Best code snippets using net.imglib2.roi.PolygonRegionOfInterest.getEdgeEnd(Showing top 4 results out of 315)

/**
 * Determine whether the given edge is horizontal. If so,
 * interpolateEdgeXAtY won't work because the X is then indeterminate.
 * 
 * @param index
 *            index of edge to examine
 * @return true if horizontal (Y coordinates of start and end are identical)
 */
public boolean isHorizontal( final int index )
{
  return getEdgeStart( index ).getDoublePosition( 1 ) == getEdgeEnd( index ).getDoublePosition( 1 );
}

/**
 * Given an edge and a Y coordinate, find its X coordinate at that Y
 * coordinate.
 * 
 * NOTE: this assumes that isHorizontal(start) is false, else the X
 * coordinate is indeterminate.
 * 
 * @param start
 *            index of the edge
 * @return the X coordinate
 */
public double interpolateEdgeXAtY( final int start, final double y )
{
  final RealLocalizable p_start = getEdgeStart( start );
  final RealLocalizable p_end = getEdgeEnd( start );
  final double x_start = p_start.getDoublePosition( 0 );
  final double y_start = p_start.getDoublePosition( 1 );
  final double x_end = p_end.getDoublePosition( 0 );
  final double y_end = p_end.getDoublePosition( 1 );
  return x_start + ( y - y_start ) * ( x_end - x_start ) / ( y_end - y_start );
}

final double y_end = getEdgeEnd( i ).getDoublePosition( 1 );
final double x_start = getEdgeStart( i ).getDoublePosition( 0 );
final double x_end = getEdgeEnd( i ).getDoublePosition( 0 );
if ( y_start == y_end )

final RealLocalizable p1 = getEdgeEnd( i );
double x0 = p0.getDoublePosition( 0 );
double y0 = p0.getDoublePosition( 1 );