Methods Summary |
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public java.lang.String | getField()
return field;
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public int[] | getTermFrequencies()
return termFreqs;
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public java.lang.String[] | getTerms()
return terms;
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public int | indexOf(java.lang.String termText)
if(terms == null)
return -1;
int res = Arrays.binarySearch(terms, termText);
return res >= 0 ? res : -1;
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public int[] | indexesOf(java.lang.String[] termNumbers, int start, int len)
// TODO: there must be a more efficient way of doing this.
// At least, we could advance the lower bound of the terms array
// as we find valid indexes. Also, it might be possible to leverage
// this even more by starting in the middle of the termNumbers array
// and thus dividing the terms array maybe in half with each found index.
int res[] = new int[len];
for (int i=0; i < len; i++) {
res[i] = indexOf(termNumbers[start+ i]);
}
return res;
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public int | size()
return terms == null ? 0 : terms.length;
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public java.lang.String | toString()
StringBuffer sb = new StringBuffer();
sb.append('{");
sb.append(field).append(": ");
if(terms != null){
for (int i=0; i<terms.length; i++) {
if (i>0) sb.append(", ");
sb.append(terms[i]).append('/").append(termFreqs[i]);
}
}
sb.append('}");
return sb.toString();
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