline(h = cutOff, lty = 2) > indMin = which.min(cverrs) > fmin = sdCV[indMin] > cutOff = fmin + cverrs[indMin] > min(which(cverrs < cutOff)) [1] 2
If we summarize, here,
- stepwise, AIC : REPUL + INCAR
- stepwise, BIC : REPUL + INCAR
- One-leave-Out CV stepwise : REPUL + INCAR
- One-leave-Out CV pairs : INCAR + PRDIA
- -fold CV stepwise : INSYS + INCAR
- -fold CV pairs : INCAR + PRDIA
- Tree : INSYS + REPUL
- Variable Importance (RF) : INSYS + INCAR
- Best GLM : REPUL + INCAR
That is the lovely part with statistical tools: there are usually multiple (valid) answers. And this is why machine learning is difficult: if there was a single answer, any machine could built up a model that works well. But obviously, it has to be more complicated…
OpenEdition suggests that you cite this post as follows:
Arthur Charpentier (July 1, 2015). Variable Selection using Cross-Validation (and Other Techniques). Freakonometrics. Retrieved November 2, 2024 from https://doi.org/10.58079/ouzz
This is a very interesting post and I would like to read more about stepwise selection using cross validation. Can you suggest any papers or textbooks? Thanks!
Hi Arthur! Just a small comment of a possible useful improvement of your site: I always read your articles in the train, I print them at work and take them with me. It would be nice if you had like a button to get a print-friendly version without the comments below and stuff. But ok it’s a luxury problem, I can keep doing copy-paste in Word 🙂 Very nice article this one, thank you and keep up the good work!
Dear Arthur, I’ve tried the codes, but I had some problem.
In doing the variable selection with BIC, I put the code
step(reg0,scope=formula(reg1),direction=”forward”,k=log(n))
But I had an error warning. I solved by making n = 7, and the analysis was fine.
Then I’ve tried the pred_i=function, but something went wrong, since, after calling this code:
AUC=rep(NA,7)
for(k in 1:7){
AUC[k]=ROC(k)$auc
cat(“Variable “,k,”(“,name_var[k],”) :”,
AUC[k],”\n”) }
I’ve got this error:
Error in order(predictions, decreasing = TRUE) :
argument 1 is not type vector
Since the code depend on past codes, I do not know how to fix this.
I’m working with R version 3.0.2 (2013-09-25) on Platform: x86_64-w64-mingw32/x64 (64-bit)
[Windows]
Thanx in advance!
that’s odd, what does ROC(7) returns ?
Tried
I’ve got the same error
There is some passage which should produce a vectorization but it does not…
It seems that ‘i’ is not defined (no value assigned to it) in both pred_i and ROC
Here my complete codes, as I took from the post:
load the data
MYOCARDE=read.table(“http://freakonometrics.free.fr/saporta.csv”,
head=TRUE,sep=”;”)
### take a look at the data
dim(MYOCARDE)
head(MYOCARDE)
str(MYOCARDE)
attach(MYOCARDE)
plot(MYOCARDE[,1:7],col=as.numeric(PRONO))
#### model without and with predictors
reg0=glm(PRONO~1,data=MYOCARDE,family=binomial)
reg1=glm(PRONO~.,data=MYOCARDE,family=binomial)
#### variable selection with Akaike Information Criterion (AIC)
step(reg0,scope=formula(reg1), direction=”forward”,k=2)
#### variable selection with Schwarz Bayesian Information Criterion (BIC)
n = 7
step(reg0,scope=formula(reg1),direction=”forward”,k=log(n))
### With those two approaches, we have the same story:
### the most important variable (or say with the highest predictive value) is REPUL.
### And we can improve the model by adding INCAR.
### AIC is asymptotically equivalent to One-Leave-Out Cross Validation (see Stone (1977)),
### while BIC is equivalent to v-fold Cross Validation (see Shao (1997)), where v = n[1 – 1/(log[n] – 1)]
### In order to select the first variable, consider 7 logistic regression, each on a single different variable.
### Each time, we estimate the model on observations and get a prediction on the remaining one
name_var=names(MYOCARDE)
pred_i=function(i,k){
fml = paste(name_var[8],”~”,name_var[k],sep=””)
reg=glm(fml,data=MYOCARDE[-i,],family=binomial)
predict(reg,newdata=MYOCARDE[i,],
type=”response”)
}
### I’m stuck here
### then for each variable , we get the ROC curve
library(AUC)
ROC=function(k){
Y=MYOCARDE[,8]==”Survival”
S=Vectorize(function(i) pred_i(i,k))
(1:length(Y))
R=roc(S,as.factor(Y))
return(list(roc=cbind(R$fpr,R$tpr),
auc=AUC::auc(R)))
}
### for each variable, we compute the area under the curve
AUC=rep(NA,7)
for(k in 1:7){
AUC[7]=ROC(k)$auc
cat(“Variable “,k,”(“,name_var[k],”) :”,
AUC[k],”\n”) }
### plot of the AUC curve
library(RColorBrewer)
CL=brewer.pal(8, “Set1″)[-7]
plot(0:1,0:1,col=”white”,xlab=””,ylab=””)
for(k in 1:7)
lines(ROC(k)$roc,type=”s”,col=CL[k])
legend(.8,.45,name_var,col=CL,lty=1,cex=.8)
ROC(7) returns “argument 1 is not a vector”
Just put “S=Vectorize(function(i) pred_i(i,k)) (1:length(Y))” into the same line in the ROC function and it wit work for you.
I think some algorithms like Random Forest or AdaBoost have an in-built procedure to select features/variables, so with these techniques it is not always necessary to make a variable selection, to obtain a good model. But of course sometimes you are just interested in the variables that have the greatest influence on the prediction.