Example usage for edu.stanford.nlp.trees Tree pathNodeToNode

List of usage examples for edu.stanford.nlp.trees Tree pathNodeToNode

Introduction

In this page you can find the example usage for edu.stanford.nlp.trees Tree pathNodeToNode.

Prototype

public List<Tree> pathNodeToNode(Tree t1, Tree t2) 

Source Link

Document

Given nodes t1 and t2 which are dominated by this node, returns a list of all the nodes on the path from t1 to t2, inclusive, or null if none found.

Usage

From source file:Anaphora_Resolution.ParseAllXMLDocuments.java

public static Tree HobbsResolve(Tree pronoun, ArrayList<Tree> forest) {
    Tree wholetree = forest.get(forest.size() - 1); // The last one is the one I am going to start from
    ArrayList<Tree> candidates = new ArrayList<Tree>();
    List<Tree> path = wholetree.pathNodeToNode(wholetree, pronoun);
    System.out.println(path);//ww  w . java 2 s .  c  o  m
    // Step 1
    Tree ancestor = pronoun.parent(wholetree); // This one locates the NP the pronoun is in, therefore we need one more "parenting" !
    // Step 2
    ancestor = ancestor.parent(wholetree);
    //System.out.println("LABEL: "+pronoun.label().value() + "\n\tVALUE: "+pronoun.firstChild());
    while (!ancestor.label().value().equals("NP") && !ancestor.label().value().equals("S"))
        ancestor = ancestor.parent(wholetree);
    Tree X = ancestor;
    path = X.pathNodeToNode(wholetree, pronoun);
    System.out.println(path);
    // Step 3
    for (Tree relative : X.children()) {
        for (Tree candidate : relative) {
            if (candidate.contains(pronoun))
                break; // I am looking to all the nodes to the LEFT (i.e. coming before) the path leading to X. contain <-> in the path
            //System.out.println("LABEL: "+relative.label().value() + "\n\tVALUE: "+relative.firstChild());
            if ((candidate.parent(wholetree) != X) && (candidate.parent(wholetree).label().value().equals("NP")
                    || candidate.parent(wholetree).label().value().equals("S")))
                if (candidate.label().value().equals("NP")) // "Propose as the antecedent any NP node that is encountered which has an NP or S node between it and X"
                    candidates.add(candidate);
        }
    }
    // Step 9 is a GOTO step 4, hence I will envelope steps 4 to 8 inside a while statement.
    while (true) { // It is NOT an infinite loop. 
        // Step 4
        if (X.parent(wholetree) == wholetree) {
            for (int q = 1; q < MAXPREVSENTENCES; ++q) {// I am looking for the prev sentence (hence we start with 1)
                if (forest.size() - 1 < q)
                    break; // If I don't have it, break
                Tree prevTree = forest.get(forest.size() - 1 - q); // go to previous tree
                // Now we look for each S subtree, in order of recency (hence right-to-left, hence opposite order of that of .children() ).
                ArrayList<Tree> backlist = new ArrayList<Tree>();
                for (Tree child : prevTree.children()) {
                    for (Tree subtree : child) {
                        if (subtree.label().value().equals("S")) {
                            backlist.add(child);
                            break;
                        }
                    }
                }
                for (int i = backlist.size() - 1; i >= 0; --i) {
                    Tree Treetovisit = backlist.get(i);
                    for (Tree relative : Treetovisit.children()) {
                        for (Tree candidate : relative) {
                            if (candidate.contains(pronoun))
                                continue; // I am looking to all the nodes to the LEFT (i.e. coming before) the path leading to X. contain <-> in the path
                            //System.out.println("LABEL: "+relative.label().value() + "\n\tVALUE: "+relative.firstChild());
                            if (candidate.label().value().equals("NP")) { // "Propose as the antecedent any NP node that you find"
                                if (!candidates.contains(candidate))
                                    candidates.add(candidate);
                            }
                        }
                    }
                }
            }
            break; // It will always come here eventually
        }
        // Step 5
        ancestor = X.parent(wholetree);
        //System.out.println("LABEL: "+pronoun.label().value() + "\n\tVALUE: "+pronoun.firstChild());
        while (!ancestor.label().value().equals("NP") && !ancestor.label().value().equals("S"))
            ancestor = ancestor.parent(wholetree);
        X = ancestor;
        // Step 6
        if (X.label().value().equals("NP")) { // If X is an NP
            for (Tree child : X.children()) { // Find the nominal nodes that X directly dominates
                if (child.label().value().equals("NN") || child.label().value().equals("NNS")
                        || child.label().value().equals("NNP") || child.label().value().equals("NNPS"))
                    if (!child.contains(pronoun))
                        candidates.add(X); // If one of them is not in the path between X and the pronoun, add X to the antecedents
            }
        }
        // Step SETTE
        for (Tree relative : X.children()) {
            for (Tree candidate : relative) {
                if (candidate.contains(pronoun))
                    continue; // I am looking to all the nodes to the LEFT (i.e. coming before) the path leading to X. contain <-> in the path
                //System.out.println("LABEL: "+relative.label().value() + "\n\tVALUE: "+relative.firstChild());
                if (candidate.label().value().equals("NP")) { // "Propose as the antecedent any NP node that you find"
                    boolean contains = false;
                    for (Tree oldercandidate : candidates) {
                        if (oldercandidate.contains(candidate)) {
                            contains = true;
                            break;
                        }
                    }
                    if (!contains)
                        candidates.add(candidate);
                }
            }
        }
        // Step 8
        if (X.label().value().equals("S")) {
            boolean right = false;
            // Now we want all branches to the RIGHT of the path pronoun -> X.
            for (Tree relative : X.children()) {
                if (relative.contains(pronoun)) {
                    right = true;
                    continue;
                }
                if (!right)
                    continue;
                for (Tree child : relative) { // Go in but do not go below any NP or S node. Go below the rest
                    if (child.label().value().equals("NP")) {
                        candidates.add(child);
                        break; // not sure if this means avoid going below NP but continuing with the rest of non-NP children. Should be since its DFS.
                    }
                    if (child.label().value().equals("S"))
                        break; // Same
                }
            }
        }
    }

    // Step 9 is a GOTO, so we use a while.

    System.out.println(pronoun + ": CHAIN IS " + candidates.toString());
    ArrayList<Integer> scores = new ArrayList<Integer>();

    for (int j = 0; j < candidates.size(); ++j) {
        Tree candidate = candidates.get(j);
        Tree parent = null;
        int parent_index = 0;
        for (Tree tree : forest) {
            if (tree.contains(candidate)) {
                parent = tree;
                break;
            }
            ++parent_index;
        }
        scores.add(0);
        if (parent_index == 0)
            scores.set(j, scores.get(j) + 100); // If in the last sentence, +100 points
        scores.set(j, scores.get(j) + syntacticScore(candidate, parent));

        if (existentialEmphasis(candidate)) // Example: "There was a dog standing outside"
            scores.set(j, scores.get(j) + 70);
        if (!adverbialEmphasis(candidate, parent))
            scores.set(j, scores.get(j) + 50);
        if (headNounEmphasis(candidate, parent))
            scores.set(j, scores.get(j) + 80);

        int sz = forest.size() - 1;
        //          System.out.println("pronoun in sentence " + sz + "(sz). Candidate in sentence "+parent_index+" (parent_index)");
        int dividend = 1;
        for (int u = 0; u < sz - parent_index; ++u)
            dividend *= 2;
        //System.out.println("\t"+dividend);
        scores.set(j, scores.get(j) / dividend);
        System.out.println(candidate + " -> " + scores.get(j));
    }
    int max = -1;
    int max_index = -1;
    for (int i = 0; i < scores.size(); ++i) {
        if (scores.get(i) > max) {
            max_index = i;
            max = scores.get(i);
        }
    }
    Tree final_candidate = candidates.get(max_index);
    System.out.println("My decision for " + pronoun + " is: " + final_candidate);
    // Decide what candidate, with both gender resolution and Lappin and Leass ranking.

    Tree pronounparent = pronoun.parent(wholetree).parent(wholetree); // 1 parent gives me the NP of the pronoun
    int pos = 0;
    for (Tree sibling : pronounparent.children()) {
        System.out.println("Sibling " + pos + ": " + sibling);
        if (sibling.contains(pronoun))
            break;
        ++pos;
    }
    System.out.println("Before setchild: " + pronounparent);
    @SuppressWarnings("unused")
    Tree returnval = pronounparent.setChild(pos, final_candidate);
    System.out.println("After setchild: " + pronounparent);

    return wholetree; // wholetree is already modified, since it contains pronounparent
}