arrays - 根据下一个对象过滤数组

Question

我正在编写一些代码来从运动传感器中过滤掉驾驶行程。我发现最好的方法是根据以下内容将子数组添加到嵌套数组:

• 检测首次发生的有把握的汽车事件
• 将以下所有运动事件添加到同一事件数组中，直到第一个有把握的观察结果表明情况并非如此。

例如

automotive confidence 2 //Add
automotive confidence 2 //Add
automotive confidence 2 //Add
walking confidence 2 //Add the sub-array to the master array and start over on the next confident automotive event. 

目前我是这样做的:

        //Remove all uncertain values.
        let confidentActivities = activities!.filter{$0.confidence.rawValue == 2}

        var needsNew = true
        var automotiveActivities:Array<Array<CMMotionActivity>> = Array() //Master array to contain subarrays of automotiveactivity arrays
        var automotiveActivitySession:Array<CMMotionActivity> = Array()
        for activity in confidentActivities {
            if activity.automotive && (!activity.cycling && !activity.running && !activity.walking){
                if needsNew {
                    needsNew = false
                }
                automotiveActivitySession.append(activity)
            } else {
                if !needsNew {
                    //If user is no longer in car, store a cpoy of the session and reset the array
                    automotiveActivities.append(Array(automotiveActivitySession))
                    automotiveActivitySession = []
                    needsNew = true
                }
            }
        }

这个解决方案不是很优雅。有没有什么方法可以使用 Swift 的 Array.filter{} 来使这种排序更漂亮？

Answer 1

过滤器不会这样做，但您可以使用 reduce。

下面的示例展示了如何将连续 “A”(表示汽车事件)的运行收集到数组数组中的数组中:

let data = ["A","A","A","B","A","A","B","A","A","A","A","B","B","B","A","B","A","A","A","A","A","A","B","A"]
var res = [[String]]()
_ = data.reduce("") { (last: String, current: String) in
    if current == "A" {
        if last != "A" {
            res.append([String]())
        }
        res[res.count-1].append(current)
    }
    return current
}
print(res)

先验值作为第一个参数传递给reduce 函数。这使得该函数可以决定是追加到当前列表还是开始一个新列表。

本次运行结果如下:

[   ["A", "A", "A"]
,   ["A", "A"]
,   ["A", "A", "A", "A"]
,   ["A"]
,   ["A", "A", "A", "A", "A", "A"]
,   ["A"]]