Question

如何在数组中计算一些时间＆＃34;是＆＃34; ？在快速

createObject

例如这里有3次谢谢;）

Answer 1

这样的事情怎么样？

//: Playground - noun: a place where people can play

import UIKit

let array = ["julien: Yes","elie: Yes", "Mark: No", "Jean: Yes"]
let results = array.map({ $0.lowercaseString.containsString("yes") })
let count = results.filter({ $0 == true }).count
count // 3

你也可以把它变成一个班轮：

let count = array.filter({ $0.lowercaseString.containsString("yes") }).count

注意：我还添加了一个区分大小写的检查，只是因为数据源不一致

Answer 2

let a = ["julien: Yes","elie: Yes", "Mark: No", "Jean: Yes"]
let filtered = a.filter { (str) -> Bool in
    return str.containsString("Yes")
}
print(filtered.count)

Answer 3

let array = ["julien: Yes","elie: Yes", "Mark: No", "Jean: Yes"]
let count = array.reduce(0) { $0 + ($1.containsString("Yes") ? 1 : 0) }
count // 3

Answer 4

您可以使用NSPredicate获取所需的输出，如下所示

var arrayTemp = NSMutableArray(objects: "julien: Yes","elie: YES", "elie: NO", "Jean: Yes")
var predicate = NSPredicate(format: "self contains[cd] %@", "Yes")
arrayTemp.filterUsingPredicate(predicate)
print("COUNT===\(arrayTemp.count)")// 3

如何计算Swift /数组中元素的出现次数？

4 个答案: