本文整理汇总了Java中com.vividsolutions.jts.geom.Coordinate.equals方法的典型用法代码示例。如果您正苦于以下问题：Java Coordinate.equals方法的具体用法？Java Coordinate.equals怎么用？Java Coordinate.equals使用的例子？那么, 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com.vividsolutions.jts.geom.Coordinate的用法示例。
在下文中一共展示了Coordinate.equals方法的2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Java代码示例。
示例1: checkRepeatedPoints2d
import com.vividsolutions.jts.geom.Coordinate; //导入方法依赖的package包/类
private static int checkRepeatedPoints2d(LineString lineString) {
int repeatedPoints = 0;
final CoordinateSequence coordSeq = lineString.getCoordinateSequence();
Coordinate nextCoord = null;
Coordinate prevCoord;
for (int i = 0; i < coordSeq.size(); ++i) {
prevCoord = nextCoord;
nextCoord = coordSeq.getCoordinate(i);
if (nextCoord.equals(prevCoord)) {
++repeatedPoints;
}
}
return repeatedPoints;
}
示例2: distancePointLine
import com.vividsolutions.jts.geom.Coordinate; //导入方法依赖的package包/类
/**
* Computes the distance from a point p to a line segment AB
*
* Note: NON-ROBUST!
*
* @param p
* the point to compute the distance for
* @param A
* one point of the line
* @param B
* another point of the line (must be different to A)
* @return the distance from p to line segment AB
*/
public static double distancePointLine(Coordinate p, Coordinate A, Coordinate B) {
// if start==end, then use pt distance
if (A.equals(B))
return p.distance(A);
// otherwise use comp.graphics.algorithms Frequently Asked Questions
// method
/*
* (1) AC dot AB r = --------- ||AB||^2 r has the following meaning: r=0
* P = A r=1 P = B r<0 P is on the backward extension of AB r>1 P is on
* the forward extension of AB 0<r<1 P is interior to AB
*/
double r = ((p.x - A.x) * (B.x - A.x) + (p.y - A.y) * (B.y - A.y))
/ ((B.x - A.x) * (B.x - A.x) + (B.y - A.y) * (B.y - A.y));
if (r <= 0.0)
return p.distance(A);
if (r >= 1.0)
return p.distance(B);
/*
* (2) (Ay-Cy)(Bx-Ax)-(Ax-Cx)(By-Ay) s = -----------------------------
* L^2
*
* Then the distance from C to P = |s|*L.
*/
double s = ((A.y - p.y) * (B.x - A.x) - (A.x - p.x) * (B.y - A.y))
/ ((B.x - A.x) * (B.x - A.x) + (B.y - A.y) * (B.y - A.y));
return Math.abs(s) * Math.sqrt(((B.x - A.x) * (B.x - A.x) + (B.y - A.y) * (B.y - A.y)));
}