c# - 性能数组乘法 Pearson

Question

我计算 Pearson correlation (平均用户/项目评级)很多次，使用我当前的代码性能非常糟糕:

public double ComputeCorrelation(double[] x, double[] y, double[] meanX, double[] meanY)
        {
            if (x.Length != y.Length)
                throw new ArgumentException("values must be the same length");

            double sumNum = 0;
            double sumDenom = 0;
            double denomX = 0;
            double denomY = 0;

            for (int a = 0; a < x.Length; a++)
            {
                sumNum += (x[a] - meanX[a]) * (y[a] - meanY[a]);
                denomX += Math.Pow(x[a] - meanX[a], 2);
                denomY += Math.Pow(y[a] - meanY[a], 2);
            }

            var sqrtDenomX = Math.Sqrt(denomX);
            var sqrtDenomY = Math.Sqrt(denomY);

            if (sqrtDenomX == 0 || sqrtDenomY == 0) return 0;

            sumDenom = Math.Sqrt(denomX) * Math.Sqrt(denomY);

            var correlation = sumNum / sumDenom;

            return correlation;
        }

我将标准 Pearson 相关系数与 MathNet.Numerics 一起使用，但这是对标准的修改，无法使用它。有没有办法加快速度？如何在时间复杂度方面对其进行优化？

Answer 1

添加一些关于 MSE 的答案——将 Pow(x,2) 更改为 diff*diff 绝对是您想要做的事情，您可能还想避免不必要的在最内层循环中进行边界检查。这可以使用 pointers in C# 来完成.

可以这样做:

    public unsafe double ComputeCorrelation(double[] x, double[] y, double[] meanX, double[] meanY)
    {
        if (x.Length != y.Length)
            throw new ArgumentException("values must be the same length");

        double sumNum = 0;
        double sumDenom = 0;
        double denomX = 0;
        double denomY = 0;
        double diffX;
        double diffY;

        int len = x.Length;

        fixed (double* xptr = &x[0], yptr = &y[0], meanXptr = &meanX[0], meanYptr = &meanY[0])
        {
            for (int a = 0; a < len; a++)
            {
                diffX = (xptr[a] - meanXptr[a]);
                diffY = (yptr[a] - meanYptr[a]);
                sumNum += diffX * diffY;
                denomX += diffX * diffX;
                denomY += diffY * diffY;
            }
        }

        var sqrtDenomX = Math.Sqrt(denomX);
        var sqrtDenomY = Math.Sqrt(denomY);

        if (sqrtDenomX == 0 || sqrtDenomY == 0) return 0;

        sumDenom = sqrtDenomX * sqrtDenomY;

        var correlation = sumNum / sumDenom;

        return correlation;
    }