Question

我正在尝试转换Matlab代码，创建带状对角矩阵。我试图转换的Matlab代码是：

N = 5;
e = ones(N-1, 1);
D = spdiags([-e 2*e -e], [-1 0 1], N-1, N-1);
D = full(D);

上述Matlab代码的输出，D =

 2    -1     0     0
-1     2    -1     0
 0    -1     2    -1
 0     0    -1     2

我转换它时最接近的C ++尝试如下：

#include<iostream>
#include<vector>

using namespace std;
vector< vector<double> > bandedMat(vector<double> &e,int N);

int main()
{
   int N = 5 ;
   vector<double> e = {-1,2,-1};
   vector< vector<double> > B = bandedMat(e,N);

   return 0;
}
vector< vector<double> > bandedMat(vector<double> &e,int N)
{
   vector< vector<double> > D(N-2, vector<double>(N,0.0));
   double val = 0.0;

  for(int i = 0; i < D.size(); i++)
  {
     for(int j = 0; j < e.size(); j++)
     {
        val = e[j];
        D[i][i+j] = val; // Put along the diagonal of matrix D. Note the    index.
     }
  }

  return D;
}

以下C ++代码的输出是D =

-1  2 -1  0  0
0  -1  2 -1  0
0  0  -1  2 -1

正如你所看到的，我的C ++版本是完全不同的，它产生D为3x5而不是Matlab版本，它产生D为4 x 4.此外，我的C ++版本的对角线有点不同。有人可以指出我怎么能得到像Matlab版本一样的精确D.

Answer 1

虽然我认为它不优雅，但至少它现在解决了我的问题。这是：

vector< vector<double> > bandedMat(vector<double> &e,int N)
{
    // Do some checking, Only tridiagonals are allowed.
    if(e.size()>3)
    {
        cout << "Only tridiagonals are allowed. Input vector to function must be only 3 elements." << endl;
        exit(EXIT_FAILURE);
    }

    vector< vector<double> > D(N-1, vector<double>(N-1,0.0));

    for(size_t i = 0; i < D.size(); i++)
    {
        for(size_t j = 0; j < D.size(); j++)
        {
            if(i == j)
            {
                D[i][j] = e[1]; // put value of e[1] as the main diagonal of D
            }
        }
    }

    int j =0; // index
    int k =0; // index

    for(size_t i =0; i < D.size()-1;i++)
    {
        D[i][j+1] = e[0]; // put value e[0] as the upper diagonal
        j = j+1;
    }

    for(size_t i =0; i < D.size()-1;i++)
    {
        D[i+1][k] = e[2]; // put value of e[2] as the lower diagonal
        k = k+1;
    }

    return D;
}

将生成类似Matlab版本的精确矩阵。如果任何人有更好的想法或更优雅的方式来做，请分享。感谢。

Answer 2

稍微优雅一些...您可以将所有三个文件放在一个循环中

for(size_t i = 0; i < D.size(); i++)
{
    if (i<D.size()) {
        D[i][i+1] = e[0];
    }
    D[i][i] = e[1]; // put value of e[1] as the main diagonal of D
    if (i>0) {
        D[i][i-1] = e[2];
    }
}

用C ++创建带状矩阵

2 个答案: