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C# Matrix.getRowDimension方法代码示例

本文整理汇总了C#中System.Matrix.getRowDimension方法的典型用法代码示例。如果您正苦于以下问题:C# Matrix.getRowDimension方法的具体用法?C# Matrix.getRowDimension怎么用?C# Matrix.getRowDimension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System.Matrix的用法示例。


在下文中一共展示了Matrix.getRowDimension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: CholeskyDecomposition

 /* ------------------------
    Constructor
  * ------------------------ */
 /** Cholesky algorithm for symmetric and positive definite matrix.
 @param  A   Square, symmetric matrix.
 @return     Structure to access L and isspd flag.
 */
 public CholeskyDecomposition(Matrix Arg)
 {
     // Initialize.
       double[][] A = Arg.getArray();
       n = Arg.getRowDimension();
       L = new double[n][];
       for(int i = 0; i < n; i++)
       {
       L[n] = new double[n];
       }
       isspd = (Arg.getColumnDimension() == n);
       // Main loop.
       for (int j = 0; j < n; j++) {
      double[] Lrowj = L[j];
      double d = 0.0;
      for (int k = 0; k < j; k++) {
     double[] Lrowk = L[k];
     double s = 0.0;
     for (int i = 0; i < k; i++) {
        s += Lrowk[i]*Lrowj[i];
     }
     Lrowj[k] = s = (A[j][k] - s)/L[k][k];
     d = d + s*s;
     isspd = isspd & (A[k][j] == A[j][k]);
      }
      d = A[j][j] - d;
      isspd = isspd & (d > 0.0);
      L[j][j] = Math.Sqrt(Math.Max(d,0.0));
      for (int k = j+1; k < n; k++) {
     L[j][k] = 0.0;
      }
       }
 }
开发者ID:Edgarwma,项目名称:Reservoir,代码行数:40,代码来源:CholeskyDecomposition.cs

示例2: SingularValueDecomposition

        /* ------------------------
           Constructor
         * ------------------------ */
        /** Construct the singular value decomposition
        @param A    Rectangular matrix
        @return     Structure to access U, S and V.
        */
        public SingularValueDecomposition(Matrix Arg)
        {
            // Derived from LINPACK code.
              // Initialize.
              double[][] A = Arg.getArrayCopy();
              m = Arg.getRowDimension();
              n = Arg.getColumnDimension();

              /* 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)];
              U = new double [m][];
              for (int i = 0; i < m; i++)
              {
              U[i] = new double[nu];
              }

              V = new double [n][];
              for(int i = 0; i < n; i++)
              {
              V[i] = new double[n];
              }

              double[] e = new double [n];
              double[] 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] = Maths.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++) {
               U[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;
//.........这里部分代码省略.........
开发者ID:Edgarwma,项目名称:Reservoir,代码行数:101,代码来源:SingularValueDecomposition.cs

示例3: LUDecomposition

        /* ------------------------
           Constructor
         * ------------------------ */
        /** LU Decomposition
        @param  A   Rectangular matrix
        @return     Structure to access L, U and piv.
        */
        public LUDecomposition(Matrix A)
        {
            // Use a "left-looking", dot-product, Crout/Doolittle algorithm.

            LU = A.getArrayCopy();
            m = A.getRowDimension();
            n = A.getColumnDimension();
            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 z = piv[p];
                    piv[p] = piv[j];
                    piv[j] = z;
                    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];
                    }
                }
            }
        }
开发者ID:Edgarwma,项目名称:Reservoir,代码行数:86,代码来源:LUDecomposition.cs

示例4: solve

        /** Solve A*X = B
        @param  B   A Matrix with as many rows as A and any number of columns.
        @return     X so that L*L'*X = B
        @exception  IllegalArgumentException  Matrix row dimensions must agree.
        @exception  RuntimeException  Matrix is not symmetric positive definite.
        */
        public Matrix solve(Matrix B)
        {
            if (B.getRowDimension() != n)
            {
                throw new IllegalArgumentException("Matrix row dimensions must agree.");
            }
            if (!isspd)
            {
                throw new RuntimeException("Matrix is not symmetric positive definite.");
            }

            // Copy right hand side.
            double[][] X = B.getArrayCopy();
            int nx = B.getColumnDimension();

            // 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, n, nx);
        }
开发者ID:Edgarwma,项目名称:Reservoir,代码行数:49,代码来源:CholeskyDecomposition.cs

示例5: B

        /** Solve A*X = B
        @param  B   A Matrix with as many rows as A and any number of columns.
        @return     X so that L*U*X = B(piv,:)
        @exception  IllegalArgumentException Matrix row dimensions must agree.
        @exception  RuntimeException  Matrix is singular.
        */
        public Matrix solve(Matrix B)
        {
            if (B.getRowDimension() != m)
            {
                throw new IllegalArgumentException("Matrix row dimensions must agree.");
            }
            //EDGAR ver pra q serve
            //if (!this.isNonsingular())
            //{
            //    throw new RuntimeException("Matrix is singular.");
            //}

            // Copy right hand side with pivoting
            int nx = B.getColumnDimension();
            Matrix Xmat = B.getMatrix(piv, 0, nx - 1);
            double[][] X = Xmat.getArray();

            // Solve L*Y = B(piv,:)
            for (int k = 0; k < n; k++)
            {
                for (int i = k + 1; i < n; i++)
                {
                    for (int j = 0; j < nx; j++)
                    {
                        X[i][j] -= X[k][j] * LU[i][k];
                    }
                }
            }
            // Solve U*X = Y;
            for (int k = n - 1; k >= 0; k--)
            {
                for (int j = 0; j < nx; j++)
                {
                    X[k][j] /= LU[k][k];
                }
                for (int i = 0; i < k; i++)
                {
                    for (int j = 0; j < nx; j++)
                    {
                        X[i][j] -= X[k][j] * LU[i][k];
                    }
                }
            }
            return Xmat;
        }
开发者ID:Edgarwma,项目名称:Reservoir,代码行数:51,代码来源:LUDecomposition.cs


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