Question

你好我是新手，我需要一些帮助。有没有办法可以将循环计数的值作为索引：

var groups = [[String]]()

groupA.append(arr[randoms[0]])
groupA.append(arr[randoms[1]])

groupB.append(arr[randoms[2]])
groupB.append(arr[randoms[3]])

groupC.append(arr[randoms[4]])
groupC.append(arr[randoms[5]])

groupD.append(arr[randoms[6]])
groupD.append(arr[randoms[7]])

groupE.append(arr[randoms[8]])
groupE.append(arr[randoms[9]])

groups.append(groupA)
groups.append(groupB)
groups.append(groupC)
groups.append(groupD)
groups.append(groupE)

我希望随机数组的索引0-10来自一个循环或者在某种程度上是动态的，有没有办法用swift实现这个？

Answer 1

我认为您正在寻找enumerated()功能......

import Foundation
let randoms = [arc4random(),arc4random(),arc4random(),arc4random(),]
randoms.enumerated().forEach {
    print($0,$1)
}

在Xcode Playground中试试。

Answer 2

你可能正在寻找这种东西。让我们摆脱random而只是 shuffle arr来生成随机顺序的实际值列表。（Stack Overflow上有很多数组重排代码，所以我不打算告诉你如何做到这一点。）然后我们也可以摆脱你的中间数组，直接处理数组的内部数组 - 数组groups：

var groups : [[String]] = Array(repeating: [], count: 5)
var shuffled = ["hey", "ho", "nonny", "no", "sis", "boom", "bah", "yo", "hi", "lo"]

var next = 0
for ix in stride(from:0, to:9, by:2) {
    for ixx in 0..<2 {
        groups[next].append(shuffled[ix+ixx])
    }
    next += 1
}

groups // [["hey", "ho"], ["nonny", "no"], ["sis", "boom"], ["bah", "yo"], ["hi", "lo"]]

我们实际上可以使用map（如Swift: what is the right way to split up a [String] resulting in a [[String]] with a given subarray size?所述）更简明扼要地表达：

let shuffled = ["hey", "ho", "nonny", "no", "sis", "boom", "bah", "yo", "hi", "lo"]
let chunkSize = 2
let groups = stride(from: 0, to: shuffled.count, by: chunkSize).map {
    Array(shuffled[$0..<min($0 + chunkSize, shuffled.count)])
}

groups // [["hey", "ho"], ["nonny", "no"], ["sis", "boom"], ["bah", "yo"], ["hi", "lo"]]

更好的是，我们可以将它作为Array的扩展：

extension Array {
    func chunked(by chunkSize:Int) -> [[Element]] {
        let groups = stride(from: 0, to: self.count, by: chunkSize).map {
            Array(self[$0..<[$0 + chunkSize, self.count].min()!])
        }
        return groups
    }
}


let shuffled = ["hey", "ho", "nonny", "no", "sis", "boom", "bah", "yo", "hi", "lo"]
let groups = shuffled.chunked(by:2)

groups // [["hey", "ho"], ["nonny", "no"], ["sis", "boom"], ["bah", "yo"], ["hi", "lo"]]

将循环计数放在数组索引中

2 个答案: