Swift 2从值中查找变量名

我有一个函数，可以找到一些预先确定的位置与当前位置最近的位置。我得到了结果值，但我想要的是名称。在本例中，它将是“Larkspur”

---

您应该使用dict

[String:CLLocationDistance]

而不是数组，这样您可以获得字符串值和距离：

let distances = [
    "Larkspur": locationCurrent.distanceFromLocation(locationLarkspur)
    // ...
]

---

您必须重写代码以找到最小距离，但它会给您想要的。

这基本上是gfpacheco对答案的扩展

代码是面向函数式编程的

1.创建一个字典，其中

• 键

  是位置的名称（一个

  字符串

  ）

• 值是位置的位置（
  CLLocation
  ）

像这样：

let coordinates = [
    "Larkspur": CLLocation(latitude: 37.944804, longitude: -122.509066),
    "Saulsalito": CLLocation(latitude: 37.856315, longitude: -122.478723),
    "SanFrancisco": CLLocation(latitude: 37.795346, longitude: -122.392711),
    "Fishermans": CLLocation(latitude: 37.809166, longitude: -122.412012),
    "Giants": CLLocation(latitude: 37.778304, longitude: -122.387754),
    "Tiburon": CLLocation(latitude: 37.872928, longitude: -122.455806),
    "Vallejo": CLLocation(latitude: 38.100116, longitude: -122.262378),
    "Alameda": CLLocation(latitude: 37.790633, longitude: -122.293910),
    "HarborBay": CLLocation(latitude: 37.736799, longitude: -122.256810),
    "JackLondon": CLLocation(latitude: 37.795287, longitude: -122.279601),
    "SouthSF": CLLocation(latitude: 37.662428, longitude: -122.377791)
]

2.定义“此处” 3.创建一个新的距离字典 4.找到最近的

let latest=distance.minElement{$0.1<$1.1}？.0

---

现在，

最近的

确实包含最近位置的名称。

这里有一个稍微不同的用法，使用自定义结构保存位置信息：

import CoreLocation

// Create a custom structure to hold place names, latitude, and longitude
struct Location {
    var name: String
    var latitude: Double
    var longitude: Double
}

let lat = 37.941053
let long = -122.483915

let locations: [Location] = [
    Location(name: "Larkspur", latitude: 37.944804, longitude: -122.509066),
    Location(name: "Saulsalito", latitude: 37.856315, longitude: -122.478723),
    Location(name: "SanFrancisco", latitude: 37.795346, longitude: -122.392711),
    Location(name: "Fishermans", latitude: 37.809166, longitude: -122.412012),
    Location(name: "Giants", latitude: 37.778304, longitude: -122.387754),
    Location(name: "Tiburon", latitude: 37.872928, longitude: -122.455806),
    Location(name: "Vallejo", latitude: 38.100116, longitude: -122.262378),
    Location(name: "Alameda", latitude: 37.790633, longitude: -122.293910),
    Location(name: "HarborBay", latitude: 37.736799, longitude: -122.256810),
    Location(name: "JackLondon", latitude: 37.795287, longitude: -122.279601),
    Location(name: "SouthSF", latitude: 37.662428, longitude: -122.377791)
]

let currentLocation = CLLocation(latitude: lat, longitude: long)

// Use map to create an array of named tuple pairs with distance and location name
let distances:[(distance: Double, name: String)] = locations.map { (CLLocation(latitude: $0.latitude, longitude: $0.longitude).distanceFromLocation(currentLocation), $0.name) }

let closest = distances.minElement {$0.distance < $1.distance}!

print("Closest location is \(closest.name) at a distance of \(closest.distance)")

导入核心位置
//创建自定义结构以保存地名、纬度和经度
结构位置{
变量名称：String
纬度：双
经度：双
}
设lat=37.941053
设long=-122.483915
出租位置：[位置]=[
位置（名称：“Larkspur”，纬度：37.944804，经度：-122.509066），
位置（名称：“索尔萨利托”，纬度：37.856315，经度：-122.478723），
位置（名称：“旧金山”，纬度：37.795346，经度：-122.392711），
地点（姓名：“渔民”，纬度：37.809166，经度：-122.412012），
位置（名称：“巨人”，纬度：37.778304，经度：-122.387754），
位置（名称：“蒂布隆”，纬度：37.872928，经度：-122.455806），
位置（名称：“Vallejo”，纬度：38.100116，经度：-122.262378），
位置（名称：“阿拉米达”，纬度：37.790633，经度：-122.293910），
位置（名称：“海港”，纬度：37.736799，经度：-122.256810），
位置（名称：“杰克伦敦”，纬度：37.795287，经度：-122.279601），
位置（名称：“南SF”，纬度：37.662428，经度：-122.377791）
]
让currentLocation=CLLocation（纬度：纬度，经度：长）
//使用“映射”创建一个命名元组对数组，该数组具有距离和位置名称
让距离：[（距离：双精度，名称：字符串）]=locations.map{（CLLocation（纬度：$0.lation，经度：$0.longitude）.distanceFromLocation（当前位置），$0.name）}
设最近距离=distance.minElement{$0.distance<$1.distance}！
打印（“最近的位置是\（最近的.名称），距离\（最近的.距离）”）

您的

最近的
计算需要工作。它总是返回
Alameda
。最接近的计算是得到最小键（总是Alameda）。您应该计算最小值，而不是返回它的键，而不是好的答案。使用自定义的
位置结构
使代码更具可读性，并允许您将数据存储在
数组
中，而不是
字典
中。不错的选择。
let here = CLLocation(latitude:  12.3, longitude:  12.3)

let distances = coordinates.enumerate().reduce([String:CLLocationDistance]()) { (var accumulator, elm) -> [String:CLLocationDistance] in
    accumulator[elm.element.0] = elm.element.1.distanceFromLocation(here)
    return accumulator
}

let closest = distances.minElement { $0.1 < $1.1}?.0

import CoreLocation

// Create a custom structure to hold place names, latitude, and longitude
struct Location {
    var name: String
    var latitude: Double
    var longitude: Double
}

let lat = 37.941053
let long = -122.483915

let locations: [Location] = [
    Location(name: "Larkspur", latitude: 37.944804, longitude: -122.509066),
    Location(name: "Saulsalito", latitude: 37.856315, longitude: -122.478723),
    Location(name: "SanFrancisco", latitude: 37.795346, longitude: -122.392711),
    Location(name: "Fishermans", latitude: 37.809166, longitude: -122.412012),
    Location(name: "Giants", latitude: 37.778304, longitude: -122.387754),
    Location(name: "Tiburon", latitude: 37.872928, longitude: -122.455806),
    Location(name: "Vallejo", latitude: 38.100116, longitude: -122.262378),
    Location(name: "Alameda", latitude: 37.790633, longitude: -122.293910),
    Location(name: "HarborBay", latitude: 37.736799, longitude: -122.256810),
    Location(name: "JackLondon", latitude: 37.795287, longitude: -122.279601),
    Location(name: "SouthSF", latitude: 37.662428, longitude: -122.377791)
]

let currentLocation = CLLocation(latitude: lat, longitude: long)

// Use map to create an array of named tuple pairs with distance and location name
let distances:[(distance: Double, name: String)] = locations.map { (CLLocation(latitude: $0.latitude, longitude: $0.longitude).distanceFromLocation(currentLocation), $0.name) }

let closest = distances.minElement {$0.distance < $1.distance}!

print("Closest location is \(closest.name) at a distance of \(closest.distance)")