Пример #1
0
 private void tallyTree(
     Tree<String> tree,
     Counter<String> symbolCounter,
     Counter<UnaryRule> unaryRuleCounter,
     Counter<BinaryRule> binaryRuleCounter) {
   if (tree.isLeaf()) return;
   if (tree.isPreTerminal()) return;
   if (tree.getChildren().size() == 1) {
     UnaryRule unaryRule = makeUnaryRule(tree);
     symbolCounter.incrementCount(tree.getLabel(), 1.0);
     unaryRuleCounter.incrementCount(unaryRule, 1.0);
   }
   if (tree.getChildren().size() == 2) {
     BinaryRule binaryRule = makeBinaryRule(tree);
     symbolCounter.incrementCount(tree.getLabel(), 1.0);
     binaryRuleCounter.incrementCount(binaryRule, 1.0);
   }
   if (tree.getChildren().size() < 1 || tree.getChildren().size() > 2) {
     throw new RuntimeException(
         "Attempted to construct a Grammar with an illegal tree: " + tree);
   }
   for (Tree<String> child : tree.getChildren()) {
     tallyTree(child, symbolCounter, unaryRuleCounter, binaryRuleCounter);
   }
 }
Пример #2
0
 private int tallySpans(Tree<String> tree, int start) {
   if (tree.isLeaf() || tree.isPreTerminal()) return 1;
   int end = start;
   for (Tree<String> child : tree.getChildren()) {
     int childSpan = tallySpans(child, end);
     end += childSpan;
   }
   String category = tree.getLabel();
   if (!category.equals("ROOT")) spanToCategories.incrementCount(end - start, category, 1.0);
   return end - start;
 }
Пример #3
0
    private static Tree<String> binarizeTree(Tree<String> tree) {
      String label = tree.getLabel();
      if (tree.isLeaf()) return new Tree<String>(label);
      if (tree.getChildren().size() == 1) {
        return new Tree<String>(
            label, Collections.singletonList(binarizeTree(tree.getChildren().get(0))));
      }
      // I think it tries to binarize a binary tree. This is silly. Just binarize the subtrees.
      if (tree.getChildren().size() == 2) {

        List<Tree<String>> children = new ArrayList<Tree<String>>();
        children.add(binarizeTree(tree.getChildren().get(0)));
        children.add(binarizeTree(tree.getChildren().get(1)));
        return new Tree<String>(label, children);
      }

      // otherwise, it's a TERNARY-or-more local tree,
      // so decompose it into a sequence of binary and unary trees.
      String intermediateLabel = "@" + label + "->";
      Tree<String> intermediateTree = binarizeTreeHelper(tree, 0, intermediateLabel);
      return new Tree<String>(label, intermediateTree.getChildren());
    }