最快的 Math.Max(value,0)

Question

我有一个算法可以将浮点数/双精度数与 0 进行大量比较。

类似的东西：

var sum = 0.0;
for(int i=0;i<data.Length;i++)
  sum += Math.Max(data[i],0);

我想知道是否有更快的方法来做到这一点？ 在我的微基准测试中，它的执行速度几乎没有提高 (5-7%)。

public static unsafe float FasterCompare(float value)
{
  var val = *(int*) &value;
  return val > 0 ? value : 0;
}

Answer 1

你当前的代码是这样的：

var sum = 0.0;
for(int i=0;i<data.Length;i++)
  sum += Math.Max(data[i],0);

Math.Max 正在这样做：

public static double Max(double val1, double val2) 
{
    if (val1 > val2)
        return val1;

    if (Double.IsNaN(val1))
        return val1;

    return val2;
}

如果您知道自己不会拥有 NaN，请在没有 NaN 检查的情况下实现自己的 Max：

public static double Max(double val1, double val2) 
{
    if (val1 > val2)
        return val1;

    return val2;
}

Answer 2

写一个小基准，这是我的结果：

你的版本：~2.3ms

优化版：~1.4ms

代码如下：

float[] data = new float[150000];
Random rnd = new Random(12345);

for (int i = 0; i < data.Length; i++)
{
    data[i] = (float)(rnd.NextDouble() * 5000.0 - 2500.0);
}

Stopwatch sw = new Stopwatch();

sw.Start();

var varsum = 0.0;  //varsum is a DOUBLE!!!!

for (int i = 0; i < data.Length; i++)
    varsum += Math.Max(data[i], 0);        //implicit conversions, float->double, int->float

sw.Stop();

Console.WriteLine("Varsum : " + varsum);

Console.WriteLine("Time it took for the original: " + sw.Elapsed.TotalMilliseconds + " ms");
   
float floatsum = 0.0f;
    
sw.Reset();
sw.Start();

floatsum = 0.0f;

for (int i = 0; i < data.Length; i++)
    if (data[i] > 0.0f)
        floatsum += data[i];

sw.Stop();

Console.WriteLine("OptimizedSum: " + floatsum);

Console.WriteLine("Time it took for \"optimized\" version: " + sw.Elapsed.TotalMilliseconds + " ms");

//Equality on floating point numbers doesn't work like this, but...
Console.WriteLine("Are these two equal? " + (floatsum == varsum).ToString());
Console.WriteLine("How close are they? " + Math.Abs(floatsum - varsum).ToString("00.0000000000000000"));

Console.ReadKey(true);

控制台应用程序，32 位进程，运行调试模式，为 AnyCPU 编译。第二个版本快 60% 左右。这可能是因为不必调用无法内联的方法。

还需要注意的是，这两个方法返回的值不同。这是因为 double 和 float 的精度不同。如果您正在使用浮点数，请继续使用浮点数。与双精度相同，不要来回转换。上面的例子给了我超过65的差异！

小提琴：https://dotnetfiddle.net/S5qmCg（小提琴版本不反映上述相同的时间值，我假设因为它的资源在服务器端有限，所以请在您自己的计算机上尝试）

Answer 3

取决于您的阵列（和您的机器）有多大。根据数组的大小，这在我的机器上要快得多：

using System;
using System.Diagnostics;
using System.Linq;

public class Program
{
    public static void Main()
    {
        float[] data = new float[1500000];
        Random rnd = new Random(12345);

        for (int i = 0; i < data.Length; i++)
        {
            data[i] = (float)(rnd.NextDouble() * 5000.0 - 2500.0);
        }

        Stopwatch sw = new Stopwatch();

        sw.Start();

        var varsum = 0.0f;  //varsum is a DOUBLE!!!! change this to 0.0f to make them equal!

        for (int i = 0; i < data.Length; i++)
            varsum += Math.Max(data[i], 0);        //implicit conversions, float->double, int->float

        sw.Stop();

        Console.WriteLine("Varsum : " + varsum);

        Console.WriteLine("Time it took for the original: " + sw.Elapsed.TotalMilliseconds + " ms");

        float floatsum = 0.0f;

        sw.Reset();
        sw.Start();

        floatsum = 0.0f;
        floatsum=data.AsParallel().Where(d=>d>0).Sum();

        sw.Stop();

        Console.WriteLine("OptimizedSum: " + floatsum);

        Console.WriteLine("Time it took for \"optimized\" version: " + sw.Elapsed.TotalMilliseconds + " ms");

        //Equality on floating point numbers doesn't work like this, but...
        Console.WriteLine("Are these two equal? " + (floatsum == varsum).ToString());
        Console.WriteLine("How close are they? " + Math.Abs(floatsum - varsum).ToString("00.0000000000000000"));

    }
}