ios - 在swift中确定数组是否包含另一个数组的一个或多个元素

Question

我有以下代码返回用户当前位置附近的地点

import UIKit
import GooglePlaces
import CoreLocation

struct GlobalVariables {
    static var acceptedEstablishments = ["bakery", "bar", "cafe", "food", "meal_takeaway", "meal_delivery", "night_club", "restaurant", "school", "university"]
}

class ViewController: UIViewController, CLLocationManagerDelegate {

    var placesClient: GMSPlacesClient!
    var locationManager: CLLocationManager!

    // Add a pair of UILabels in Interface Builder, and connect the outlets to these variables.
    @IBOutlet var nameLabel: UILabel!
    @IBOutlet var addressLabel: UILabel!


    override func viewDidLoad() {
        super.viewDidLoad()
        placesClient = GMSPlacesClient.shared()

        locationManager = CLLocationManager()
        locationManager.delegate = self
        locationManager.requestWhenInUseAuthorization()
    }

    func locationManager(_ manager: CLLocationManager, didChangeAuthorization status: CLAuthorizationStatus) {
        if status == .authorizedWhenInUse {
            locationManager.desiredAccuracy = kCLLocationAccuracyBest
            locationManager.startUpdatingLocation()
        }
    }

    // Add a UIButton in Interface Builder, and connect the action to this function.
    @IBAction func getCurrentPlace(_ sender: UIButton) {

        placesClient.currentPlace(callback: { (placeLikelihoodList, error) -> Void in
            if let error = error {
                print("Pick Place error: \(error.localizedDescription)")
                return
            }

            if let placeLikelihoodList = placeLikelihoodList {
                for likelihood in placeLikelihoodList.likelihoods {
                    let place = likelihood.place
                    // only return places that are relevant to me
                    for placeType in place.types {
                        if (GlobalVariables.acceptedEstablishments.contains(placeType)) {
                            print("Current place name: \(place.name)")
                            print("Place type: \(placeType)")
                        }
                    }

                }
            }
        })
    }
}

place.types 在我底部的回调函数中为每个地点实例返回一个字符串数组，它看起来像这样:

["health", "point_of_interest", "establishment"]

我有一个全局字符串数组，其中还包含 bakery、bar 等标签。

当用户按下按钮时，回调函数被触发并根据附近的位置返回位置。

输出看起来像这样:

Current place name: LOCAL SUPERMARKET
Place type: food
Current place name: LOCAL GRILL
Place type: cafe
Current place name: LOCAL GRILL
Place type: food
Current place name: LOCAL SCHOOL
Place type: school
Current place name: LOCAL TAKEAWAY
Place type: meal_takeaway
Current place name: LOCAL TAKEAWAY
Place type: restaurant
Current place name: LOCAL TAKEAWAY
Place type: food

同一个机构被重复多次，因为一个机构有多个与之关联的标签。

例如:

LOCAL TAKEAWAY 的 place.types 返回数组是:["meal_takeaway", "restaurant", "food"]

并且因为我的 GlobalVariables.acceptedEstablishments 数组包含所有这三个字符串，所以 print 命令将被执行三次。

如果 place.types 数组包含一个或多个匹配的字符串，如何修改这段代码，使其只显示一次建立？我似乎无法找到解决方案。

Answer 1

关键是使用Set，不会有重复项。 Set 是一个集合类，可以像数组一样使用。你可以遍历它，它有count、map、filter、contains等

let acceptedPlaces: Set = ["home", "pub", "hospital"]
let availablePlaces: Set = ["home", "pub", "mountains"]
let inBoth = acceptedPlaces.intersection(availablePlaces) // ["home", "pub"]

您可以轻松地从 Arraylet someSet = Set(someArray) 创建 Set，反之亦然 let someArray = 数组(someSet)。您可能还想看看 Set 的以下函数:union、substract、isSuperSet、是子集。