Question

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

decimal TargetNumber = 70.40;

这里，最近的值是77.00，如何找到最接近的十进制数组的索引？

注意：它应该保持与我需要的最接近值的精确索引相同的顺序。在这里，指数比价值重要

Answer 1

使用LINQ很长一段时间，如果找到完全匹配，你可以提前停止检查。 minIndex保留索引，如果数组为空，则保留-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;
    }
}

为了好玩，我制作了一个完全通用的版本，它要求你传递一个委托来计算距离因子，它还有一个可选参数来定义＆＃34;最小距离＆＃34;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;
    }
}

Runnable example

Answer 2

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

Answer 3

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

Answer 4

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，而且只能在array内迭代一次。

编辑：如果找到完全匹配的话，这个版本就会突破早期的，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;

C＃ - 从十进制数组中查找最接近的索引

4 个答案: