3

我是 C# 的新手，我正在努力学习这种语言。

你们能给我一个提示，我如何将数组与从中选择最低值的值进行比较？

像：

Double[] w = { 1000, 2000, 3000, 4000, 5000 };

double min = double.MaxValue;
double max = double.MinValue;

foreach (double value in w)
{
    if (value < min)
        min = value;
    if (value > max)
        max = value;
}

Console.WriteLine(" min:", min);

给我 的最低值w，我现在该如何比较？

如果我有：

int p = 1001 + 2000;  // 3001

我现在如何与数组列表进行比较并找出 (3000) 值是最接近我的“搜索值”的值？

4

4 回答 4

13

你可以用一些简单的数学来做到这一点，并且有不同的方法。

LINQ

Double searchValue = ...;

Double nearest = w.Select(p => new { Value = p, Difference = Math.Abs(p - searchValue) })
                  .OrderBy(p => p.Difference)
                  .First().Value;

手动

Double[] w = { 1000, 2000, 3000, 4000, 5000 };

Double searchValue = 3001;
Double currentNearest = w[0];
Double currentDifference = Math.Abs(currentNearest - searchValue);

for (int i = 1; i < w.Length; i++)
{
    Double diff = Math.Abs(w[i] - searchValue);
    if (diff < currentDifference)
    {
        currentDifference = diff;
        currentNearest = w[i];
    }
}
于 2012-06-05T09:20:40.650 回答
1 
Double[] w = { 1000, 2000, 3000, 4000, 5000 };
var minimumValueFromArray = w.Min();

生产

1000，正如预期的那样，因为我们执行Enumerable.Min。

Enumerable.Max也是如此，计算最大值：

Double[] w = { 1000, 2000, 3000, 4000, 5000 };
var maximumValueFromArray = w.Max();

考虑到您正在与double.MinValueand进行比较double.MaxValue，我假设您只想从数组中选择最小值和最大值。

如果这不是您要搜索的内容，请澄清。

于 2012-06-05T09:15:54.383 回答
0

根据您的代码，您可以以非常简单的方式实现这一目标

    Double[] w = { 1000, 2000, 3000, 4000, 5000 }; // it has to be sorted

    double search = 3001;
    double lowerClosest = 0;
    double upperClosest = 0;


        for (int i = 1; i < w.Length; i++)
        {
            if (w[i] > search)
            {
                upperClosest = w[i];
                break; // interrupts your foreach
            }

        }
        for (int i = w.Length-1; i >=0; i--)
        {
            if (w[i] <= search)
            {
                lowerClosest = w[i];
                break; // interrupts your foreach
            }

        }

    Console.WriteLine(" lowerClosest:{0}", lowerClosest);
    Console.WriteLine(" upperClosest:{0}", upperClosest);
    if (upperClosest - search > search - lowerClosest)
        Console.WriteLine(" Closest:{0}", lowerClosest);
    else
        Console.WriteLine(" Closest:{0}", upperClosest);

    Console.ReadLine();

根据您的搜索值的位置，这将小于 O(n)

于 2013-07-22T08:40:26.547 回答
-2 
                Performance wise custom code will be more use full. 

                List<int> results;
                int targetNumber = 0;
                int nearestValue=0;
                if (results.Any(ab => ab == targetNumber ))
                {
                    nearestValue= results.FirstOrDefault<int>(i => i == targetNumber );
                }
                else
                {
                    int greaterThanTarget = 0;
                    int lessThanTarget = 0;
                    if (results.Any(ab => ab > targetNumber ))
                    {
                        greaterThanTarget = results.Where<int>(i => i > targetNumber ).Min();
                    }
                    if (results.Any(ab => ab < targetNumber ))
                    {
                        lessThanTarget = results.Where<int>(i => i < targetNumber ).Max();
                    }

                    if (lessThanTarget == 0 )
                    {
                        nearestValue= greaterThanTarget;
                    }
                    else if (greaterThanTarget == 0)
                    {
                        nearestValue= lessThanTarget;
                    }
                    else if (targetNumber - lessThanTarget < greaterThanTarget - targetNumber )
                    {
                        nearestValue= lessThanTarget;
                    }
                    else
                    {
                            nearestValue= greaterThanTarget;
                    }
                }
于 2014-01-07T08:06:23.967 回答