我正在制作一个程序，我需要在其中计算线性回归，但我卡在矩阵的反转上。
我有
double[,] location = new double[3,3];
然后它充满了数字，但后来我不知道，如何像线性代数一样计算它的逆矩阵。我在互联网上搜索了一个解决方案，但有一些 Matrix 类我不知道如何将我的double[,]转换为。
那么，你知道一些优雅的反double[,]方法吗，比如线性代数中矩阵的反演？
这里有一个工作示例，只需将整个代码复制到控制台项目中并运行它即可。我从这个链接中获取了它 https://jamesmccaffrey.wordpress.com/2015/03/06/inverting-a-matrix-using-c/
using System;
using System.Collections.Generic;
using System.Linq;
namespace matrixExample
{
class Program
{
static void Main(string[] args)
{
double[][] m = new double[][] { new double[] { 7, 2, 1 }, new double[] { 0, 3, -1 }, new double[] { -3, 4, 2 } };
double[][] inv = MatrixInverse(m);
//printing the inverse
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 3; j++)
Console.Write(Math.Round(inv[i][j], 1).ToString().PadLeft(5, ' ') + "|");
Console.WriteLine();
}
}
static double[][] MatrixCreate(int rows, int cols)
{
double[][] result = new double[rows][];
for (int i = 0; i < rows; ++i)
result[i] = new double[cols];
return result;
}
static double[][] MatrixIdentity(int n)
{
// return an n x n Identity matrix
double[][] result = MatrixCreate(n, n);
for (int i = 0; i < n; ++i)
result[i][i] = 1.0;
return result;
}
static double[][] MatrixProduct(double[][] matrixA, double[][] matrixB)
{
int aRows = matrixA.Length; int aCols = matrixA[0].Length;
int bRows = matrixB.Length; int bCols = matrixB[0].Length;
if (aCols != bRows)
throw new Exception("Non-conformable matrices in MatrixProduct");
double[][] result = MatrixCreate(aRows, bCols);
for (int i = 0; i < aRows; ++i) // each row of A
for (int j = 0; j < bCols; ++j) // each col of B
for (int k = 0; k < aCols; ++k) // could use k less-than bRows
result[i][j] += matrixA[i][k] * matrixB[k][j];
return result;
}
static double[][] MatrixInverse(double[][] matrix)
{
int n = matrix.Length;
double[][] result = MatrixDuplicate(matrix);
int[] perm;
int toggle;
double[][] lum = MatrixDecompose(matrix, out perm,
out toggle);
if (lum == null)
throw new Exception("Unable to compute inverse");
double[] b = new double[n];
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < n; ++j)
{
if (i == perm[j])
b[j] = 1.0;
else
b[j] = 0.0;
}
double[] x = HelperSolve(lum, b);
for (int j = 0; j < n; ++j)
result[j][i] = x[j];
}
return result;
}
static double[][] MatrixDuplicate(double[][] matrix)
{
// allocates/creates a duplicate of a matrix.
double[][] result = MatrixCreate(matrix.Length, matrix[0].Length);
for (int i = 0; i < matrix.Length; ++i) // copy the values
for (int j = 0; j < matrix[i].Length; ++j)
result[i][j] = matrix[i][j];
return result;
}
static double[] HelperSolve(double[][] luMatrix, double[] b)
{
// before calling this helper, permute b using the perm array
// from MatrixDecompose that generated luMatrix
int n = luMatrix.Length;
double[] x = new double[n];
b.CopyTo(x, 0);
for (int i = 1; i < n; ++i)
{
double sum = x[i];
for (int j = 0; j < i; ++j)
sum -= luMatrix[i][j] * x[j];
x[i] = sum;
}
x[n - 1] /= luMatrix[n - 1][n - 1];
for (int i = n - 2; i >= 0; --i)
{
double sum = x[i];
for (int j = i + 1; j < n; ++j)
sum -= luMatrix[i][j] * x[j];
x[i] = sum / luMatrix[i][i];
}
return x;
}
static double[][] MatrixDecompose(double[][] matrix, out int[] perm, out int toggle)
{
// Doolittle LUP decomposition with partial pivoting.
// rerturns: result is L (with 1s on diagonal) and U;
// perm holds row permutations; toggle is +1 or -1 (even or odd)
int rows = matrix.Length;
int cols = matrix[0].Length; // assume square
if (rows != cols)
throw new Exception("Attempt to decompose a non-square m");
int n = rows; // convenience
double[][] result = MatrixDuplicate(matrix);
perm = new int[n]; // set up row permutation result
for (int i = 0; i < n; ++i) { perm[i] = i; }
toggle = 1; // toggle tracks row swaps.
// +1 -greater-than even, -1 -greater-than odd. used by MatrixDeterminant
for (int j = 0; j < n - 1; ++j) // each column
{
double colMax = Math.Abs(result[j][j]); // find largest val in col
int pRow = j;
//for (int i = j + 1; i less-than n; ++i)
//{
// if (result[i][j] greater-than colMax)
// {
// colMax = result[i][j];
// pRow = i;
// }
//}
// reader Matt V needed this:
for (int i = j + 1; i < n; ++i)
{
if (Math.Abs(result[i][j]) > colMax)
{
colMax = Math.Abs(result[i][j]);
pRow = i;
}
}
// Not sure if this approach is needed always, or not.
if (pRow != j) // if largest value not on pivot, swap rows
{
double[] rowPtr = result[pRow];
result[pRow] = result[j];
result[j] = rowPtr;
int tmp = perm[pRow]; // and swap perm info
perm[pRow] = perm[j];
perm[j] = tmp;
toggle = -toggle; // adjust the row-swap toggle
}
// --------------------------------------------------
// This part added later (not in original)
// and replaces the 'return null' below.
// if there is a 0 on the diagonal, find a good row
// from i = j+1 down that doesn't have
// a 0 in column j, and swap that good row with row j
// --------------------------------------------------
if (result[j][j] == 0.0)
{
// find a good row to swap
int goodRow = -1;
for (int row = j + 1; row < n; ++row)
{
if (result[row][j] != 0.0)
goodRow = row;
}
if (goodRow == -1)
throw new Exception("Cannot use Doolittle's method");
// swap rows so 0.0 no longer on diagonal
double[] rowPtr = result[goodRow];
result[goodRow] = result[j];
result[j] = rowPtr;
int tmp = perm[goodRow]; // and swap perm info
perm[goodRow] = perm[j];
perm[j] = tmp;
toggle = -toggle; // adjust the row-swap toggle
}
// --------------------------------------------------
// if diagonal after swap is zero . .
//if (Math.Abs(result[j][j]) less-than 1.0E-20)
// return null; // consider a throw
for (int i = j + 1; i < n; ++i)
{
result[i][j] /= result[j][j];
for (int k = j + 1; k < n; ++k)
{
result[i][k] -= result[i][j] * result[j][k];
}
}
} // main j column loop
return result;
}
}
}
我想这就是你要找的：
static double[][] MatrixInverse(double[][] matrix)
{
// assumes determinant is not 0
// that is, the matrix does have an inverse
int n = matrix.Length;
double[][] result = MatrixCreate(n, n); // make a copy of matrix
for (int i = 0; i < n; ++i)
for (int j = 0; j < n; ++j)
result[i][j] = matrix[i][j];
double[][] lum; // combined lower & upper
int[] perm;
int toggle;
toggle = MatrixDecompose(matrix, out lum, out perm);
double[] b = new double[n];
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < n; ++j)
if (i == perm[j])
b[j] = 1.0;
else
b[j] = 0.0;
double[] x = Helper(lum, b); //
for (int j = 0; j < n; ++j)
result[j][i] = x[j];
}
return result;
}
有关参考，请参阅测试运行 - 使用 C# 进行矩阵反转。