本文整理汇总了C#中VectorD.Clone方法的典型用法代码示例。如果您正苦于以下问题：C# VectorD.Clone方法的具体用法？C# VectorD.Clone怎么用？C# VectorD.Clone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VectorD的用法示例。
在下文中一共展示了VectorD.Clone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: SolveByGEPP
/// <summary>
/// 部分ピボット選択付きガウス消去法により連立一次方程式の解を求める
/// </summary>
/// <param name="a"></param>
/// <param name="b"></param>
/// <returns></returns>
private static VectorD SolveByGEPP(MatrixD a, VectorD b)
{
int r = 0;
int c = 0;
var a2 = a.Clone();
var b2 = b.Clone();
while (r < a2.Rows - 1)
{
var maxA = a2.GetColumn(c).Where((x, i) => i >= r).Max();
var idxMaxA = a2.GetRowIndex((x, i) => (i >= r) && (x[c] == maxA));
if (idxMaxA != r)
{
a2.SwapRow(r, idxMaxA);
b2.Swap(r, idxMaxA);
}
for (int i = r + 1; i < a2.Rows; i++)
{
var x = a2[i, c] * -1.0;
a2[i, c] = 0;
for (int j = c + 1; j < a2.Columns; j++)
{
a2[i, j] += a2[r, j] * (x / a2[r, c]);
}
b2[i] += b2[r] * (x / a2[r, c]);
}
r++;
c++;
}
var n = b2.Length;
var result = new VectorD(new double[n]);
for (int k = n - 1; k >= 0; k--)
{
var sum = 0.0;
for (int j = k + 1; j < n; j++)
{
sum += a2[k, j] * result[j];
}
result[k] = (b2[k] - sum) / a2[k, k];
}
return result;
}示例2: SolveByGE
/// <summary>
/// ガウス消去法により連立一次方程式の解を求める
/// </summary>
/// <param name="a"></param>
/// <param name="b"></param>
/// <returns></returns>
private static VectorD SolveByGE(MatrixD a, VectorD b)
{
int r = 0;
int c = 0;
var a2 = a.Clone();
var b2 = b.Clone();
while (r < a2.Rows - 1)
{
for (int i = r + 1; i < a2.Rows; i++)
{
var x = a2[i, c] * -1.0;
a2[i, c] = 0;
for (int j = c + 1; j < a2.Columns; j++)
{
a2[i, j] += a2[r, j] * (x / a2[r, c]);
}
b2[i] += b2[r] * (x / a2[r, c]);
}
r++;
c++;
}
var n = b2.Length;
var result = new VectorD(new double[n]);
for (int k = n - 1; k >= 0; k--)
{
var sum = 0.0;
for (int j = k + 1; j < n; j++)
{
sum += a2[k, j] * result[j];
}
result[k] = (b2[k] - sum) / a2[k, k];
}
return result;
}示例3: Solve
/// <summary>
/// Solves the specified linear system of equations <i>Ax=b</i>.
/// </summary>
/// <param name="matrixA">The matrix A.</param>
/// <param name="initialX">
/// The initial guess for x. If this value is <see langword="null"/>, a zero vector will be used
/// as initial guess.
/// </param>
/// <param name="vectorB">The vector b.</param>
/// <returns>The solution vector x.</returns>
/// <exception cref="ArgumentNullException">
/// <paramref name="matrixA"/> is <see langword="null"/>.
/// </exception>
/// <exception cref="ArgumentNullException">
/// <paramref name="vectorB"/> is <see langword="null"/>.
/// </exception>
/// <exception cref="ArgumentException">
/// <paramref name="matrixA"/> is not a square matrix.
/// </exception>
/// <exception cref="ArgumentException">
/// The number of elements of <paramref name="initialX"/> does not match.
/// </exception>
public override VectorD Solve(MatrixD matrixA, VectorD initialX, VectorD vectorB)
{
// TODO: We can possible improve the method by reordering after each step.
// This can be done randomly or we sort by the "convergence" of the elements.
// See book Physics-Based Animation.
NumberOfIterations = 0;
if (matrixA == null)
throw new ArgumentNullException("matrixA");
if (vectorB == null)
throw new ArgumentNullException("vectorB");
if (matrixA.IsSquare == false)
throw new ArgumentException("Matrix A must be a square matrix.", "matrixA");
if (matrixA.NumberOfRows != vectorB.NumberOfElements)
throw new ArgumentException("The number of rows of A and b do not match.");
if (initialX != null && initialX.NumberOfElements != vectorB.NumberOfElements)
throw new ArgumentException("The number of elements of the initial guess for x and b do not match.");
VectorD xOld = initialX ?? new VectorD(vectorB.NumberOfElements);
VectorD xNew = new VectorD(vectorB.NumberOfElements);
bool isConverged = false;
// Make iterations until max iteration count or the result has converged.
for (int i = 0; i < MaxNumberOfIterations && !isConverged; i++)
{
for (int j = 0; j < vectorB.NumberOfElements; j++)
{
double delta = 0;
for (int k = 0; k < j; k++)
delta += matrixA[j, k] * xNew[k];
for (int k = j + 1; k < vectorB.NumberOfElements; k++)
delta += matrixA[j, k] * xOld[k];
xNew[j] = (vectorB[j] - delta) / matrixA[j, j];
}
// Test convergence
isConverged = VectorD.AreNumericallyEqual(xOld, xNew, Epsilon);
xOld = xNew.Clone();
NumberOfIterations = i + 1;
}
return xNew;
}示例4: Solve
/// <summary>
/// Solves the specified linear system of equations <i>Ax=b</i>.
/// </summary>
/// <param name="matrixA">The matrix A.</param>
/// <param name="initialX">
/// The initial guess for x. If this value is <see langword="null"/>, a zero vector will be used
/// as initial guess.
/// </param>
/// <param name="vectorB">The vector b.</param>
/// <returns>The solution vector x.</returns>
/// <exception cref="ArgumentNullException">
/// <paramref name="matrixA"/> is <see langword="null"/>.
/// </exception>
/// <exception cref="ArgumentNullException">
/// <paramref name="vectorB"/> is <see langword="null"/>.
/// </exception>
/// <exception cref="ArgumentException">
/// <paramref name="matrixA"/> is not a square matrix.
/// </exception>
/// <exception cref="ArgumentException">
/// The number of elements of <paramref name="initialX"/> does not match.
/// </exception>
public override VectorD Solve(MatrixD matrixA, VectorD initialX, VectorD vectorB)
{
NumberOfIterations = 0;
if (matrixA == null)
throw new ArgumentNullException("matrixA");
if (vectorB == null)
throw new ArgumentNullException("vectorB");
if (matrixA.IsSquare == false)
throw new ArgumentException("Matrix A must be a square matrix.", "matrixA");
if (matrixA.NumberOfRows != vectorB.NumberOfElements)
throw new ArgumentException("The number of rows of A and b do not match.");
if (initialX != null && initialX.NumberOfElements != vectorB.NumberOfElements)
throw new ArgumentException("The number of elements of the initial guess for x and b do not match.");
VectorD xOld = initialX ?? new VectorD(vectorB.NumberOfElements);
VectorD xNew = new VectorD(vectorB.NumberOfElements);
bool isConverged = false;
// Make iterations until max iteration count or the result has converged.
for (int i = 0; i < MaxNumberOfIterations && !isConverged; i++)
{
for (int j=0; j<vectorB.NumberOfElements; j++)
{
double delta = 0;
for (int k=0; k < j; k++)
delta += matrixA[j, k] * xNew[k];
for (int k=j+1; k < vectorB.NumberOfElements; k++)
delta += matrixA[j, k] * xOld[k];
delta = (vectorB[j] - delta) / matrixA[j, j];
xNew[j] = xOld[j] + RelaxationFactor * (delta - xOld[j]);
}
// Test convergence
isConverged = VectorD.AreNumericallyEqual(xOld, xNew, Epsilon);
xOld = xNew.Clone();
NumberOfIterations = i + 1;
}
return xNew;
}示例5: CloneTest
public void CloneTest()
{
VectorD v = new VectorD(new List<double>(new double[] { 1, 2, 3, 4, 5 }));
VectorD clonedVector = v.Clone();
Assert.AreEqual(v, clonedVector);
}