本文整理汇总了C#中System.Matrix.GetArrayCopy方法的典型用法代码示例。如果您正苦于以下问题：C# Matrix.GetArrayCopy方法的具体用法？C# Matrix.GetArrayCopy怎么用？C# Matrix.GetArrayCopy使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类System.Matrix的用法示例。
在下文中一共展示了Matrix.GetArrayCopy方法的5个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C#代码示例。
示例1: QRDecomposition
/// <summary>
/// QR Decomposition, computed by Householder reflections.
/// </summary>
/// <param name="A">Structure to access R and the Householder vectors and compute Q.</param>
public QRDecomposition(Matrix A)
{
// Initialize.
QR = A.GetArrayCopy();
m = A.Rows;
n = A.Cols;
Rdiag = new double[n];
// Main loop.
for (int k = 0; k < n; k++)
{
// Compute 2-norm of k-th column without under/overflow.
double nrm = 0;
for (int i = k; i < m; i++)
{
nrm = EncogMath.Hypot(nrm, QR[i][k]);
}
if (nrm != 0.0)
{
// Form k-th Householder vector.
if (QR[k][k] < 0)
{
nrm = -nrm;
}
for (int i = k; i < m; i++)
{
QR[i][k] /= nrm;
}
QR[k][k] += 1.0;
// Apply transformation to remaining columns.
for (int j = k + 1; j < n; j++)
{
double s = 0.0;
for (int i = k; i < m; i++)
{
s += QR[i][k] * QR[i][j];
}
s = -s / QR[k][k];
for (int i = k; i < m; i++)
{
QR[i][j] += s * QR[i][k];
}
}
}
Rdiag[k] = -nrm;
}
}示例2: Solve
/// <summary>
/// Least squares solution of A*X = B
/// </summary>
/// <param name="B">A Matrix with as many rows as A and any number of columns.</param>
/// <returns>that minimizes the two norm of Q*R*X-B.</returns>
public Matrix Solve(Matrix B)
{
if (B.Rows != m)
{
throw new MatrixError(
"Matrix row dimensions must agree.");
}
if (!this.IsFullRank() )
{
throw new MatrixError("Matrix is rank deficient.");
}
// Copy right hand side
int nx = B.Cols;
double[][] X = B.GetArrayCopy();
// Compute Y = transpose(Q)*B
for (int k = 0; k < n; k++)
{
for (int j = 0; j < nx; j++)
{
double s = 0.0;
for (int i = k; i < m; i++)
{
s += QR[i][k] * X[i][j];
}
s = -s / QR[k][k];
for (int i = k; i < m; i++)
{
X[i][j] += s * QR[i][k];
}
}
}
// Solve R*X = Y;
for (int k = n - 1; k >= 0; k--)
{
for (int j = 0; j < nx; j++)
{
X[k][j] /= Rdiag[k];
}
for (int i = 0; i < k; i++)
{
for (int j = 0; j < nx; j++)
{
X[i][j] -= X[k][j] * QR[i][k];
}
}
}
return (new Matrix(X).GetMatrix(0, n - 1, 0, nx - 1));
}示例3: SingularValueDecomposition
/// <summary>
/// Construct the singular value decomposition
/// </summary>
/// <param name="Arg">Rectangular matrix</param>
public SingularValueDecomposition(Matrix Arg)
{
// Derived from LINPACK code.
// Initialize.
double[][] A = Arg.GetArrayCopy();
m = Arg.Rows;
n = Arg.Cols;
/*
* Apparently the failing cases are only a proper subset of (m<n), so
* let's not throw error. Correct fix to come later? if (m<n) { throw
* new IllegalArgumentException("Jama SVD only works for m >= n"); }
*/
int nu = Math.Min(m, n);
s = new double[Math.Min(m + 1, n)];
umatrix = EngineArray.AllocateDouble2D(m, nu);
vmatrix = EngineArray.AllocateDouble2D(n, n);
var e = new double[n];
var work = new double[m];
bool wantu = true;
bool wantv = true;
// Reduce A to bidiagonal form, storing the diagonal elements
// in s and the super-diagonal elements in e.
int nct = Math.Min(m - 1, n);
int nrt = Math.Max(0, Math.Min(n - 2, m));
for (int k = 0; k < Math.Max(nct, nrt); k++)
{
if (k < nct)
{
// Compute the transformation for the k-th column and
// place the k-th diagonal in s[k].
// Compute 2-norm of k-th column without under/overflow.
s[k] = 0;
for (int i = k; i < m; i++)
{
s[k] = EncogMath.Hypot(s[k], A[i][k]);
}
if (s[k] != 0.0)
{
if (A[k][k] < 0.0)
{
s[k] = -s[k];
}
for (int i = k; i < m; i++)
{
A[i][k] /= s[k];
}
A[k][k] += 1.0;
}
s[k] = -s[k];
}
for (int j = k + 1; j < n; j++)
{
if ((k < nct) & (s[k] != 0.0))
{
// Apply the transformation.
double t = 0;
for (int i = k; i < m; i++)
{
t += A[i][k]*A[i][j];
}
t = -t/A[k][k];
for (int i = k; i < m; i++)
{
A[i][j] += t*A[i][k];
}
}
// Place the k-th row of A into e for the
// subsequent calculation of the row transformation.
e[j] = A[k][j];
}
if (wantu & (k < nct))
{
// Place the transformation in U for subsequent back
// multiplication.
for (int i = k; i < m; i++)
{
umatrix[i][k] = A[i][k];
}
}
if (k < nrt)
{
// Compute the k-th row transformation and place the
// k-th super-diagonal in e[k].
// Compute 2-norm without under/overflow.
e[k] = 0;
for (int i = k + 1; i < n; i++)
{
e[k] = EncogMath.Hypot(e[k], e[i]);
}
//.........这里部分代码省略.........
示例4: LUDecomposition
/// <summary>
/// LU Decomposition
/// </summary>
/// <param name="A">Rectangular matrix</param>
public LUDecomposition(Matrix A)
{
// Use a "left-looking", dot-product, Crout/Doolittle algorithm.
LU = A.GetArrayCopy();
m = A.Rows;
n = A.Cols;
piv = new int[m];
for (int i = 0; i < m; i++)
{
piv[i] = i;
}
pivsign = 1;
double[] LUrowi;
double[] LUcolj = new double[m];
// Outer loop.
for (int j = 0; j < n; j++)
{
// Make a copy of the j-th column to localize references.
for (int i = 0; i < m; i++)
{
LUcolj[i] = LU[i][j];
}
// Apply previous transformations.
for (int i = 0; i < m; i++)
{
LUrowi = LU[i];
// Most of the time is spent in the following dot product.
int kmax = Math.Min(i, j);
double s = 0.0;
for (int k = 0; k < kmax; k++)
{
s += LUrowi[k] * LUcolj[k];
}
LUrowi[j] = LUcolj[i] -= s;
}
// Find pivot and exchange if necessary.
int p = j;
for (int i = j + 1; i < m; i++)
{
if (Math.Abs(LUcolj[i]) > Math.Abs(LUcolj[p]))
{
p = i;
}
}
if (p != j)
{
for (int k = 0; k < n; k++)
{
double t = LU[p][k];
LU[p][k] = LU[j][k];
LU[j][k] = t;
}
int temp = piv[p];
piv[p] = piv[j];
piv[j] = temp;
pivsign = -pivsign;
}
// Compute multipliers.
if (j < m & LU[j][j] != 0.0)
{
for (int i = j + 1; i < m; i++)
{
LU[i][j] /= LU[j][j];
}
}
}
}示例5: Solve
/// <summary>
/// Solve A*X = B.
/// </summary>
/// <param name="b">A Matrix with as many rows as A and any number of columns.</param>
/// <returns>X so that L*L'*X = b.</returns>
public Matrix Solve(Matrix b)
{
if (b.Rows != n)
{
throw new MatrixError(
"Matrix row dimensions must agree.");
}
if (!isspd)
{
throw new MatrixError(
"Matrix is not symmetric positive definite.");
}
// Copy right hand side.
double[][] x = b.GetArrayCopy();
int nx = b.Cols;
// Solve L*Y = B;
for (int k = 0; k < n; k++)
{
for (int j = 0; j < nx; j++)
{
for (int i = 0; i < k; i++)
{
x[k][j] -= x[i][j] * l[k][i];
}
x[k][j] /= l[k][k];
}
}
// Solve L'*X = Y;
for (int k = n - 1; k >= 0; k--)
{
for (int j = 0; j < nx; j++)
{
for (int i = k + 1; i < n; i++)
{
x[k][j] -= x[i][j] * l[i][k];
}
x[k][j] /= l[k][k];
}
}
return new Matrix(x);
}