将GMSVisibleRegion转换为CLRegion或MKCoordinateRegion

Question

我正在使用GoogleMaps软件开发工具包，目前我正在尝试将GMSVisibleRegion转换为CLRegion。

GMSVisibleRegion定义为：

typedef struct {
  CLLocationCoordinate2D nearLeft;
  CLLocationCoordinate2D nearRight;
  CLLocationCoordinate2D farLeft;
  CLLocationCoordinate2D farRight;
} GMSVisibleRegion;

执行此操作的最快方法是什么？

不幸的是，很难理解开发人员命名“近”和“远”是什么意思。我认为这条评论也很有用：

/**
 * Returns the region (four location coordinates) that is visible according to
 * the projection.
 *
 * The visible region can be non-rectangular. The result is undefined if the
 * projection includes points that do not map to anywhere on the map (e.g.,
 * camera sees outer space).
 */
 - (GMSVisibleRegion)visibleRegion;

非常感谢!

编辑:好的，我的第一步是创建一个GMSVisibleRegion的MKCoordinateRegion。

我建议使用以下代码将GMSVisibleRegion转换为MKCoordinateRegion。任何反对意见。

+ (MKCoordinateRegion)regionForCenter:(CLLocationCoordinate2D)center andGMSVisibleRegion:(GMSVisibleRegion)visibleRegion
{
    CLLocationDegrees latitudeDelta = visibleRegion.farLeft.latitude - visibleRegion.nearLeft.latitude;
    CLLocationDegrees longitudeDelta = visibleRegion.farRight.longitude - visibleRegion.farLeft.longitude;
    MKCoordinateSpan span = MKCoordinateSpanMake(latitudeDelta, longitudeDelta);

    return MKCoordinateRegionMake(center, span);
}

Answer 1

我猜“near”是指屏幕底部的视图角落，而“far”是指屏幕顶部的角落。这是因为如果您倾斜了视图，则底部角距离摄影机最近，而顶部角距离摄影机最远。

将其转换为CLRegion的一种方法可能是使用相机的目标作为中心，然后计算从最大距离到四个角的半径。这可能不是该区域上最紧密的拟合圆，但由于圆无论如何都不能拟合视图的四边形，所以它可能足够接近。

下面是一个辅助函数，用于计算两个CLLocationCoordinate值之间以米为单位的距离：

double getDistanceMetresBetweenLocationCoordinates(
    CLLocationCoordinate2D coord1, 
    CLLocationCoordinate2D coord2)
{
    CLLocation* location1 = 
        [[CLLocation alloc] 
            initWithLatitude: coord1.latitude 
            longitude: coord1.longitude];
    CLLocation* location2 = 
        [[CLLocation alloc] 
            initWithLatitude: coord2.latitude 
            longitude: coord2.longitude];

    return [location1 distanceFromLocation: location2];
}

然后，可以这样计算CLRegion：

GMSMapView* mapView = ...;
...
CLLocationCoordinate2D centre = mapView.camera.target;
GMSVisibleRegion* visibleRegion = mapView.projection.visibleRegion;

double nearLeftDistanceMetres = 
    getDistanceMetresBetweenLocationCoordinates(centre, visibleRegion.nearLeft);
double nearRightDistanceMetres = 
    getDistanceMetresBetweenLocationCoordinates(centre, visibleRegion.nearRight);
double farLeftDistanceMetres = 
    getDistanceMetresBetweenLocationCoordinates(centre, visibleRegion.farLeft);
double farRightDistanceMetres = 
    getDistanceMetresBetweenLocationCoordinates(centre, visibleRegion.farRight);
double radiusMetres = 
    MAX(nearLeftDistanceMetres, 
    MAX(nearRightDistanceMetres, 
    MAX(farLeftDistanceMetres, farRightDistanceMetres)));

CLRegion region = [[CLRegion alloc] 
    initCircularRegionWithCenter: centre radius: radius identifier: @"id"];

更新：

关于您的MKCoordinateRegion更新，您的示例代码可能无法工作。如果地图旋转了90度，那么farLeft和nearLeft将具有相同的纬度，farRight和farLeft将具有相同的经度，因此您的纬度和经度差值将为零。

您需要遍历farLeft、farRight、nearLeft、nearRight中的所有四个，计算每一个的纬度和经度的最小和最大值，然后从中计算增量。

Google Maps SDK for iOS包含一个helper类，它已经为您完成了其中的一些工作-- GMSCoordinateBounds。它可以使用GMSVisibleRegion进行初始化

GMSMapView* mapView = ...;
....
GMSVisibleRegion visibleRegion = mapView.projection.visibleRegion;
GMSCoordinateBounds bounds = 
    [[GMSCoordinateBounds alloc] initWithRegion: visibleRegion];

然后，GMSCoordinateBounds具有定义边界的northEast和southWest属性。因此，您可以按如下方式计算增量：

CLLocationDegrees latitudeDelta = 
    bounds.northEast.latitude - bounds.southWest.latitude;
CLLocationDegrees longitudeDelta = 
    bounds.northEast.longitude - bounds.southWest.longitude;

您还可以从边界计算中心，从而计算MKCoordinateRegion

CLLocationCoordinate2D centre = CLLocationCoordinate2DMake(
    (bounds.southWest.latitude + bounds.northEast.latitude) / 2,
    (bounds.southWest.longitude + bounds.northEast.longitude) / 2);
MKCoordinateSpan span = MKCoordinateSpanMake(latitudeDelta, longitudeDelta);
return MKCoordinateRegionMake(centre, span);

Answer 2

纯粹主义者附录

如果你想要绝对严谨，你需要在国际日期变更线周围做一个修正。这在大多数应用程序中都是浪费精力的，但这个问题最近一直给我带来很大的痛苦，所以我想把它扔到社区帽子里去

基于Druce的更新(恐怕我不能发表评论)。

GMSMapView* mapView = ...;
....
GMSVisibleRegion visibleRegion = mapView.projection.visibleRegion;
GMSCoordinateBounds bounds = 
    [[GMSCoordinateBounds alloc] initWithRegion: visibleRegion];

Latitude不需要对它做任何操作

CLLocationDegrees latitudeDelta = 
bounds.northEast.latitude - bounds.southWest.latitude;

根据协议，一个跨越国际日期变更线的地区可能在日本有southWest角(+140经度)，在阿拉斯加有northEast角(-150经度)。加起来再除以2得出的是地球另一边的一个点。

northEast.longitude小于southWest.longitude的特殊情况需要处理

CLLocationCoordinate2D centre;
CLLocationDegrees longitudeDelta;

if(bounds.northEast.longitude >= bounds.southWest.longitude) {
//Standard case
    centre = CLLocationCoordinate2DMake(
             (bounds.southWest.latitude + bounds.northEast.latitude) / 2,
             (bounds.southWest.longitude + bounds.northEast.longitude) / 2);
    longitudeDelta = bounds.northEast.longitude - bounds.southWest.longitude;
} else {
//Region spans the international dateline
    centre = CLLocationCoordinate2DMake(
             (bounds.southWest.latitude + bounds.northEast.latitude) / 2,
             (bounds.southWest.longitude + bounds.northEast.longitude + 360) / 2);
    longitudeDelta = bounds.northEast.longitude + 360 
                    - bounds.southWest.longitude;
}
MKCoordinateSpan span = MKCoordinateSpanMake(latitudeDelta, longitudeDelta);
return MKCoordinateRegionMake(centre, span);

Answer 3

对于任何基于目前提供的所有答案和更正来寻找样板代码的人，这里是作为GMSMapView上的一个类别实现的region：

//
//  GMSMapViewExtensions.h
//

#import <Foundation/Foundation.h>
#import <MapKit/MapKit.h>
#import <GoogleMaps/GoogleMaps.h>

@interface GMSMapView (GMSMapViewExtensions)

@end

和

//
//  GMSMapViewExtensions.m
//

#import "GMSMapViewExtensions.h"

@implementation GMSMapView (GMSMapViewExtensions)

- (MKCoordinateRegion) region {
    GMSVisibleRegion visibleRegion = self.projection.visibleRegion;
    GMSCoordinateBounds * bounds = [[GMSCoordinateBounds alloc] initWithRegion: visibleRegion];

    CLLocationDegrees latitudeDelta = bounds.northEast.latitude - bounds.southWest.latitude;

    CLLocationCoordinate2D centre;
    CLLocationDegrees longitudeDelta;

    if (bounds.northEast.longitude >= bounds.southWest.longitude) {
        // Standard case
        centre = CLLocationCoordinate2DMake(
            (bounds.southWest.latitude + bounds.northEast.latitude) / 2,
            (bounds.southWest.longitude + bounds.northEast.longitude) / 2);
        longitudeDelta = bounds.northEast.longitude - bounds.southWest.longitude;
    } else {
        // Region spans the international dateline
        centre = CLLocationCoordinate2DMake(
            (bounds.southWest.latitude + bounds.northEast.latitude) / 2,
            (bounds.southWest.longitude + bounds.northEast.longitude + 360) / 2);
        longitudeDelta = bounds.northEast.longitude + 360 - bounds.southWest.longitude;
    }

    MKCoordinateSpan span = MKCoordinateSpanMake(latitudeDelta, longitudeDelta);
    return MKCoordinateRegionMake(centre, span);
}


- (MKMapRect)visibleMapRect {
    MKCoordinateRegion region = [self region];
    MKMapPoint a = MKMapPointForCoordinate(CLLocationCoordinate2DMake(
        region.center.latitude + region.span.latitudeDelta / 2,
        region.center.longitude - region.span.longitudeDelta / 2));
     MKMapPoint b = MKMapPointForCoordinate(CLLocationCoordinate2DMake(
        region.center.latitude - region.span.latitudeDelta / 2,
        region.center.longitude + region.span.longitudeDelta / 2));
     return MKMapRectMake(MIN(a.x, b.x), MIN(a.y, b.y), ABS(a.x - b.x), ABS(a.y - b.y));
}

@end

使用示例：

GMSMapView * mapView = .... // init code
MKCoordinateRegion mapRegion = mapView.region;