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我正在完成一个应用程序，希望我需要向用户显示他与大约 500 个坐标之间的距离。

使用 CLLocation 方法计算，效果很好，但在 iPhone 4 中完成每个位置的计算大约需要 1 分钟。

最好的方法是什么？使用跨度？还有其他更快的方法吗？

谢谢大家，

瑞

4

4 回答 4

6

我认为 Sahgal 是对的，这里有一些代码，也许它会对你有所帮助。

+(CGFloat)calculateDistanceBetweenSource:(CLLocationCoordinate2D)firstCoords andDestination:(CLLocationCoordinate2D)secondCoords 
{

    // this radius is in KM => if miles are needed it is calculated during setter of Place.distance

    double nRadius = 6371;

    // Get the difference between our two points

    // then convert the difference into radians

    double nDLat = (firstCoords.latitude - secondCoords.latitude)* (M_PI/180);
    double nDLon = (firstCoords.longitude - secondCoords.longitude)* (M_PI/180);

    double nLat1 =  secondCoords.latitude * (M_PI/180);
    double nLat2 =  secondCoords.latitude * (M_PI/180);

    double nA = pow ( sin(nDLat/2), 2 ) + cos(nLat1) * cos(nLat2) * pow ( sin(nDLon/2), 2 );

    double nC = 2 * atan2( sqrt(nA), sqrt( 1 - nA ));

    double nD = nRadius * nC;

    NSLog(@"Distance is %f",nD);

    return nD; // converts to miles or not (if en_) => implicit in method
}
于 2012-01-16T09:41:56.917 回答
2

我看过其他答案，不知道他们是否正确，但我认为有更好的解决方案：

（来自文档）：

- (CLLocationDistance)distanceFromLocation:(const CLLocation *)location

你可以像这样使用它：

- (CLLocationDistance) DistanceBetweenCoordinate:(CLLocationCoordinate2D)originCoordinate andCoordinate:(CLLocationCoordinate2D)destinationCoordinate {

        CLLocation *originLocation = [[CLLocation alloc] initWithLatitude:originCoordinate.latitude longitude:originCoordinate.longitude];
        CLLocation *destinationLocation = [[CLLocation alloc] initWithLatitude:destinationCoordinate.latitude longitude:destinationCoordinate.longitude];
        CLLocationDistance distance = [originLocation distanceFromLocation:destinationLocation];
        [originLocation release];
        [destinationLocation release];

        return distance;
    }
于 2012-02-03T12:45:32.143 回答
1

这是代码..

-(NSString *)findDistanceBetweenTwoLatLon
{
    int intEarthRadius = 3963;

    double dblLat1 = DegreesToRadians(firstLatitude);
    double dblLon1 = DegreesToRadians(firstLongitude);

    double dblLat2 = DegreesToRadians(secondLatitude);
    double dblLon2 = DegreesToRadians(secondLongitude);

    float fltLat = dblLat2 - dblLat1;
    float fltLon = dblLon2 - dblLon1;

    double a = sin(fltLat/2) * sin(fltLat/2) + cos(dblLat2) * cos(dblLat2) * sin(fltLon/2) * sin(fltLon/2) ;
    double c = 2 * atan2(sqrt(a), sqrt(1-a));
    double d = intEarthRadius * c;

    double dMeters = d * kOneMileMeters;

    NSString *strDistance = [NSString stringWithFormat:@"%1.2f meters",dMeters];

    return strDistance;
}

定义所有这个宏..

度数为弧度

#define DegreesToRadians(degrees) (degrees * M_PI / 180)

其中 M_PI

#define M_PI   3.14159265358979323846264338327950288 
#define kOneMileMeters 1609.344
于 2012-01-16T09:42:34.700 回答
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您可以尝试另一种方法。您可以获取两个点的经纬度（假设经纬度查找不会花费那么多时间）并且有公式可以使用经纬度计算距离。

于 2012-01-16T09:24:58.857 回答