Swift-按值对Array的元素进行分组（2 x 2、3 x 3等） [英] Swift - Group elements of Array by value (2 by 2, 3 by 3, etc...)

问题描述

编辑：我不是在要求函数对事件进行计数。我要一个函数来统计发生次数 2乘2 ， 3乘3 ， 10乘10 等，这是我的问题

I'm not asking a function to count an occurence. I'm asking a function to count an occurence 2 by 2, 3 by 3, 10 by 10, etc... this is my problem

我有一系列分数，可以说：

I have an array of scores, lets say:

[2,2,3,4,4,4,4,5,6,6,8,8,8,9,10,10]

我希望有一个函数可以将此 Array 转换为 Dictionary [Int：Int] （）具有类似的内容：

I would like to have a function that transform this Array into a Dictionary [Int: Int]() to have something like that:

func groupArrayBy(array: Array<Int>, coef: Int) -> Array<Int, Int>{
   // if coef = 2 -> 2 by 2, count occurence
   // Transform the array to:
   // [2: 3, 4: 5, 6: 2, 8: 4, 10: 2]
   // return Dictionary
}

（当coef = 3时，将是： [2：7，5：3，8：6] -> 3 3）

(With coef = 3, it would be: [2: 7, 5: 3, 8: 6] -> 3 by 3)

我对此一无所获。可能吗？

I found nothing about that. Is it even possible ?

推荐答案

这是我的版本，其中我根据数组的第一个值和过滤范围coef 变量，根据结果，我将那些已经计数的元素切掉，并在循环中再次过滤较小的数组。此解决方案要求输入数组以升序排序

Here is my version where I filter on a range based on first value of the array and the coef variable, based on the result I slice away those elements already counted and filter again on the smaller array in a loop. This solution requires the input array to be sorted in ascending order

func group(_ array: [Int], coef: Int) -> [Int: Int] {
    var result:[Int:Int] = [:]

    var start = array[0]
    var end = start + coef - 1
    var arr  = array

    while start <= array[array.count - 1] {
       let count = arr.filter({ $0 >= start && $0 <= end}).count

       result[start] = count
       start = end + 1
       end = start + coef - 1
       arr = Array(arr[count...])
    }
    return result
}

这是上述函数的递归版本

And here is a recursive version of the above function

func group(_ array: [Int], coef: Int) -> [Int: Int] {
    var result:[Int:Int] = [:]
    if array.isEmpty { return result }

    let end = array[0] + coef - 1
    let count = array.filter({ $0 >= array[0] && $0 <= end}).count
    result[array[0]] = count
    result = result.merging(group(Array(array[count...]), coef: coef)) { $1 }
    return result
}

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