Lines Matching refs:frequency
57 * A string with a frequency.
64 public WeightedString(String word, int frequency) {
66 mFrequency = frequency;
71 * A group of characters, with a frequency, shortcuts, bigrams, and children.
78 * in the frequency, where NOT_A_TERMINAL (= -1) means this is not a terminal and any other
79 * value is the frequency of this terminal. A terminal may have non-null shortcuts and/or
93 final ArrayList<WeightedString> bigrams, final int frequency) {
95 mFrequency = frequency;
101 final ArrayList<WeightedString> bigrams, final int frequency, final Node children) {
103 mFrequency = frequency;
162 * This method adds a word to the dictionary with the given frequency. Optional
167 * @param frequency the frequency of the word, in the range [0..255].
170 public void add(String word, int frequency, ArrayList<WeightedString> bigrams) {
179 add(getCodePoints(word), frequency, bigrams);
207 * @param frequency the frequency of the word, in the range [0..255].
210 private void add(int[] word, int frequency, ArrayList<WeightedString> bigrams) {
211 assert(frequency >= 0 && frequency <= 255);
234 Arrays.copyOfRange(word, charIndex, word.length), bigrams, frequency);
248 bigrams, frequency, currentGroup.mChildren);
257 bigrams, frequency);
263 // Exact same word. Check the frequency is 0 or -1, and update.
264 if (0 != frequency) {
266 throw new RuntimeException("This word already exists with frequency "
271 currentGroup.mBigrams, frequency, currentGroup.mChildren);
288 bigrams, frequency, newChildren);
295 word.length), bigrams, frequency);
441 // tails that share the same frequency. Though it would result in the above loss of
474 // if (aGroup.frequency != bGroup.frequency) return false;
516 // s.append(g.frequency);
