Swift：什么是分割[String]的正确方法，导致[[String]]具有给定的子阵列大小？ [英] Swift: what is the right way to split up a [String] resulting in a [[String]] with a given subarray size?

问题描述

从一个大的[String]和一个给定的子阵列大小开始，我可以将这个数组拆分成更小的数组的最佳方法是什么？ （最后一个数组将小于给定的子阵列大小。）

Starting with a large [String] and a given subarray size, what is the best way I could go about splitting up this array into smaller arrays? (The last array will be smaller than the given subarray size).

具体示例：

分割[1，2，3，4，5，6，7]，最大分割尺寸为2

Split up ["1","2","3","4","5","6","7"] with max split size 2

代码会产生[[1，2]，[3，4]，[5，6]，[7 ]]

The code would produce [["1","2"],["3","4"],["5","6"],["7"]]

显然我可以手动更多地做这件事，但我感觉像是快速的东西，如map（）或reduce（ ）可能会做我想要的非常漂亮。

Obviously I could do this a little more manually, but I feel like in swift something like map() or reduce() may do what I want really beautifully.

推荐答案

我不会称之为漂亮，但这是一个使用<$ c的方法$ c> map ：

I wouldn't call it beautiful, but here's a method using map:

let numbers = ["1","2","3","4","5","6","7"]
let splitSize = 2
let chunks = numbers.startIndex.stride(to: numbers.count, by: splitSize).map {
  numbers[$0 ..< $0.advancedBy(splitSize, limit: numbers.endIndex)]
}

stride（to：by：）方法为您提供每个块的第一个元素的索引，因此您可以使用 advancedBy（距离：限制:) 。

The stride(to:by:) method gives you the indices for the first element of each chunk, so you can map those indices to a slice of the source array using advancedBy(distance:limit:).

一种更实用的方法就是递归数组，如下所示： / p>

A more "functional" approach would simply be to recurse over the array, like so:

func chunkArray<T>(s: [T], splitSize: Int) -> [[T]] {
    if countElements(s) <= splitSize {
        return [s]
    } else {
        return [Array<T>(s[0..<splitSize])] + chunkArray(Array<T>(s[splitSize..<s.count]), splitSize)
    }
}

这篇关于Swift：什么是分割[String]的正确方法，导致[[String]]具有给定的子阵列大小？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！