822 lines
34 KiB
Java
822 lines
34 KiB
Java
/*
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* Copyright (C) 2011 The Android Open Source Project
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*
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* Licensed under the Apache License, Version 2.0 (the "License"); you may not
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* use this file except in compliance with the License. You may obtain a copy of
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* the License at
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*
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* http://www.apache.org/licenses/LICENSE-2.0
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*
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* Unless required by applicable law or agreed to in writing, software
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* distributed under the License is distributed on an "AS IS" BASIS, WITHOUT
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* WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the
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* License for the specific language governing permissions and limitations under
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* the License.
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*/
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package com.android.inputmethod.latin.makedict;
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import com.android.inputmethod.latin.Constants;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.Collections;
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import java.util.HashMap;
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import java.util.Iterator;
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import java.util.LinkedList;
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/**
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* A dictionary that can fusion heads and tails of words for more compression.
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*/
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public class FusionDictionary implements Iterable<Word> {
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private static final boolean DBG = MakedictLog.DBG;
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/**
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* A node of the dictionary, containing several CharGroups.
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*
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* A node is but an ordered array of CharGroups, which essentially contain all the
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* real information.
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* This class also contains fields to cache size and address, to help with binary
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* generation.
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*/
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public static class Node {
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ArrayList<CharGroup> mData;
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// To help with binary generation
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int mCachedSize = Integer.MIN_VALUE;
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int mCachedAddress = Integer.MIN_VALUE;
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int mCachedParentAddress = 0;
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public Node() {
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mData = new ArrayList<CharGroup>();
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}
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public Node(ArrayList<CharGroup> data) {
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mData = data;
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}
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}
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/**
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* A string with a frequency.
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*
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* This represents an "attribute", that is either a bigram or a shortcut.
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*/
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public static class WeightedString {
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public final String mWord;
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public int mFrequency;
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public WeightedString(String word, int frequency) {
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mWord = word;
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mFrequency = frequency;
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}
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@Override
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public int hashCode() {
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return Arrays.hashCode(new Object[] { mWord, mFrequency });
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}
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@Override
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public boolean equals(Object o) {
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if (o == this) return true;
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if (!(o instanceof WeightedString)) return false;
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WeightedString w = (WeightedString)o;
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return mWord.equals(w.mWord) && mFrequency == w.mFrequency;
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}
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}
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/**
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* A group of characters, with a frequency, shortcut targets, bigrams, and children.
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*
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* This is the central class of the in-memory representation. A CharGroup is what can
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* be seen as a traditional "trie node", except it can hold several characters at the
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* same time. A CharGroup essentially represents one or several characters in the middle
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* of the trie trie; as such, it can be a terminal, and it can have children.
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* In this in-memory representation, whether the CharGroup is a terminal or not is represented
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* in the frequency, where NOT_A_TERMINAL (= -1) means this is not a terminal and any other
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* value is the frequency of this terminal. A terminal may have non-null shortcuts and/or
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* bigrams, but a non-terminal may not. Moreover, children, if present, are null.
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*/
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public static class CharGroup {
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public static final int NOT_A_TERMINAL = -1;
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final int mChars[];
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ArrayList<WeightedString> mShortcutTargets;
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ArrayList<WeightedString> mBigrams;
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int mFrequency; // NOT_A_TERMINAL == mFrequency indicates this is not a terminal.
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Node mChildren;
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boolean mIsNotAWord; // Only a shortcut
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boolean mIsBlacklistEntry;
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// The two following members to help with binary generation
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int mCachedSize;
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int mCachedAddress;
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public CharGroup(final int[] chars, final ArrayList<WeightedString> shortcutTargets,
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final ArrayList<WeightedString> bigrams, final int frequency,
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final boolean isNotAWord, final boolean isBlacklistEntry) {
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mChars = chars;
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mFrequency = frequency;
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mShortcutTargets = shortcutTargets;
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mBigrams = bigrams;
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mChildren = null;
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mIsNotAWord = isNotAWord;
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mIsBlacklistEntry = isBlacklistEntry;
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}
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public CharGroup(final int[] chars, final ArrayList<WeightedString> shortcutTargets,
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final ArrayList<WeightedString> bigrams, final int frequency,
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final boolean isNotAWord, final boolean isBlacklistEntry, final Node children) {
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mChars = chars;
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mFrequency = frequency;
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mShortcutTargets = shortcutTargets;
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mBigrams = bigrams;
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mChildren = children;
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mIsNotAWord = isNotAWord;
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mIsBlacklistEntry = isBlacklistEntry;
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}
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public void addChild(CharGroup n) {
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if (null == mChildren) {
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mChildren = new Node();
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}
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mChildren.mData.add(n);
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}
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public boolean isTerminal() {
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return NOT_A_TERMINAL != mFrequency;
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}
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public boolean hasSeveralChars() {
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assert(mChars.length > 0);
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return 1 < mChars.length;
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}
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/**
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* Adds a word to the bigram list. Updates the frequency if the word already
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* exists.
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*/
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public void addBigram(final String word, final int frequency) {
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if (mBigrams == null) {
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mBigrams = new ArrayList<WeightedString>();
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}
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WeightedString bigram = getBigram(word);
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if (bigram != null) {
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bigram.mFrequency = frequency;
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} else {
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bigram = new WeightedString(word, frequency);
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mBigrams.add(bigram);
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}
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}
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/**
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* Gets the shortcut target for the given word. Returns null if the word is not in the
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* shortcut list.
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*/
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public WeightedString getShortcut(final String word) {
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// TODO: Don't do a linear search
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if (mShortcutTargets != null) {
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final int size = mShortcutTargets.size();
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for (int i = 0; i < size; ++i) {
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WeightedString shortcut = mShortcutTargets.get(i);
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if (shortcut.mWord.equals(word)) {
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return shortcut;
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}
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}
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}
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return null;
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}
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/**
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* Gets the bigram for the given word.
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* Returns null if the word is not in the bigrams list.
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*/
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public WeightedString getBigram(final String word) {
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// TODO: Don't do a linear search
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if (mBigrams != null) {
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final int size = mBigrams.size();
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for (int i = 0; i < size; ++i) {
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WeightedString bigram = mBigrams.get(i);
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if (bigram.mWord.equals(word)) {
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return bigram;
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}
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}
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}
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return null;
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}
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/**
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* Updates the CharGroup with the given properties. Adds the shortcut and bigram lists to
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* the existing ones if any. Note: unigram, bigram, and shortcut frequencies are only
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* updated if they are higher than the existing ones.
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*/
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public void update(final int frequency, final ArrayList<WeightedString> shortcutTargets,
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final ArrayList<WeightedString> bigrams,
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final boolean isNotAWord, final boolean isBlacklistEntry) {
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if (frequency > mFrequency) {
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mFrequency = frequency;
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}
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if (shortcutTargets != null) {
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if (mShortcutTargets == null) {
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mShortcutTargets = shortcutTargets;
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} else {
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final int size = shortcutTargets.size();
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for (int i = 0; i < size; ++i) {
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final WeightedString shortcut = shortcutTargets.get(i);
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final WeightedString existingShortcut = getShortcut(shortcut.mWord);
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if (existingShortcut == null) {
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mShortcutTargets.add(shortcut);
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} else if (existingShortcut.mFrequency < shortcut.mFrequency) {
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existingShortcut.mFrequency = shortcut.mFrequency;
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}
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}
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}
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}
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if (bigrams != null) {
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if (mBigrams == null) {
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mBigrams = bigrams;
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} else {
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final int size = bigrams.size();
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for (int i = 0; i < size; ++i) {
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final WeightedString bigram = bigrams.get(i);
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final WeightedString existingBigram = getBigram(bigram.mWord);
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if (existingBigram == null) {
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mBigrams.add(bigram);
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} else if (existingBigram.mFrequency < bigram.mFrequency) {
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existingBigram.mFrequency = bigram.mFrequency;
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}
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}
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}
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}
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mIsNotAWord = isNotAWord;
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mIsBlacklistEntry = isBlacklistEntry;
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}
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}
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/**
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* Options global to the dictionary.
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*
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* There are no options at the moment, so this class is empty.
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*/
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public static class DictionaryOptions {
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public final boolean mGermanUmlautProcessing;
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public final boolean mFrenchLigatureProcessing;
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public final HashMap<String, String> mAttributes;
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public DictionaryOptions(final HashMap<String, String> attributes,
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final boolean germanUmlautProcessing, final boolean frenchLigatureProcessing) {
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mAttributes = attributes;
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mGermanUmlautProcessing = germanUmlautProcessing;
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mFrenchLigatureProcessing = frenchLigatureProcessing;
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}
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}
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public final DictionaryOptions mOptions;
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public final Node mRoot;
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public FusionDictionary(final Node root, final DictionaryOptions options) {
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mRoot = root;
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mOptions = options;
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}
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public void addOptionAttribute(final String key, final String value) {
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mOptions.mAttributes.put(key, value);
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}
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/**
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* Helper method to convert a String to an int array.
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*/
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static private int[] getCodePoints(final String word) {
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// TODO: this is a copy-paste of the contents of StringUtils.toCodePointArray,
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// which is not visible from the makedict package. Factor this code.
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final char[] characters = word.toCharArray();
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final int length = characters.length;
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final int[] codePoints = new int[Character.codePointCount(characters, 0, length)];
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int codePoint = Character.codePointAt(characters, 0);
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int dsti = 0;
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for (int srci = Character.charCount(codePoint);
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srci < length; srci += Character.charCount(codePoint), ++dsti) {
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codePoints[dsti] = codePoint;
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codePoint = Character.codePointAt(characters, srci);
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}
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codePoints[dsti] = codePoint;
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return codePoints;
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}
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/**
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* Helper method to add a word as a string.
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*
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* This method adds a word to the dictionary with the given frequency. Optional
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* lists of bigrams and shortcuts can be passed here. For each word inside,
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* they will be added to the dictionary as necessary.
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*
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* @param word the word to add.
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* @param frequency the frequency of the word, in the range [0..255].
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* @param shortcutTargets a list of shortcut targets for this word, or null.
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* @param isNotAWord true if this should not be considered a word (e.g. shortcut only)
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*/
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public void add(final String word, final int frequency,
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final ArrayList<WeightedString> shortcutTargets, final boolean isNotAWord) {
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add(getCodePoints(word), frequency, shortcutTargets, isNotAWord,
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false /* isBlacklistEntry */);
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}
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/**
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* Helper method to add a blacklist entry as a string.
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*
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* @param word the word to add as a blacklist entry.
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* @param shortcutTargets a list of shortcut targets for this word, or null.
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* @param isNotAWord true if this is not a word for spellcheking purposes (shortcut only or so)
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*/
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public void addBlacklistEntry(final String word,
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final ArrayList<WeightedString> shortcutTargets, final boolean isNotAWord) {
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add(getCodePoints(word), 0, shortcutTargets, isNotAWord, true /* isBlacklistEntry */);
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}
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/**
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* Sanity check for a node.
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*
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* This method checks that all CharGroups in a node are ordered as expected.
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* If they are, nothing happens. If they aren't, an exception is thrown.
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*/
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private void checkStack(Node node) {
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ArrayList<CharGroup> stack = node.mData;
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int lastValue = -1;
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for (int i = 0; i < stack.size(); ++i) {
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int currentValue = stack.get(i).mChars[0];
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if (currentValue <= lastValue)
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throw new RuntimeException("Invalid stack");
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else
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lastValue = currentValue;
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}
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}
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/**
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* Helper method to add a new bigram to the dictionary.
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*
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* @param word1 the previous word of the context
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* @param word2 the next word of the context
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* @param frequency the bigram frequency
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*/
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public void setBigram(final String word1, final String word2, final int frequency) {
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CharGroup charGroup = findWordInTree(mRoot, word1);
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if (charGroup != null) {
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final CharGroup charGroup2 = findWordInTree(mRoot, word2);
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if (charGroup2 == null) {
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add(getCodePoints(word2), 0, null, false /* isNotAWord */,
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false /* isBlacklistEntry */);
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}
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charGroup.addBigram(word2, frequency);
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} else {
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throw new RuntimeException("First word of bigram not found");
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}
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}
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/**
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* Add a word to this dictionary.
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*
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* The shortcuts, if any, have to be in the dictionary already. If they aren't,
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* an exception is thrown.
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*
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* @param word the word, as an int array.
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* @param frequency the frequency of the word, in the range [0..255].
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* @param shortcutTargets an optional list of shortcut targets for this word (null if none).
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* @param isNotAWord true if this is not a word for spellcheking purposes (shortcut only or so)
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* @param isBlacklistEntry true if this is a blacklisted word, false otherwise
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*/
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private void add(final int[] word, final int frequency,
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final ArrayList<WeightedString> shortcutTargets,
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final boolean isNotAWord, final boolean isBlacklistEntry) {
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assert(frequency >= 0 && frequency <= 255);
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if (word.length >= Constants.Dictionary.MAX_WORD_LENGTH) {
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MakedictLog.w("Ignoring a word that is too long: word.length = " + word.length);
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return;
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}
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Node currentNode = mRoot;
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int charIndex = 0;
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CharGroup currentGroup = null;
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int differentCharIndex = 0; // Set by the loop to the index of the char that differs
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int nodeIndex = findIndexOfChar(mRoot, word[charIndex]);
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while (CHARACTER_NOT_FOUND != nodeIndex) {
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currentGroup = currentNode.mData.get(nodeIndex);
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differentCharIndex = compareArrays(currentGroup.mChars, word, charIndex);
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if (ARRAYS_ARE_EQUAL != differentCharIndex
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&& differentCharIndex < currentGroup.mChars.length) break;
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if (null == currentGroup.mChildren) break;
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charIndex += currentGroup.mChars.length;
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if (charIndex >= word.length) break;
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currentNode = currentGroup.mChildren;
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nodeIndex = findIndexOfChar(currentNode, word[charIndex]);
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}
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if (-1 == nodeIndex) {
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// No node at this point to accept the word. Create one.
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final int insertionIndex = findInsertionIndex(currentNode, word[charIndex]);
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final CharGroup newGroup = new CharGroup(
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Arrays.copyOfRange(word, charIndex, word.length),
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shortcutTargets, null /* bigrams */, frequency, isNotAWord, isBlacklistEntry);
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currentNode.mData.add(insertionIndex, newGroup);
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if (DBG) checkStack(currentNode);
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} else {
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// There is a word with a common prefix.
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if (differentCharIndex == currentGroup.mChars.length) {
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if (charIndex + differentCharIndex >= word.length) {
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// The new word is a prefix of an existing word, but the node on which it
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// should end already exists as is. Since the old CharNode was not a terminal,
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// make it one by filling in its frequency and other attributes
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currentGroup.update(frequency, shortcutTargets, null, isNotAWord,
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isBlacklistEntry);
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} else {
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// The new word matches the full old word and extends past it.
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// We only have to create a new node and add it to the end of this.
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final CharGroup newNode = new CharGroup(
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Arrays.copyOfRange(word, charIndex + differentCharIndex, word.length),
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shortcutTargets, null /* bigrams */, frequency, isNotAWord,
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isBlacklistEntry);
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currentGroup.mChildren = new Node();
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currentGroup.mChildren.mData.add(newNode);
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}
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} else {
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if (0 == differentCharIndex) {
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// Exact same word. Update the frequency if higher. This will also add the
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// new shortcuts to the existing shortcut list if it already exists.
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currentGroup.update(frequency, shortcutTargets, null,
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currentGroup.mIsNotAWord && isNotAWord,
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currentGroup.mIsBlacklistEntry || isBlacklistEntry);
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} else {
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// Partial prefix match only. We have to replace the current node with a node
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// containing the current prefix and create two new ones for the tails.
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Node newChildren = new Node();
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final CharGroup newOldWord = new CharGroup(
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Arrays.copyOfRange(currentGroup.mChars, differentCharIndex,
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currentGroup.mChars.length), currentGroup.mShortcutTargets,
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currentGroup.mBigrams, currentGroup.mFrequency,
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currentGroup.mIsNotAWord, currentGroup.mIsBlacklistEntry,
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currentGroup.mChildren);
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newChildren.mData.add(newOldWord);
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final CharGroup newParent;
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if (charIndex + differentCharIndex >= word.length) {
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newParent = new CharGroup(
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Arrays.copyOfRange(currentGroup.mChars, 0, differentCharIndex),
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shortcutTargets, null /* bigrams */, frequency,
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isNotAWord, isBlacklistEntry, newChildren);
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} else {
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newParent = new CharGroup(
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Arrays.copyOfRange(currentGroup.mChars, 0, differentCharIndex),
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null /* shortcutTargets */, null /* bigrams */, -1,
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false /* isNotAWord */, false /* isBlacklistEntry */, newChildren);
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final CharGroup newWord = new CharGroup(Arrays.copyOfRange(word,
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charIndex + differentCharIndex, word.length),
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shortcutTargets, null /* bigrams */, frequency,
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isNotAWord, isBlacklistEntry);
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final int addIndex = word[charIndex + differentCharIndex]
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> currentGroup.mChars[differentCharIndex] ? 1 : 0;
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newChildren.mData.add(addIndex, newWord);
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}
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currentNode.mData.set(nodeIndex, newParent);
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}
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if (DBG) checkStack(currentNode);
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}
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}
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}
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private static int ARRAYS_ARE_EQUAL = 0;
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/**
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* Custom comparison of two int arrays taken to contain character codes.
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*
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* This method compares the two arrays passed as an argument in a lexicographic way,
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* with an offset in the dst string.
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|
* This method does NOT test for the first character. It is taken to be equal.
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* I repeat: this method starts the comparison at 1 <> dstOffset + 1.
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* The index where the strings differ is returned. ARRAYS_ARE_EQUAL = 0 is returned if the
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* strings are equal. This works BECAUSE we don't look at the first character.
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*
|
|
* @param src the left-hand side string of the comparison.
|
|
* @param dst the right-hand side string of the comparison.
|
|
* @param dstOffset the offset in the right-hand side string.
|
|
* @return the index at which the strings differ, or ARRAYS_ARE_EQUAL = 0 if they don't.
|
|
*/
|
|
private static int compareArrays(final int[] src, final int[] dst, int dstOffset) {
|
|
// We do NOT test the first char, because we come from a method that already
|
|
// tested it.
|
|
for (int i = 1; i < src.length; ++i) {
|
|
if (dstOffset + i >= dst.length) return i;
|
|
if (src[i] != dst[dstOffset + i]) return i;
|
|
}
|
|
if (dst.length > src.length) return src.length;
|
|
return ARRAYS_ARE_EQUAL;
|
|
}
|
|
|
|
/**
|
|
* Helper class that compares and sorts two chargroups according to their
|
|
* first element only. I repeat: ONLY the first element is considered, the rest
|
|
* is ignored.
|
|
* This comparator imposes orderings that are inconsistent with equals.
|
|
*/
|
|
static private class CharGroupComparator implements java.util.Comparator<CharGroup> {
|
|
@Override
|
|
public int compare(CharGroup c1, CharGroup c2) {
|
|
if (c1.mChars[0] == c2.mChars[0]) return 0;
|
|
return c1.mChars[0] < c2.mChars[0] ? -1 : 1;
|
|
}
|
|
}
|
|
final static private CharGroupComparator CHARGROUP_COMPARATOR = new CharGroupComparator();
|
|
|
|
/**
|
|
* Finds the insertion index of a character within a node.
|
|
*/
|
|
private static int findInsertionIndex(final Node node, int character) {
|
|
final ArrayList<CharGroup> data = node.mData;
|
|
final CharGroup reference = new CharGroup(new int[] { character },
|
|
null /* shortcutTargets */, null /* bigrams */, 0, false /* isNotAWord */,
|
|
false /* isBlacklistEntry */);
|
|
int result = Collections.binarySearch(data, reference, CHARGROUP_COMPARATOR);
|
|
return result >= 0 ? result : -result - 1;
|
|
}
|
|
|
|
private static int CHARACTER_NOT_FOUND = -1;
|
|
|
|
/**
|
|
* Find the index of a char in a node, if it exists.
|
|
*
|
|
* @param node the node to search in.
|
|
* @param character the character to search for.
|
|
* @return the position of the character if it's there, or CHARACTER_NOT_FOUND = -1 else.
|
|
*/
|
|
private static int findIndexOfChar(final Node node, int character) {
|
|
final int insertionIndex = findInsertionIndex(node, character);
|
|
if (node.mData.size() <= insertionIndex) return CHARACTER_NOT_FOUND;
|
|
return character == node.mData.get(insertionIndex).mChars[0] ? insertionIndex
|
|
: CHARACTER_NOT_FOUND;
|
|
}
|
|
|
|
/**
|
|
* Helper method to find a word in a given branch.
|
|
*/
|
|
public static CharGroup findWordInTree(Node node, final String s) {
|
|
int index = 0;
|
|
final StringBuilder checker = DBG ? new StringBuilder() : null;
|
|
|
|
CharGroup currentGroup;
|
|
do {
|
|
int indexOfGroup = findIndexOfChar(node, s.codePointAt(index));
|
|
if (CHARACTER_NOT_FOUND == indexOfGroup) return null;
|
|
currentGroup = node.mData.get(indexOfGroup);
|
|
|
|
if (s.length() - index < currentGroup.mChars.length) return null;
|
|
int newIndex = index;
|
|
while (newIndex < s.length() && newIndex - index < currentGroup.mChars.length) {
|
|
if (currentGroup.mChars[newIndex - index] != s.codePointAt(newIndex)) return null;
|
|
newIndex++;
|
|
}
|
|
index = newIndex;
|
|
|
|
if (DBG) checker.append(new String(currentGroup.mChars, 0, currentGroup.mChars.length));
|
|
if (index < s.length()) {
|
|
node = currentGroup.mChildren;
|
|
}
|
|
} while (null != node && index < s.length());
|
|
|
|
if (index < s.length()) return null;
|
|
if (!currentGroup.isTerminal()) return null;
|
|
if (DBG && !s.equals(checker.toString())) return null;
|
|
return currentGroup;
|
|
}
|
|
|
|
/**
|
|
* Helper method to find out whether a word is in the dict or not.
|
|
*/
|
|
public boolean hasWord(final String s) {
|
|
if (null == s || "".equals(s)) {
|
|
throw new RuntimeException("Can't search for a null or empty string");
|
|
}
|
|
return null != findWordInTree(mRoot, s);
|
|
}
|
|
|
|
/**
|
|
* Recursively count the number of character groups in a given branch of the trie.
|
|
*
|
|
* @param node the parent node.
|
|
* @return the number of char groups in all the branch under this node.
|
|
*/
|
|
public static int countCharGroups(final Node node) {
|
|
final int nodeSize = node.mData.size();
|
|
int size = nodeSize;
|
|
for (int i = nodeSize - 1; i >= 0; --i) {
|
|
CharGroup group = node.mData.get(i);
|
|
if (null != group.mChildren)
|
|
size += countCharGroups(group.mChildren);
|
|
}
|
|
return size;
|
|
}
|
|
|
|
/**
|
|
* Recursively count the number of nodes in a given branch of the trie.
|
|
*
|
|
* @param node the node to count.
|
|
* @return the number of nodes in this branch.
|
|
*/
|
|
public static int countNodes(final Node node) {
|
|
int size = 1;
|
|
for (int i = node.mData.size() - 1; i >= 0; --i) {
|
|
CharGroup group = node.mData.get(i);
|
|
if (null != group.mChildren)
|
|
size += countNodes(group.mChildren);
|
|
}
|
|
return size;
|
|
}
|
|
|
|
// Recursively find out whether there are any bigrams.
|
|
// This can be pretty expensive especially if there aren't any (we return as soon
|
|
// as we find one, so it's much cheaper if there are bigrams)
|
|
private static boolean hasBigramsInternal(final Node node) {
|
|
if (null == node) return false;
|
|
for (int i = node.mData.size() - 1; i >= 0; --i) {
|
|
CharGroup group = node.mData.get(i);
|
|
if (null != group.mBigrams) return true;
|
|
if (hasBigramsInternal(group.mChildren)) return true;
|
|
}
|
|
return false;
|
|
}
|
|
|
|
/**
|
|
* Finds out whether there are any bigrams in this dictionary.
|
|
*
|
|
* @return true if there is any bigram, false otherwise.
|
|
*/
|
|
// TODO: this is expensive especially for large dictionaries without any bigram.
|
|
// The up side is, this is always accurate and correct and uses no memory. We should
|
|
// find a more efficient way of doing this, without compromising too much on memory
|
|
// and ease of use.
|
|
public boolean hasBigrams() {
|
|
return hasBigramsInternal(mRoot);
|
|
}
|
|
|
|
// Historically, the tails of the words were going to be merged to save space.
|
|
// However, that would prevent the code to search for a specific address in log(n)
|
|
// time so this was abandoned.
|
|
// The code is still of interest as it does add some compression to any dictionary
|
|
// that has no need for attributes. Implementations that does not read attributes should be
|
|
// able to read a dictionary with merged tails.
|
|
// Also, the following code does support frequencies, as in, it will only merges
|
|
// tails that share the same frequency. Though it would result in the above loss of
|
|
// performance while searching by address, it is still technically possible to merge
|
|
// tails that contain attributes, but this code does not take that into account - it does
|
|
// not compare attributes and will merge terminals with different attributes regardless.
|
|
public void mergeTails() {
|
|
MakedictLog.i("Do not merge tails");
|
|
return;
|
|
|
|
// MakedictLog.i("Merging nodes. Number of nodes : " + countNodes(root));
|
|
// MakedictLog.i("Number of groups : " + countCharGroups(root));
|
|
//
|
|
// final HashMap<String, ArrayList<Node>> repository =
|
|
// new HashMap<String, ArrayList<Node>>();
|
|
// mergeTailsInner(repository, root);
|
|
//
|
|
// MakedictLog.i("Number of different pseudohashes : " + repository.size());
|
|
// int size = 0;
|
|
// for (ArrayList<Node> a : repository.values()) {
|
|
// size += a.size();
|
|
// }
|
|
// MakedictLog.i("Number of nodes after merge : " + (1 + size));
|
|
// MakedictLog.i("Recursively seen nodes : " + countNodes(root));
|
|
}
|
|
|
|
// The following methods are used by the deactivated mergeTails()
|
|
// private static boolean isEqual(Node a, Node b) {
|
|
// if (null == a && null == b) return true;
|
|
// if (null == a || null == b) return false;
|
|
// if (a.data.size() != b.data.size()) return false;
|
|
// final int size = a.data.size();
|
|
// for (int i = size - 1; i >= 0; --i) {
|
|
// CharGroup aGroup = a.data.get(i);
|
|
// CharGroup bGroup = b.data.get(i);
|
|
// if (aGroup.frequency != bGroup.frequency) return false;
|
|
// if (aGroup.alternates == null && bGroup.alternates != null) return false;
|
|
// if (aGroup.alternates != null && !aGroup.equals(bGroup.alternates)) return false;
|
|
// if (!Arrays.equals(aGroup.chars, bGroup.chars)) return false;
|
|
// if (!isEqual(aGroup.children, bGroup.children)) return false;
|
|
// }
|
|
// return true;
|
|
// }
|
|
|
|
// static private HashMap<String, ArrayList<Node>> mergeTailsInner(
|
|
// final HashMap<String, ArrayList<Node>> map, final Node node) {
|
|
// final ArrayList<CharGroup> branches = node.data;
|
|
// final int nodeSize = branches.size();
|
|
// for (int i = 0; i < nodeSize; ++i) {
|
|
// CharGroup group = branches.get(i);
|
|
// if (null != group.children) {
|
|
// String pseudoHash = getPseudoHash(group.children);
|
|
// ArrayList<Node> similarList = map.get(pseudoHash);
|
|
// if (null == similarList) {
|
|
// similarList = new ArrayList<Node>();
|
|
// map.put(pseudoHash, similarList);
|
|
// }
|
|
// boolean merged = false;
|
|
// for (Node similar : similarList) {
|
|
// if (isEqual(group.children, similar)) {
|
|
// group.children = similar;
|
|
// merged = true;
|
|
// break;
|
|
// }
|
|
// }
|
|
// if (!merged) {
|
|
// similarList.add(group.children);
|
|
// }
|
|
// mergeTailsInner(map, group.children);
|
|
// }
|
|
// }
|
|
// return map;
|
|
// }
|
|
|
|
// private static String getPseudoHash(final Node node) {
|
|
// StringBuilder s = new StringBuilder();
|
|
// for (CharGroup g : node.data) {
|
|
// s.append(g.frequency);
|
|
// for (int ch : g.chars) {
|
|
// s.append(Character.toChars(ch));
|
|
// }
|
|
// }
|
|
// return s.toString();
|
|
// }
|
|
|
|
/**
|
|
* Iterator to walk through a dictionary.
|
|
*
|
|
* This is purely for convenience.
|
|
*/
|
|
public static class DictionaryIterator implements Iterator<Word> {
|
|
|
|
private static class Position {
|
|
public Iterator<CharGroup> pos;
|
|
public int length;
|
|
public Position(ArrayList<CharGroup> groups) {
|
|
pos = groups.iterator();
|
|
length = 0;
|
|
}
|
|
}
|
|
final StringBuilder mCurrentString;
|
|
final LinkedList<Position> mPositions;
|
|
|
|
public DictionaryIterator(ArrayList<CharGroup> root) {
|
|
mCurrentString = new StringBuilder();
|
|
mPositions = new LinkedList<Position>();
|
|
final Position rootPos = new Position(root);
|
|
mPositions.add(rootPos);
|
|
}
|
|
|
|
@Override
|
|
public boolean hasNext() {
|
|
for (Position p : mPositions) {
|
|
if (p.pos.hasNext()) {
|
|
return true;
|
|
}
|
|
}
|
|
return false;
|
|
}
|
|
|
|
@Override
|
|
public Word next() {
|
|
Position currentPos = mPositions.getLast();
|
|
mCurrentString.setLength(mCurrentString.length() - currentPos.length);
|
|
|
|
do {
|
|
if (currentPos.pos.hasNext()) {
|
|
final CharGroup currentGroup = currentPos.pos.next();
|
|
currentPos.length = currentGroup.mChars.length;
|
|
for (int i : currentGroup.mChars)
|
|
mCurrentString.append(Character.toChars(i));
|
|
if (null != currentGroup.mChildren) {
|
|
currentPos = new Position(currentGroup.mChildren.mData);
|
|
mPositions.addLast(currentPos);
|
|
}
|
|
if (currentGroup.mFrequency >= 0)
|
|
return new Word(mCurrentString.toString(), currentGroup.mFrequency,
|
|
currentGroup.mShortcutTargets, currentGroup.mBigrams,
|
|
currentGroup.mIsNotAWord, currentGroup.mIsBlacklistEntry);
|
|
} else {
|
|
mPositions.removeLast();
|
|
currentPos = mPositions.getLast();
|
|
mCurrentString.setLength(mCurrentString.length() - mPositions.getLast().length);
|
|
}
|
|
} while (true);
|
|
}
|
|
|
|
@Override
|
|
public void remove() {
|
|
throw new UnsupportedOperationException("Unsupported yet");
|
|
}
|
|
|
|
}
|
|
|
|
/**
|
|
* Method to return an iterator.
|
|
*
|
|
* This method enables Java's enhanced for loop. With this you can have a FusionDictionary x
|
|
* and say : for (Word w : x) {}
|
|
*/
|
|
@Override
|
|
public Iterator<Word> iterator() {
|
|
return new DictionaryIterator(mRoot.mData);
|
|
}
|
|
}
|