Question

我是C＃的新手，我正在努力学习这门语言。

你们能给我一个提示，我可以比较一个数组与从中选择最低值的数值吗？

喜欢：

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w的最低值，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，发现（3000）值是我的“Searchvalue”的最接近的值？

Answer 1

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];
    }
}

Answer 2

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

生成

正如预期的那样，

1000导致我们执行Enumerable.Min。

同样适用于Enumerable.Max，以确定最大值：

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

考虑到您正在与double.MinValue和double.MaxValue进行比较，我假设您只想从数组中选择最小和最大的值。

如果这不是您要搜索的内容，请澄清。

Answer 3

根据您的代码，您可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();

根据搜索值的位置，小于O（n）

Answer 4

                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;
                    }
                }

将值与数组进行比较并获得最接近的值

4 个答案:

LINQ

手动