本文整理汇总了C#中Lucene.Net.Search.Similarity.Coord方法的典型用法代码示例。如果您正苦于以下问题：C# Similarity.Coord方法的具体用法？C# Similarity.Coord怎么用？C# Similarity.Coord使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类Lucene.Net.Search.Similarity的用法示例。
在下文中一共展示了Similarity.Coord方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C#代码示例。
示例1: ConjunctionScorer
public ConjunctionScorer(Similarity similarity, Scorer[] scorers)
: base(similarity)
{
this.scorers = scorers;
coord = similarity.Coord(scorers.Length, scorers.Length);
for (int i = 0; i < scorers.Length; i++)
{
if (scorers[i].NextDoc() == NO_MORE_DOCS)
{
// If even one of the sub-scorers does not have any documents, this
// scorer should not attempt to do any more work.
lastDoc = NO_MORE_DOCS;
return ;
}
}
// Sort the array the first time...
// We don't need to sort the array in any future calls because we know
// it will already start off sorted (all scorers on same doc).
// note that this comparator is not consistent with equals!
System.Array.Sort(scorers, new AnonymousClassComparator(this));
// NOTE: doNext() must be called before the re-sorting of the array later on.
// The reason is this: assume there are 5 scorers, whose first docs are 1,
// 2, 3, 5, 5 respectively. Sorting (above) leaves the array as is. Calling
// doNext() here advances all the first scorers to 5 (or a larger doc ID
// they all agree on).
// However, if we re-sort before doNext() is called, the order will be 5, 3,
// 2, 1, 5 and then doNext() will stop immediately, since the first scorer's
// docs equals the last one. So the invariant that after calling doNext()
// all scorers are on the same doc ID is broken.
if (DoNext() == NO_MORE_DOCS)
{
// The scorers did not agree on any document.
lastDoc = NO_MORE_DOCS;
return ;
}
// If first-time skip distance is any predictor of
// scorer sparseness, then we should always try to skip first on
// those scorers.
// Keep last scorer in it's last place (it will be the first
// to be skipped on), but reverse all of the others so that
// they will be skipped on in order of original high skip.
int end = scorers.Length - 1;
int max = end >> 1;
for (int i = 0; i < max; i++)
{
Scorer tmp = scorers[i];
int idx = end - i - 1;
scorers[i] = scorers[idx];
scorers[idx] = tmp;
}
}