我有一个简单的见面会

class Meetup {
  var title: String
  var date: String
}


从meetup.com获取的一系列见面聚会称为meetups

我想按日期在字典中组织这些见面会:[String, [Meetup]]其中字符串是日期。

这是我的实现

func buildDateMeetupDict(meetups: [Meetup]) -> [String, [Meetup]] {

    var dateMeetupDict = [String: [Meetup]]()

    for meetup in meetups {
        for var meetupsByDay in dateMeetupDict {
            if meetupsByDay.day == meetup.day {
                meetupsByDay.meetupArray.append(meetup)
            } else {
                let newMeetupDay = [meetup.day, [meetup]]
                dateMeetupDict.append(newMeetupDay)
            }
        }
    }
    return dateMeetupDict
}


它可以工作,但效率极低,而且感觉和外观都一样贫民窟。

如何有效地从数组中的对象中提取属性并基于该属性构建索引?

最佳答案

我几乎会像您正在做的那样做。毕竟,您只循环一次遍历数组。

我认为您可能想更清楚地表达算法。每次聚会的选择是:


如果该键不存在,则创建它并使其值成为其中包含此Meetup的数组;
如果键确实存在,请将此聚会添加到其值数组中。


我认为我们可以很明确地说如下:

// here's a test class
// [Note: I used `id` instead of `date`, but it's still just a string...]
class Meetup : CustomStringConvertible {
    var id: String
    var title: String
    init(id:String, title:String) {
        self.id = id; self.title = title
    }
    var description: String {
        return "\(self.id)/\(self.title)"
    }
}

// here's a test array of Meetups
let meetups : [Meetup] = [
    Meetup(id:"one", title:"Howdy"),
    Meetup(id:"two", title:"Hello"),
    Meetup(id:"two", title:"Bonjour"),
    Meetup(id:"one", title:"Namaste")
]

// and here's our actual code!
var dict = [String:[Meetup]]()
for meetup in meetups {
    let val = dict[meetup.id]
    dict[meetup.id] = val == nil ? [meetup] : val! + [meetup]
}


现在让我们证明它是可行的:

print(dict) // ["one": [one/Howdy, one/Namaste], "two": [two/Hello, two/Bonjour]]


因此,我们最后得到了一个词典,该词典的键是原始的id(您的日期),并且每个id的值都是具有该id的Meetups的数组。

10-08 07:45