10

我有一个任意的CGPath，我想找到它的地理中心。我可以得到路径边界框，CGPathGetPathBoundingBox然后找到该框的中心。但是有没有更好的方法来找到路径的中心？

喜欢看代码的人更新：这里是使用 Adam 在答案中建议的平均点法的代码（不要错过下面答案中更好的技术）......

    BOOL moved = NO; // the first coord should be a move, the rest add lines
    CGPoint total = CGPointZero;
    for (NSDictionary *coord in [polygon objectForKey:@"coordinates"]) {
        CGPoint point = CGPointMake([(NSNumber *)[coord objectForKey:@"x"] floatValue], 
                                    [(NSNumber *)[coord objectForKey:@"y"] floatValue]);
        if (moved) {
            CGContextAddLineToPoint(context, point.x, point.y);
            // calculate totals of x and y to help find the center later
            // skip the first "move" point since it is repeated at the end in this data
            total.x = total.x + point.x;
            total.y = total.y + point.y;
        } else {
            CGContextMoveToPoint(context, point.x, point.y);
            moved = YES; // we only move once, then we add lines
        }
    }

    // the center is the average of the total points
    CGPoint center = CGPointMake(total.x / ([[polygon objectForKey:@"coordinates"] count]-1), total.y / ([[polygon objectForKey:@"coordinates"] count]-1));

如果你有更好的想法，请分享！

4

6 回答 6

9

该技术有效，但您在问题中输入的代码无效。AFAICS，仅适用于您只做直线多边形的少数情况，并且您有一个点列表，并且您还没有制作 CGPath 对象。

我需要为任意 CGPath 对象做这件事。使用亚当（其他亚当）的建议和苹果的 CGPathApply，我想出了这个，效果很好：

{
            float dataArray[3] = { 0, 0, 0 };
            CGPathApply( (CGPathRef) YOUR_PATH, dataArray, pathApplierSumCoordinatesOfAllPoints);

            float averageX = dataArray[0] / dataArray[2];
            float averageY = dataArray[1]  / dataArray[2];
            CGPoint centerOfPath = CGPointMake(averageX, averageY);
}

static void pathApplierSumCoordinatesOfAllPoints(void* info, const CGPathElement* element)
{
    float* dataArray = (float*) info;
    float xTotal = dataArray[0];
    float yTotal = dataArray[1];
    float numPoints = dataArray[2];


    switch (element->type)
    {
        case kCGPathElementMoveToPoint:
        {
            /** for a move to, add the single target point only */

            CGPoint p = element->points[0];
            xTotal += p.x;
            yTotal += p.y;
            numPoints += 1.0;

        }
            break;
        case kCGPathElementAddLineToPoint:
        {
            /** for a line to, add the single target point only */

            CGPoint p = element->points[0];
            xTotal += p.x;
            yTotal += p.y;
            numPoints += 1.0;

        }
            break;
        case kCGPathElementAddQuadCurveToPoint:
            for( int i=0; i<2; i++ ) // note: quad has TWO not THREE
            {
                /** for a curve, we add all ppints, including the control poitns */
                CGPoint p = element->points[i];
                xTotal += p.x;
                yTotal += p.y;
                numPoints += 1.0;
            }
            break;
        case kCGPathElementAddCurveToPoint:         
            for( int i=0; i<3; i++ ) // note: cubic has THREE not TWO
            {
                /** for a curve, we add all ppints, including the control poitns */
                CGPoint p = element->points[i];
                xTotal += p.x;
                yTotal += p.y;
                numPoints += 1.0;
            }
            break;
        case kCGPathElementCloseSubpath:
            /** for a close path, do nothing */
            break;
    }

    //NSLog(@"new x=%2.2f, new y=%2.2f, new num=%2.2f", xTotal, yTotal, numPoints);
    dataArray[0] = xTotal;
    dataArray[1] = yTotal;
    dataArray[2] = numPoints;
}
于 2011-12-24T20:34:42.583 回答
6

对我来说，路径中所有点的简单平均值对于我正在处理的某些多边形来说是不够的。

我使用该区域实现了它（参见维基百科、多边形的质心和Paul Bourke 的页面）。它可能不是最有效的实现，但它对我有用。

请注意，它仅适用于封闭的、不相交的多边形。假设顶点按照它们沿多边形周长出现的顺序进行编号，并且假设最后一个点与第一个点相同。

CGPoint GetCenterPointOfCGPath (CGPathRef aPath)
{
    // Convert path to an array
    NSMutableArray* a = [NSMutableArray new];
    CGPathApply(aPath, (__bridge void *)(a), convertToListOfPoints);
    return centroid(a);
}

static void convertToListOfPoints(void* info, const CGPathElement* element)
{
    NSMutableArray* a = (__bridge NSMutableArray*) info;

    switch (element->type)
    {
        case kCGPathElementMoveToPoint:
        {
            [a addObject:[NSValue valueWithCGPoint:element->points[0]]];
        }
        break;
        case kCGPathElementAddLineToPoint:
        {
            [a addObject:[NSValue valueWithCGPoint:element->points[0]]];
        }
        break;
        case kCGPathElementAddQuadCurveToPoint:
        {
            for (int i=0; i<2; i++)
                [a addObject:[NSValue valueWithCGPoint:element->points[i]]];
        }
        break;
        case kCGPathElementAddCurveToPoint:
        {
            for (int i=0; i<3; i++)
                [a addObject:[NSValue valueWithCGPoint:element->points[i]]];
        }
        break;
        case kCGPathElementCloseSubpath:
        break;
    }
}

double polygonArea(NSMutableArray* points) {
    int i,j;
    double area = 0;
    int N = [points count];

    for (i=0;i<N;i++) {
        j = (i + 1) % N;
        CGPoint pi =  [(NSValue*)[points objectAtIndex:i] CGPointValue];
        CGPoint pj =  [(NSValue*)[points objectAtIndex:j] CGPointValue];
        area += pi.x * pj.y;
        area -= pi.y * pj.x;
    }

    area /= 2;
    return area;
}

CGPoint centroid(NSMutableArray* points) {
    double cx = 0, cy = 0;
    double area = polygonArea(points);

    int i, j, n = [points count];

    double factor = 0;
    for (i = 0; i < n; i++) {
        j = (i + 1) % n;
        CGPoint pi =  [(NSValue*)[points objectAtIndex:i] CGPointValue];
        CGPoint pj =  [(NSValue*)[points objectAtIndex:j] CGPointValue];
        factor = (pi.x * pj.y - pj.x * pi.y);
        cx += (pi.x + pj.x) * factor;
        cy += (pi.y + pj.y) * factor;
    }

    cx *= 1 / (6.0f * area);
    cy *= 1 / (6.0f * area);

    return CGPointMake(cx, cy);
}
于 2013-10-04T09:18:30.593 回答
2

路径中点的所有 x 和所有 y 的简单平均值是否给出了您想要的点？为 x 计算一个值，为 y 计算一个值。我做了一个速写，这个方法给出了一个可信的答案。

参见维基百科，找到一组有限点的质心。

如果不是，您可能需要先找到该区域 - 请参阅Paul Bourke 的页面。

于 2011-08-25T06:56:36.680 回答
1

更新了亚当对 swift4 版本的回答：

extension CGPath {
    func findCenter() -> CGPoint {
        class Context {
            var sumX: CGFloat = 0
            var sumY: CGFloat = 0
            var points = 0
        }

        var context = Context()

        apply(info: &context) { (context, element) in
            guard let context = context?.assumingMemoryBound(to: Context.self).pointee else {
                return
            }
            switch element.pointee.type {
            case .moveToPoint, .addLineToPoint:
                let point = element.pointee.points[0]
                context.sumX += point.x
                context.sumY += point.y
                context.points += 1
            case .addQuadCurveToPoint:
                let controlPoint = element.pointee.points[0]
                let point = element.pointee.points[1]
                context.sumX += point.x + controlPoint.x
                context.sumY += point.y + controlPoint.y
                context.points += 2
            case .addCurveToPoint:
                let controlPoint1 = element.pointee.points[0]
                let controlPoint2 = element.pointee.points[1]
                let point = element.pointee.points[2]
                context.sumX += point.x + controlPoint1.x + controlPoint2.x
                context.sumY += point.y + controlPoint1.y + controlPoint2.y
                context.points += 3
            case .closeSubpath:
                break
            }
        }

        return CGPoint(x: context.sumX / CGFloat(context.points),
                y: context.sumY / CGFloat(context.points))
    }
}

但要小心，CGPath 可能有额外的移动命令，因为点数会破坏这个逻辑

于 2018-11-15T12:51:19.083 回答
-1

这是质心，来吧：

-(CLLocationCoordinate2D)getCentroidFor:(GMSMutablePath *)rect
{
  CLLocationCoordinate2D coord = [rect coordinateAtIndex:0];
  double minX = coord.longitude;
  double maxX = coord.longitude;
  double minY = coord.latitude;
  double maxY = coord.latitude;
  for (int i = 1; i < rect.count; i++)
  {
    CLLocationCoordinate2D coord = [rect coordinateAtIndex:i];
    if (minX > coord.longitude)
      minX = coord.longitude;
    if (maxX < coord.longitude)
      maxX = coord.longitude;
    if (minY > coord.latitude)
      minY = coord.latitude;
    if (maxY < coord.latitude)
      maxY = coord.latitude;
  }

  CLLocationDegrees centerX = minX + ((maxX - minX) / 2);
  CLLocationDegrees centerY = minY + ((maxY - minY) / 2); 
    return CLLocationCoordinate2DMake(centerY, centerX);
}
于 2015-07-01T06:00:58.520 回答
-3

找到 CGPath 的边界框并取其中心。

CGRect boundingBox = CGPathGetBoundingBox(my_path); my_center_point = ccp(boundingBox.origin.x+boundingBox.size.width/2, boundingBox.origin.y+boundingBox.size.height/2);

于 2013-10-01T10:42:46.583 回答