我使用Lutz Roeder的Mapack矩陣庫在C#中重新實現了一個。也許這個,或者Java版本,對你有用。
/// <summary>
/// The difference between 1 and the smallest exactly representable number
/// greater than one. Gives an upper bound on the relative error due to
/// rounding of floating point numbers.
/// </summary>
const double MACHEPS = 2E-16;
// NOTE: Code for pseudoinverse is from:
// http://the-lost-beauty.blogspot.com/2009/04/moore-penrose-pseudoinverse-in-jama.html
/// <summary>
/// Computes the Moore–Penrose pseudoinverse using the SVD method.
/// Modified version of the original implementation by Kim van der Linde.
/// </summary>
/// <param name="x"></param>
/// <returns>The pseudoinverse.</returns>
public static Matrix MoorePenrosePsuedoinverse(Matrix x)
{
if (x.Columns > x.Rows)
return MoorePenrosePsuedoinverse(x.Transpose()).Transpose();
SingularValueDecomposition svdX = new SingularValueDecomposition(x);
if (svdX.Rank < 1)
return null;
double[] singularValues = svdX.Diagonal;
double tol = Math.Max(x.Columns, x.Rows) * singularValues[0] * MACHEPS;
double[] singularValueReciprocals = new double[singularValues.Length];
for (int i = 0; i < singularValues.Length; ++i)
singularValueReciprocals[i] = Math.Abs(singularValues[i]) < tol ? 0 : (1.0/singularValues[i]);
Matrix u = svdX.GetU();
Matrix v = svdX.GetV();
int min = Math.Min(x.Columns, u.Columns);
Matrix inverse = new Matrix(x.Columns, x.Rows);
for (int i = 0; i < x.Columns; i++)
for (int j = 0; j < u.Rows; j++)
for (int k = 0; k < min; k++)
inverse[i, j] += v[i, k] * singularValueReciprocals[k] * u[j, k];
return inverse;
}
你可能想利用這個問題要麼http://math.stackexchange.com/或http://dsp.stackexchange.com/ – Ali