本文整理了Java中org.locationtech.jts.algorithm.Orientation.index()方法的一些代码示例，展示了Orientation.index()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。Orientation.index()方法的具体详情如下：
包路径：org.locationtech.jts.algorithm.Orientation
类名称：Orientation
方法名：index

Orientation.index介绍

[英]Returns the orientation index of the direction of the point q relative to a directed infinite line specified by p1-p2. The index indicates whether the point lies to the #LEFT or #RIGHTof the line, or lies on it #COLLINEAR. The index also indicates the orientation of the triangle formed by the three points ( #COUNTERCLOCKWISE, #CLOCKWISE, or #STRAIGHT )
[中]返回点q相对于p1-p2指定的有向无限线的方向索引。该索引指示该点是位于直线的#左侧还是#右侧，还是位于其上#共线。该指数还表示由三个点（逆时针、顺时针或直线）构成的三角形的方向

代码示例

代码示例来源：origin: locationtech/jts

private boolean isInsideCorner(Coordinate queryPt, Coordinate base, Coordinate p1, Coordinate p2)
{
  return Orientation.index(base, p1, queryPt) == Orientation.CLOCKWISE
      && Orientation.index(base, p2, queryPt) == Orientation.COUNTERCLOCKWISE;
}

代码示例来源：origin: locationtech/jts

private boolean isConcave(Coordinate p0, Coordinate p1, Coordinate p2)
 {
  int orientation = Orientation.index(p0, p1, p2);
  boolean isConcave = (orientation == angleOrientation);
  return isConcave;
 }
}

代码示例来源：origin: locationtech/jts

/**
 * Determines the orientation index of a {@link Coordinate} relative to this segment.
 * The orientation index is as defined in {@link Orientation#computeOrientation}.
 *
 * @param p the coordinate to compare
 *
 * @return 1 (LEFT) if <code>p</code> is to the left of this segment
 * @return -1 (RIGHT) if <code>p</code> is to the right of this segment
 * @return 0 (COLLINEAR) if <code>p</code> is collinear with this segment
 * 
 * @see Orientation#computeOrientation(Coordinate, Coordinate, Coordinate)
 */
public int orientationIndex(Coordinate p)
{
 return Orientation.index(p0, p1, p);
}

代码示例来源：origin: locationtech/jts

/**
 * Returns an array of pts such that p0 - p[0] - [p1] is CW.
 * 
 * @param p0
 * @param p1
 * @param p2
 * @return
 */
private static Coordinate[] orientCorner(Coordinate p0, Coordinate p1, Coordinate p2)
{
  Coordinate[] orient;
  // TODO: not sure if determining orientation is necessary?
 if (Orientation.CLOCKWISE == Orientation.index(p0, p1, p2)) {
   orient = new Coordinate[] {p1, p2 };
 }
 else {
   orient = new Coordinate[] { p2, p1 };
 }
 
 return orient;
}

代码示例来源：origin: locationtech/jts

/**
 * 
 * @param pt
 * @param cornerBase the two vertices defining the 
 * @param corner the two vertices defining the arms of the corner, oriented CW
 * @return the quadrant the pt lies in
 */
private static int quadrant(Coordinate pt, Coordinate cornerBase, Coordinate[] corner)
{
  if (Orientation.index(cornerBase, corner[0], pt) == Orientation.CLOCKWISE) {
    if (Orientation.index(cornerBase, corner[1], pt) == Orientation.COUNTERCLOCKWISE) {
      return 0;
    }
    else
      return 3;
  }
  else {
    if (Orientation.index(cornerBase, corner[1], pt) == Orientation.COUNTERCLOCKWISE) {
      return 1;
    }
    else
      return 2; 		
  }
}

代码示例来源：origin: locationtech/jts

/**
 * Implements the total order relation:
 * <p>
 *    a has a greater angle with the positive x-axis than b
 * <p>
 * Using the obvious algorithm of simply computing the angle is not robust,
 * since the angle calculation is obviously susceptible to roundoff.
 * A robust algorithm is:
 * - first compare the quadrant.  If the quadrants
 * are different, it it trivial to determine which vector is "greater".
 * - if the vectors lie in the same quadrant, the computeOrientation function
 * can be used to decide the relative orientation of the vectors.
 */
public int compareDirection(EdgeEnd e)
{
 if (dx == e.dx && dy == e.dy)
  return 0;
 // if the rays are in different quadrants, determining the ordering is trivial
 if (quadrant > e.quadrant) return 1;
 if (quadrant < e.quadrant) return -1;
 // vectors are in the same quadrant - check relative orientation of direction vectors
 // this is > e if it is CCW of e
 return Orientation.index(e.p0, e.p1, p1);
}

代码示例来源：origin: locationtech/jts

/**
 * Returns 1 if this DirectedEdge has a greater angle with the
 * positive x-axis than b", 0 if the DirectedEdges are collinear, and -1 otherwise.
 * <p>
 * Using the obvious algorithm of simply computing the angle is not robust,
 * since the angle calculation is susceptible to roundoff. A robust algorithm
 * is:
 * <ul>
 * <li>first compare the quadrants. If the quadrants are different, it it
 * trivial to determine which vector is "greater".
 * <li>if the vectors lie in the same quadrant, the robust
 * {@link Orientation#computeOrientation(Coordinate, Coordinate, Coordinate)}
 * function can be used to decide the relative orientation of the vectors.
 * </ul>
 */
public int compareDirection(DirectedEdge e)
{
 // if the rays are in different quadrants, determining the ordering is trivial
 if (quadrant > e.quadrant) return 1;
 if (quadrant < e.quadrant) return -1;
 // vectors are in the same quadrant - check relative orientation of direction vectors
 // this is > e if it is CCW of e
 return Orientation.index(e.p0, e.p1, p1);
}

代码示例来源：origin: locationtech/jts

/**
 *@return    whether the three coordinates are collinear and c2 lies between
 *      c1 and c3 inclusive
 */
private boolean isBetween(Coordinate c1, Coordinate c2, Coordinate c3) {
 if (Orientation.index(c1, c2, c3) != 0) {
  return false;
 }
 if (c1.x != c3.x) {
  if (c1.x <= c2.x && c2.x <= c3.x) {
   return true;
  }
  if (c3.x <= c2.x && c2.x <= c1.x) {
   return true;
  }
 }
 if (c1.y != c3.y) {
  if (c1.y <= c2.y && c2.y <= c3.y) {
   return true;
  }
  if (c3.y <= c2.y && c2.y <= c1.y) {
   return true;
  }
 }
 return false;
}

代码示例来源：origin: locationtech/jts

private static double distanceToSeg(Coordinate p, Coordinate p0, Coordinate p1)
  {
    distSeg.p0 = p0;
    distSeg.p1 = p1;
    double segDist = distSeg.distance(p);
    
    // robust calculation of zero distance
    if (Orientation.index(p0, p1, p) == Orientation.COLLINEAR)
      segDist = 0.0;
    
    return segDist;
  }
}

代码示例来源：origin: locationtech/jts

private boolean isShallowConcavity(Coordinate p0, Coordinate p1, Coordinate p2, double distanceTol)
{
 int orientation = Orientation.index(p0, p1, p2);
 boolean isAngleToSimplify = (orientation == angleOrientation);
 if (! isAngleToSimplify)
  return false;
 
 double dist = Distance.pointToSegment(p1, p0, p2);
 return dist < distanceTol;
}

代码示例来源：origin: locationtech/jts

private static boolean isPointNearButNotOnSeg(Coordinate p, Coordinate p0, Coordinate p1, double distTol)
{
  // don't rely on segment distance algorithm to correctly compute zero distance
  // on segment
  if (Orientation.index(p0, p1, p) == Orientation.COLLINEAR)
    return false;
  // compute actual distance
  distSeg.p0 = p0;
  distSeg.p1 = p1;
  double segDist = distSeg.distance(p);
  if (segDist > distTol)
    return false;
  return true;
}

代码示例来源：origin: locationtech/jts

/**
 * Tests whether the orientations around a triangle of points
 * are all equal (as is expected if the orientation predicate is correct)
 * 
 * @param pts an array of three points
 * @return true if all the orientations around the triangle are equal
 */
public static boolean isAllOrientationsEqual(Coordinate[] pts)
{
 int[] orient = new int[3];
 orient[0] = Orientation.index(pts[0], pts[1], pts[2]);
 orient[1] = Orientation.index(pts[1], pts[2], pts[0]);
 orient[2] = Orientation.index(pts[2], pts[0], pts[1]);
 return orient[0] == orient[1] && orient[0] == orient[2];
}

代码示例来源：origin: locationtech/jts

/**
 * Uses the Graham Scan algorithm to compute the convex hull vertices.
 * 
 * @param c a list of points, with at least 3 entries
 * @return a Stack containing the ordered points of the convex hull ring
 */
private Stack grahamScan(Coordinate[] c) {
 Coordinate p;
 Stack ps = new Stack();
 ps.push(c[0]);
 ps.push(c[1]);
 ps.push(c[2]);
 for (int i = 3; i < c.length; i++) {
  p = (Coordinate) ps.pop();
  // check for empty stack to guard against robustness problems
  while (
    ! ps.empty() && 
    Orientation.index((Coordinate) ps.peek(), p, c[i]) > 0) {
   p = (Coordinate) ps.pop();
  }
  ps.push(p);
  ps.push(c[i]);
 }
 ps.push(c[0]);
 return ps;
}

代码示例来源：origin: locationtech/jts

public void computeIntersection(Coordinate p, Coordinate p1, Coordinate p2) {
 isProper = false;
 // do between check first, since it is faster than the orientation test
 if (Envelope.intersects(p1, p2, p)) {
  if ((Orientation.index(p1, p2, p) == 0)
    && (Orientation.index(p2, p1, p) == 0)) {
   isProper = true;
   if (p.equals(p1) || p.equals(p2)) {
    isProper = false;
   }
   result = POINT_INTERSECTION;
   return;
  }
 }
 result = NO_INTERSECTION;
}

代码示例来源：origin: locationtech/jts

public static int orientationIndex(Geometry segment, Geometry ptGeom) {
 if (segment.getNumPoints() != 2 || ptGeom.getNumPoints() != 1) {
  throw new IllegalArgumentException("A must have two points and B must have one");
 }
 Coordinate[] segPt = segment.getCoordinates();
 
 Coordinate p = ptGeom.getCoordinate();
 int index = Orientation.index(segPt[0], segPt[1], p);
 return index;
}

代码示例来源：origin: locationtech/jts

/**
 * The coordinate pairs match if they define line segments lying in the same direction.
 * E.g. the segments are parallel and in the same quadrant
 * (as opposed to parallel and opposite!).
 */
private boolean matchInSameDirection(Coordinate p0, Coordinate p1, Coordinate ep0, Coordinate ep1)
{
 if (! p0.equals(ep0))
  return false;
 if (Orientation.index(p0, p1, ep1) == Orientation.COLLINEAR
    && Quadrant.quadrant(p0, p1) == Quadrant.quadrant(ep0, ep1) )
  return true;
 return false;
}

代码示例来源：origin: locationtech/jts

public void testOrientationIndexRobust() throws Exception 
{ 
 Coordinate p0 = new Coordinate(219.3649559090992, 140.84159161824724); 
 Coordinate p1 = new Coordinate(168.9018919682399, -5.713787599646864); 
 Coordinate p = new Coordinate(186.80814046338352, 46.28973405831556); 
 int orient = Orientation.index(p0, p1, p); 
 int orientInv = Orientation.index(p1, p0, p); 
 assert(orient != orientInv); 
} 
/**

代码示例来源：origin: locationtech/jts

public void testIsCCW() {
 assertEquals(1, Orientation.index(new Coordinate(-123456789, -40), new Coordinate(0, 0), new Coordinate(381039468754763d, 123456789)));
}

代码示例来源：origin: locationtech/jts

public void testIsCCW2() {
 assertEquals(0, Orientation.index(new Coordinate(10, 10), new Coordinate(20, 20), new Coordinate(0, 0)));
}

代码示例来源：origin: locationtech/jts

public void testA() {
 Coordinate p1 = new Coordinate(-123456789, -40);
 Coordinate p2 = new Coordinate(381039468754763d, 123456789);
 Coordinate q  = new Coordinate(0, 0);
 LineString l = new GeometryFactory().createLineString(new Coordinate[] {p1, p2});
 Point p = new GeometryFactory().createPoint(q);
 assertEquals(false, l.intersects(p));
 assertEquals(false, PointLocation.isOnLine(q, new Coordinate[] {p1, p2}));
 assertEquals(-1, Orientation.index(p1, p2, q));
}