从十进制数组中找到最近值的索引

Question

decimal[] array = new decimal[5]{80.23,60.20,88.01,77.00,20.45};
decimal TargetNumber = 70.40;

这里，最接近的值是77.00，如何找到最接近的十进制数组的索引?

注意:它应该保持相同的顺序，因为我需要最近值的确切索引。这里，Index比value

Answer 1

int nearestIndex = Array.IndexOf(array, array.OrderBy(number => Math.Abs(number - TargetNumber)).First());         

Answer 2

与使用LINQ相比，这样做的一个好处是，如果您找到了精确匹配，您可以提前停止检查。minIndex保存索引，如果数组为空，则将保存-1。

decimal minDistance = 0; //0 is fine here it is never read, it is just to make the compiler happy.
int minIndex = -1;
for(int i = 0; i < array.Length; i++)
{
    var distance = Math.Abs(TargetNumber - array[i]);
    if(minIndex == -1 || distance < minDistance)
    {
        minDistance = distance;
        minIndex = i;
        //Optional, stop testing if we find a exact match.
        if(minDistance == 0)
            break;
    }
}

---

只是为了好玩，我做了一个完全通用的版本，它需要你传递一个委托来计算距离因子，它也有一个可选的参数来定义"最小距离"，它需要停止检查更多的结果。

using System;
using System.Collections.Generic;
public class Program
{
    public static void Main()
    {
        decimal[] array = new decimal[5]{80.23M,80.40M,80.80M,80.00M,20.45M};
        decimal TargetNumber = 70.40M;
        var result = FindClosestIndex(TargetNumber, array, (target, element) => Math.Abs(target - element)); //Optionally add in a "(distance) => distance == 0" at the end to enable early termination.
        Console.WriteLine(result);
    }
    public static int FindClosestIndex<T,U>(T target, IEnumerable<T> elements, Func<T,T,U> distanceCalculator, Func<U, bool> earlyTermination = null) where U : IComparable<U>
    {
        U minDistance = default(U);
        int minIndex = -1;
        using(var enumerator = elements.GetEnumerator())
        for(int i = 0; enumerator.MoveNext(); i++)
        {
            var distance = distanceCalculator(enumerator.Current, target);
            if(minIndex == -1 || minDistance.CompareTo(distance) > 0)
            {
                minDistance = distance;
                minIndex = i;
            }
            if(earlyTermination != null && earlyTermination(minDistance))
                break;
        }
        return minIndex;
    }
}

可运行示例

Answer 3

您可以使用Enumerable.Aggregate:

int nearestIndex = array
    .Select((x, i) => new { Diff = Math.Abs(x - TargetNumber), Index = i })
    .Aggregate((x, y) =>  x.Diff < y.Diff ? x : y)
    .Index;

这只是不需要任何排序，它只迭代一次array。

---

编辑:这个版本如果找到一个完全匹配的提前跳出循环，就像@ScottChamberlein建议的那样:

int exactMatch = 0;
var nearestIndex = array
   .Select((x, i) => new { Diff = Math.Abs(x - TargetNumber), Index = i })
   .TakeWhile(x => x.Diff != 0M || exactMatch++ == 0)
   .Aggregate((x, y) =>  x.Diff < y.Diff ? x : y)
   .Index;

Answer 4

int index = array.Select((x,i) => new {Index=i, Distance = Math.Abs(TargetNumber - x)}).OrderBy(x => x.Distance).First().Index;