Classifier - Software Development Case Studies in Java - page 77

If there is only one category, that is the category of the set, otherwise the
category of the set cannot be determined.
private static String findCategory(Map items){
// no category if the set is empty
if (items.size() ## 0)
return "?";
// computes the frequency of each category
Map catFreq # new HashMap();
Iterator it # items.keySet().iterator();
String category # "";
while (it.hasNext()){
Item item # (Item)it.next();
category # (String)items.get(item);
Integer count # (Integer)catFreq.get(category);
if (count ## null )
catFreq.put(category, new Integer(1));
else
catFreq.put(category, new Integer(1 ! count.intValue()));
}
// if only one category is present it is the set's category
if (catFreq.keySet().size() ## 1)
return category;
// otherwise it is not possible to assign a category
return "?";
}
The selectSplit() method determines the best split feature for a set of items.
The best split feature is the one implying the highest information gain.
private static String selectSplit(Map items,Map features){
Iterator attr # features.keySet().iterator();
String split # null ;
double maxGain # 0.0;
while (attr.hasNext()){
String candidate # (String)attr.next();
FeatureType type # (FeatureType)features.get(candidate);
double gain # evaluateSplitGain(items,
candidate,type.allowedValues());
if (gain>maxGain){
maxGain # gain;
split # candidate;
}
}
return split;
}