The HPSPLIT Procedure

A common use of classification trees is to predict the likelihood that a mortgage applicant will default on a loan. The data set Hmeq, which is in the Sampsio library that SAS provides, contains observations for 5,960 mortgage applicants. A variable named Bad indicates whether the applicant paid or defaulted on the loan that was given.

This example uses Hmeq to build a tree model that is used to score the data and can be used to score data about new applicants. Table 61.6 describes the variables in Hmeq.

The response variable for the tree model is Bad, a class variable that has two values (0 indicates payment of loan, and 1 indicates default). The other variables are predictor variables for the model. The following statements display the first 10 observations of the data set:

/* Convert variable names to mixed case */
data hmeq;
   length Bad Loan MortDue Value 8 Reason Job $7
          YoJ Derog Delinq CLAge nInq CLNo DebtInc 8;
   set sampsio.hmeq;
run;

proc print data=hmeq(obs=10); run;

Output 61.4.1 shows the partial listing of Hmeq.

The following statements use the HPSPLIT procedure to create a classification tree:

ods graphics on;

proc hpsplit data=hmeq maxdepth=5;
   class Bad Delinq Derog Job nInq Reason;
   model Bad(event='1') = Delinq Derog Job nInq Reason CLAge CLNo
               DebtInc Loan MortDue Value YoJ;
   prune costcomplexity;
   partition fraction(validate=0.3 seed=123);
   code file='hpsplexc.sas';
   rules file='rules.txt';
run;

The MAXDEPTH= option specifies the maximum depth of the tree to be grown.

Specifying Bad to the left of the equal sign in the MODEL statement indicates that it is the response variable. Because Bad is a binary response, the EVENT= option specifies that the calculations for sensitivity, specificity, AUC, and ROC be based on the level Bad=1.

By default, the entropy metric is used to grow the tree. The PRUNE statement requests cost-complexity pruning.

The PARTITION statement specifies that the observations in Hmeq be logically partitioned into disjoint subsets for model training and validation. Observations are randomly selected for the validation subset with the probability 0.3; the remaining observations are selected for the training subset.

The FILE= option in the CODE statement requests that SAS DATA step score code be saved to a file named hpsplexc.sas, and the FILE= option in the RULES statement requests that the node rules be saved to a file named rules.txt.

The tree diagram in Output 61.4.2 provides an overview of the full tree.

Output 61.4.2: Overview Diagram of Final Tree

You can see from this diagram that the observations in terminal nodes 4, 9, C, and E are assigned a prediction of Bad=0 and those in terminal nodes 6, 8, A, and D are assigned a prediction of Bad=1. You can easily see that node 4 contains the most observations, as indicated by thickness of the link from its parent node.

The tree diagram in Output 61.4.3 is a detailed view of the top portion of the tree. You can use the PLOTS= option in the PROC HPSPLIT statement to control which nodes are displayed.

Output 61.4.3: Detailed Tree Diagram

By default, this view provides detailed splitting information about the first three levels of the tree, including the splitting variable and splitting values. The splitting rule above each node determines which observations from the parent node are included in the node. The first row of the table inside the node provides the node identifier. The second row of the table provides the number of training and validation observations, separated by a slash. The third row shows the predicted response for observations in that node if classification was occurring at that point, along with the proportion of training and validation observations with that observed response. Note that the legend shows what actual value of the response variable is represented by the value shown in the node. For example, in node 6, all 36 observations in the training data and all 18 observations in the validation data have an observed response value of Bad=1, as indicated by the value 2 shown on the third line.

Because the response is categorical, the confusion matrices for the training data and for the validation data are displayed in the table shown in Output 61.4.4.

For the training data, there are 73 observations where the model correctly predicted Bad=1 and 2,135 observations where the model correctly predicted Bad=0. The values in the off-diagonal entries of the matrices show how many times the model misclassified observations.

When you are modeling a binary response variable, a plot of the ROC curve is displayed as shown in Output 61.4.5.

Output 61.4.5: ROC Plot

This plot summarizes the performance of the tree model in terms of sensitivity and specificity.

Output 61.4.6 displays the "Tree Performance" table, which provides several impurity measures and fit statistics for the final tree.

The last three columns of this table contain statistics related to the ROC curve plotted in Output 61.4.5. This model does not classify the observations with Bad=1 (the event of interest) very well, as you can see in the confusion matrix and from the low sensitivity reported in this table. This is often the case with a relatively rare event. The AUC measures the area under the ROC curve. A model that fits the data perfectly would have an AUC of 1. These three columns are included only for a binary response variable.

Output 61.4.7 displays the pruning plot.

Output 61.4.7: Pruning Plot

This plot displays misclassification rates for the training and validation data as the tree is pruned. The tree with eight leaves is selected as the final tree because it has the lowest misclassification rate for the validation data.

data scored;
   set hmeq;
   %include 'hpsplexc.sas';
run;

*------------------------------------------------------------*                  
NODE = 4                                                                        
*------------------------------------------------------------*                  
MISSING(Delinq) OR (Delinq IS ONE OF 0, 1, 8, 10, 11, 12, 13, 15)               
AND MISSING(DebtInc) OR (DebtInc < 43.843383)                                   
    PREDICTED VALUE IS 0                                                        
    PREDICTED 0 = 0.9439( 2037/2158)                                            
    PREDICTED 1 = 0.05607( 121/2158)                                            
*------------------------------------------------------------*                  
NODE = 13                                                                       
*------------------------------------------------------------*                  
MISSING(CLNo) OR (CLNo < 30.08)                                                 
AND (Delinq IS ONE OF 4, 5, 6, 7)                                               
AND MISSING(Value) OR (Value < 129275.32)                                       
AND (Delinq IS ONE OF 2, 3, 4, 5, 6, 7)                                         
AND MISSING(DebtInc) OR (DebtInc < 43.843383)                                   
    PREDICTED VALUE IS 1                                                        
    PREDICTED 0 = 0( 0/11)                                                      
    PREDICTED 1 = 1( 11/11)                                                     
*------------------------------------------------------------*                  
NODE = 14                                                                       
*------------------------------------------------------------*                  
(CLNo >= 30.08)                                                                 
AND (Delinq IS ONE OF 4, 5, 6, 7)                                               
AND MISSING(Value) OR (Value < 129275.32)                                       
AND (Delinq IS ONE OF 2, 3, 4, 5, 6, 7)                                         
AND MISSING(DebtInc) OR (DebtInc < 43.843383)                                   
    PREDICTED VALUE IS 0                                                        
    PREDICTED 0 = 1( 5/5)                                                       
    PREDICTED 1 = 0( 0/5)                                                       
*------------------------------------------------------------*                  
NODE = 9                                                                        
*------------------------------------------------------------*                  
(Reason IS HomeImp)                                                             
AND (DebtInc < 46.7104)                                                         
AND (DebtInc >= 43.843383)                                                      
    PREDICTED VALUE IS 0                                                        
    PREDICTED 0 = 0.8182( 9/11)                                                 
    PREDICTED 1 = 0.1818( 2/11)