按距当前位置的距离对数组进行排序

Question

我正在尝试按距当前位置的距离对数组进行排序。首先，您会看到organizationObject其中包含不同值的，然后我们得到的位置是一组不同位置的数组。之所以如此，array是因为一个组织可以有多个位置。然后在ViewController中，我创建了一组测试对象并将其追加到数组中。我的问题是如何orgArray根据distanceFromLocation 对此进行排序？如果位置数组中有多个位置，则应采用最接近的位置。

OrganizationObject

class OrganizationObject {
    var id: Int
    var name: String
    var image: UIImage
    var locations: [CLLocationCoordinate2D]
    init(id: Int, name: String, image: UIImage, locations: [CLLocationCoordinate2D]) {
        self.id = id
        self.name = name
        self.image = image
        self.locations = locations
    }

}

将测试对象追加到数组。在viewDidLoad中

orgArray.append(OrganizationObject(id: 0, name: "Statens Museum For Kunst", image: UIImage(named: "statensmuseum.jpg")!, locations: [CLLocationCoordinate2D(latitude: 55.6888127, longitude: 12.578330300000061)]))
orgArray.append(OrganizationObject(id: 0, name: "7 eleven", image: UIImage(named: "7eleven.jpg")!, locations: [CLLocationCoordinate2D(latitude: 58.334682, longitude: 8.231820900000002)]))
orgArray.append(OrganizationObject(id: 0, name: "Kongens have", image: UIImage(named: "kongenshave.jpg")!, locations: [CLLocationCoordinate2D(latitude: 55.6852905, longitude:12.579845200000022)]))
orgArray.append(OrganizationObject(id: 0, name: "Magasin du nord", image: UIImage(named: "magasin.jpg")!, locations: [CLLocationCoordinate2D(latitude: 50.6456604, longitude: 3.053486600000042), CLLocationCoordinate2D(latitude: 55.7835017, longitude: 12.370985799999971)]))

Answer 1

这是一个解决方案。这是一个更复杂一点，要排序功能的简单调用了几个原因：首先，需要找到该组织最近的位置，在问题中提到，其次是内部使用的半正矢计算的事实CLLocation的distanceFromLocation:方法如果天真地使用它会减慢速度。

由于这些原因，我创建了一个特殊的对象进行排序，因此我可以使用成员字典来记忆对的调用结果distanceFromLocation。这不会对测试数据产生影响，但是如果您需要处理大量地点，那将很重要。

附带说明-如果OrganizationObject存储CLLocation而不是-可能会使事情变得简单一些CLLocationCoordinate-尽管这是一个相当小的问题。

这是代码：

class OrganizationSorter {

  var memoizedValues = [Int:CLLocationDistance]()

  private func shortestDistanceToOrganizationFromLocation(organization:OrganizationObject,location:CLLocation) -> CLLocationDistance? {

    let memoizedValue = memoizedValues[organization.id] //Check whether we've done this calculation before, if so return the result from last time
    if memoizedValue != nil {
        return memoizedValue
    }

    //There should probably be some code around here to check
    //that the organization object has at least one location
    //I'm assuming it does to simplify things

    var shortestDistance : CLLocationDistance? = nil
    let locations = organization.locations
    if locations.count > 0 {
      for coord in locations {
        let loc = CLLocation(latitude: coord.latitude, longitude: coord.longitude)
        let dist = loc.distanceFromLocation(location)

        if shortestDistance == nil || shortestDistance > dist {
          shortestDistance = dist
        }
      }
    }

    if shortestDistance != nil {
      memoizedValues[organization.id] = shortestDistance
    }

    return shortestDistance
  }

  func sortOrganizationsByDistanceFromLocation(orgArray:[OrganizationObject],location:CLLocation) -> [OrganizationObject] {
    let sortedArray = orgArray.sort { (a:OrganizationObject, b:OrganizationObject) -> Bool in
      let dist1 = self.shortestDistanceToOrganizationFromLocation(a, location: location)
      let dist2 = self.shortestDistanceToOrganizationFromLocation(b, location: location)
      return dist1 < dist2     
    }
    memoizedValues.removeAll() //reset memoized values in case object is used twice
    return sortedArray
  }
}

我已经使用哥本哈根的克里斯蒂安堡宫的位置作为测试位置，对您的示例数据进行了测试，并获得了以下顺序：

Kongens have
Statens Museum For Kunst
Magasin du nord
7 eleven

似乎与给定的坐标相匹配-看起来，最近的Magasin du Nord坐标是在哥本哈根郊区的一个城镇中的某个地方（另一个在里尔吗？），还有11个在瑞典。

这是使用类的方式（使用原始问题中的测试数据，并更改id了OrganizationObjects中的值，以使它们不全为0（否则代码将不起作用）。

let location = CLLocation(latitude: 55.676251, longitude: 12.580570) //Christiansborg Palace, chosen since it is relatively near the other locations, to make it obvious whether results are sensible or not

let orgSorter = OrganizationSorter()

let sortedLocations = orgSorter.sortOrganizationsByDistanceFromLocation(orgArray, location: location)

for org in orgArray {
  print(org.name)
}

print("\n")

for org in sortedLocations {
  print(org.name)
}