我有个简单的聚会课
class Meetup {
var title: String
var date: String
}以及从meetup.com获取的一个名为meetups的会议数组。
我想在字典中按日期组织这些会议:[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
}它能工作,但效率极低,感觉和看上去都一样。
如何从数组中的对象中提取属性,并根据该属性高效地构建索引?
发布于 2016-02-21 03:18:16
我会和你做的差不多。毕竟,您只在数组中循环一次。
我想你应该把算法表达得更清楚一点。你对每次聚会的选择是:
我想我们可以很清楚地这样说:
// 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]
}现在让我们来证明它是有效的:
print(dict) // ["one": [one/Howdy, one/Namaste], "two": [two/Hello, two/Bonjour]]因此,我们最终得到了一个字典,它的键是原始的ids (您的日期),每个id的值是一个带有该id的Meetups数组。
https://stackoverflow.com/questions/35531738
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