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arrays - 在Swift中合并值数组的最有效方法是什么

问题描述

我有一个稍微理论上的问题，但有一个非常实际的要求。

我试图找到基于多种因素合并 n 个大小相等的数组的最有效方法。这需要高性能。数组都是 512 字节，将以 40fps 进行评估。

规则（按顺序）： 活动必须为真 最高优先级总是获胜 最大大小

结果数组是 512 x UInt8 值。

struct Slot {
   var size: UInt8
   var priority: UInt8
   var active: Bool
}

let source1 = [Slot(size: 0, priority: 100, active: true), Slot(size: 20, priority: 150, active: true), Slot(size: 150, priority: 100, active: true)]
let source2 = [Slot(size: 10, priority: 150, active: false), Slot(size: 40, priority: 100, active: true), Slot(size: 200, priority: 100, active: true)]
let source3 = [Slot(size: 20, priority: 200, active: false), Slot(size: 10, priority: 100, active: false), Slot(size: 30, priority: 100, active: true)]


// result expected [0, 20, 200]

我强调这完全与性能有关，生成的代码有多漂亮并不重要。

这是我正在寻找的 for in 循环：


struct Slot {
   var size: UInt8
   var priority: UInt8
   var active: Bool
}

let source1 = [Slot(size: 0, priority: 100, active: true), Slot(size: 20, priority: 150, active: true), Slot(size: 150, priority: 100, active: true)]
let source2 = [Slot(size: 10, priority: 150, active: false), Slot(size: 40, priority: 100, active: true), Slot(size: 200, priority: 100, active: true)]
let source3 = [Slot(size: 20, priority: 200, active: false), Slot(size: 10, priority: 100, active: false), Slot(size: 30, priority: 100, active: true)]

let sources = [source1, source2, source3]

var resultingArray = [UInt8]()

// loop through each array index
for index in 0...2 {

   let slots = sources.map { $0[index] }
   let activeSlots = slots.filter { $0.active }

   let sortedSlots = activeSlots.sorted {
      if $0.priority != $1.priority {
         return $0.priority > $1.priority
      } else {
        return $0.size > $1.size
      }
   }

    resultingArray.append(sortedSlots[0].size)

}

print(resultingArray)

标签： arraysswiftperformance

解决方案

因此，根据我对您评论的理解，您可以做的最好的事情是线性遍历每个数组以找到“获胜”插槽，然后提取大小。无需排序。

extension Slot: Comparable {
    static func < (lhs: Slot, rhs: Slot) -> Bool {
        !lhs.active && rhs.active ||
        lhs.active == rhs.active && (
            lhs.priority < rhs.priority ||
            lhs.priority == rhs.priority && lhs.size < rhs.size
        )
    }
}

let sizes = zip(source1, zip(source2, source3))
        .map { max(max($0, $1.0), $1.1).size }

其中，使用给定的值将产生：

[0, 20, 200]

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