Question

需要一些帮助。我有一些颜色的反射值，需要将它们转换为L * a * b * C * h值。

我找到了一些声称可以做到这一点的视觉基本来源，但我不知道将它转换成C sharp的visual basic。

任何人都可以提供帮助吗？

http://www.vbforums.com/showthread.php?p=821074

示例数据反射值：0.35380,0.44130,0.50230,0.51650,0.52210,0.52780,0.53110,0.53350,0.53630,0.53900,0.54130,0.54330,0.54500,0.54630,0.54690,0.54680,0.54640,0.54710,0.54940,0.55240,0.55330,0.55250,0.55200,0.55190 ，0.55220,0.55310,0.55260,0.55340,0.55500,0.555540,0.55400

Answer 1

我设法回答了我自己的问题！

double[] d65Vals_X = new double[31] { 0.137, 0.676, 1.603, 2.451, 3.418, 3.699, 3.064, 1.933, 0.802, 0.156, 0.039, 0.347, 1.070, 2.170, 3.397, 4.732, 6.070, 7.311, 8.291, 8.634, 8.672, 7.930, 6.446, 4.669, 3.095, 1.859, 1.056, 0.570, 0.274, .0121, 0.058 };
        double[] d65Vals_Y = new double[31] { 0.014, 0.069, 0.168, 0.300, 0.554, 0.890, 1.290, 1.838, 2.520, 3.226, 4.320, 5.621, 6.907, 8.059, 8.668, 8.855, 8.581, 7.951, 7.106, 6.004, 5.079, 4.065, 2.999, 2.042, 1.290, 0.746, 0.417, 0.223, 0.107, 0.047, 0.023 };
        double[] d65Vals_Z = new double[31] { 0.612, 3.110, 7.627, 12.095, 17.537, 19.888, 17.695, 13.000, 7.699, 3.938, 2.046, 1.049, 0.544, 0.278, 0.122, 0.035, 0.001, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000 };

        private double FXYZ(double[] REF, int A){
            double XYZref = 0;
            double illXYZ = 0;
            int i;

            for (i = 0; i < 31; i++)
            {
                if (A == 0){
                    XYZref = XYZref + (REF[i] * d65Vals_X[i]);
                    illXYZ = illXYZ + d65Vals_X[i];
                } else if (A == 1){
                    XYZref = XYZref + (REF[i] * d65Vals_Y[i]);
                    illXYZ = illXYZ + d65Vals_Y[i];
                } else {
                    XYZref = XYZref + (REF[i] * d65Vals_Z[i]);
                    illXYZ = illXYZ + d65Vals_Z[i];
                }                
            }
            return (XYZref / (100 * illXYZ));
        }
        private void btnCalculate_Click(object sender, EventArgs e)
        {
            int i;

            double L, A, B;

            double[] reflexVals = new double[31];
            Console.Write("Reflex Data: ");
            for (i = 0; i < 31; i++)
            {
                TextBox reflex = this.Controls["txtReflex_" + (i + 1)] as TextBox;                
                reflexVals[i] = double.Parse(reflex.Text);
                Console.Write(reflexVals[i].ToString() + ", ");
            }
            Console.Write("\n");

            double[] FFXYZ = new double[3];
            double[] XYZ = new double[3];
            double CR = (1 / 3);
            CR = 0.33333333333333333333333333333333;
            for (i = 0; i < 3; i++)
            {
                XYZ[i] = FXYZ(reflexVals, i);
                Console.WriteLine("XYZ[" + i + "] = " + XYZ[i]);
                if (XYZ[i] > 0.008856)
                {

                    FFXYZ[i] = Math.Pow(XYZ[i], CR);
                    Console.WriteLine("FFXYZ[" + i + "] = " + Math.Pow(XYZ[i], CR));
                }
                else
                {
                    FFXYZ[i] = (7.787 * XYZ[i]) + (16 / 116);
                    Console.WriteLine("FFXYZ[" + i + "] = (7.787 * " + XYZ[i] + ") + (16 / 116)");
                }
            }

            if (XYZ[1] > 0.00856)
            {
                L = (116 * FFXYZ[1]) - 16;
                Console.WriteLine("L = (116 * " + FFXYZ[1] + ") - 16");
            }
            else
            {
                L = 903.3 * XYZ[1];
                Console.WriteLine("L = 903.3 * " + XYZ[1] + ")");
            }

            A = 500 * (FFXYZ[0] - FFXYZ[1]);
            B = 200 * (FFXYZ[1] - FFXYZ[2]);

            Console.WriteLine("A = 500 * (" + FFXYZ[0] + " - " + FFXYZ[1] + ")");
            Console.WriteLine("A = 200 * (" + FFXYZ[1] + " - " + FFXYZ[2] + ")");

            txtCalcL.Text = L.ToString();
            txtCalcA.Text = A.ToString();
            txtCalcB.Text = B.ToString();
        }

将颜色反射率数据转换为L A B C和H值

1 个答案: