/*
* Copyright 2004 The Apache Software Foundation
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*/
using System;
namespace Lucene.Net.Index{
public class SegmentTermVector : TermFreqVector
{
private System.String field;
private System.String[] terms;
private int[] termFreqs;
internal SegmentTermVector(System.String field, System.String[] terms, int[] termFreqs)
{
this.field = field;
this.terms = terms;
this.termFreqs = termFreqs;
}
/// <summary> </summary>
/// <returns> The number of the field this vector is associated with
/// </returns>
public virtual System.String GetField()
{
return field;
}
public override System.String ToString()
{
System.Text.StringBuilder sb = new System.Text.StringBuilder();
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();
}
public virtual int Size()
{
return terms == null?0:terms.Length;
}
public virtual System.String[] GetTerms()
{
return terms;
}
public virtual int[] GetTermFrequencies()
{
return termFreqs;
}
public virtual int IndexOf(System.String termText)
{
if (terms == null)
return - 1;
int res = System.Array.BinarySearch(terms, termText);
return res >= 0?res:- 1;
}
public virtual int[] IndexesOf(System.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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