本文整理了Java中opennlp.tools.util.Span.intersects()方法的一些代码示例，展示了Span.intersects()的具体用法。这些代码示例主要来源于Github/Stackoverflow/Maven等平台，是从一些精选项目中提取出来的代码，具有较强的参考意义，能在一定程度帮忙到你。Span.intersects()方法的具体详情如下：
包路径：opennlp.tools.util.Span
类名称：Span
方法名：intersects

Span.intersects介绍

[英]Returns true if the specified span intersects with this span.
[中]如果指定的跨度与此跨度相交，则返回true。

代码示例

代码示例来源：origin: apache/opennlp

if (lastSpan.intersects(span)) {
 it.remove();
 span = lastSpan;

代码示例来源：origin: apache/opennlp

/**
 * Test for {@link Span#intersects(Span)}.
 */
@Test
public void testIntersects() {
 Span a = new Span(10, 50);
 Span b = new Span(40, 100);
 Assert.assertTrue(a.intersects(b));
 Assert.assertTrue(b.intersects(a));
 Span c = new Span(10, 20);
 Span d = new Span(40, 50);
 Assert.assertFalse(c.intersects(d));
 Assert.assertFalse(d.intersects(c));
 Assert.assertTrue(b.intersects(d));
}

代码示例来源：origin: org.apache.opennlp/opennlp-tools

if (lastSpan.intersects(span)) {
 it.remove();
 span = lastSpan;

代码示例来源：origin: ai.idylnlp/idylnlp-opennlp-tools-1.8.3

if (lastSpan.intersects(span)) {
 it.remove();
 span = lastSpan;

代码示例来源：origin: cogroo/cogroo4

public boolean canAddOmission(int start, int end){
  Span s = new Span(start, end);
  for (Omission o : omissions) {
    if(o.getSpan().intersects(s)) {
      return false;
    }
  }
  return true;
}

代码示例来源：origin: org.cogroo.lang.pt_br/cogroo-addon-pt_br

public boolean canAddOmission(int start, int end){
  Span s = new Span(start, end);
  for (Omission o : omissions) {
    if(o.getSpan().intersects(s)) {
      return false;
    }
  }
  return true;
}

代码示例来源：origin: spinscale/elasticsearch-opennlp-plugin

ai--;
} else if (prev.getSpan().intersects(curr.getSpan())) {
  if (prev.getProb() > curr.getProb()) {
    deleteCurr = true;

代码示例来源：origin: org.cogroo/cogroo-gc

public int compare(Mistake m1, Mistake m2) {
 // First we check if they overlap. If they don't we simply use the start position
 Span a = new Span(m1.getStart(), m1.getEnd());
 Span b = new Span(m2.getStart(), m2.getEnd());
 if(!a.intersects(b)) {
  return a.compareTo(b);
 }
 
 // they intersect, so we should sort using the priority. The higher the
 // number, the higher th priority
 if(m1.getRulePriority() > m2.getRulePriority()) {
  return -1;
 } else if(m1.getRulePriority() < m2.getRulePriority()) {
  return 1;
 } else {
  // equal priority! so we try to use the rule id
  if(m2.getRuleIdentifier().startsWith("xml:")) {
   Integer id1 = new Integer(m1.getRuleIdentifier().substring(4));
   Integer id2 = new Integer(m2.getRuleIdentifier().substring(4));
   
   return id1.compareTo(id2);
  }
  return m1.getRuleIdentifier().compareTo(m2.getRuleIdentifier());
 }
}

代码示例来源：origin: cogroo/cogroo4

public int compare(Mistake m1, Mistake m2) {
 // First we check if they overlap. If they don't we simply use the start position
 Span a = new Span(m1.getStart(), m1.getEnd());
 Span b = new Span(m2.getStart(), m2.getEnd());
 if(!a.intersects(b)) {
  return a.compareTo(b);
 }
 
 // they intersect, so we should sort using the priority. The higher the
 // number, the higher th priority
 if(m1.getRulePriority() > m2.getRulePriority()) {
  return -1;
 } else if(m1.getRulePriority() < m2.getRulePriority()) {
  return 1;
 } else {
  // equal priority! so we try to use the rule id
  if(m2.getRuleIdentifier().startsWith("xml:")) {
   Integer id1 = new Integer(m1.getRuleIdentifier().substring(4));
   Integer id2 = new Integer(m2.getRuleIdentifier().substring(4));
   
   return id1.compareTo(id2);
  }
  return m1.getRuleIdentifier().compareTo(m2.getRuleIdentifier());
 }
}