基于下一个对象筛选数组

Question

我正在写一些代码来过滤来自运动传感器的驾驶行程。我发现最好的方法是基于以下内容将子数组添加到嵌套数组中：

• 检测首次发生的自信汽车事件
• 将以下所有运动事件添加到相同的事件数组中，直到第一个确信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。有什么方法可以使用Swift的Array.filter{}使这种排序更漂亮吗？

Answer 1

Filter不会这么做，但您可以使用reduce。

下面的示例显示了如何将连续"A"（表示汽车事件）的运行收集到阵列内部的阵列中：

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)

先前的值作为第一个参数传递给reduce的函数。这使得函数可以决定是附加到当前列表还是启动新列表。

这次运行的结果如下：

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

Answer 2

如果您想要一个漂亮的解决方案，您可以使用split来实现这一点。你只需要为它提供一个应该被视为分隔符的条件。在你的情况下，这将是任何非汽车运动事件。

let arr = ["A","A","A","B","A","A","B","A","A","C","A","B","D","B","A","B","A","E","A","A","F","A","B","A","B"]
let split = arr.split {$0 != "A"} // insert your condition for whether the given element should be considered a 'seperator'

这里的$0是数组中某个元素的匿名闭包参数（当它遍历时）。为了使命名更加明确，您可以始终扩展闭包，尽管它看起来不那么优雅。例如：

let split = arr.split {element in
    return element != "A"
}

这将返回一个ArraySlices数组，如下所示：

[
    ArraySlice(["A", "A", "A"]),
    ArraySlice(["A", "A"]),
    ArraySlice(["A", "A"]),
    ArraySlice(["A"]),
    ArraySlice(["A"]),
    ArraySlice(["A"]),
    ArraySlice(["A", "A"]),
    ArraySlice(["A"]),
    ArraySlice(["A"])
]

如果您希望它们是显式Arrays，您可以在之后简单地使用map：

let split = arr.split {$0 != "A"}.map{Array($0)}

退货：

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