在Eigen中将向量分配给矩阵列 [英] Assigning a vector to a matrix column in Eigen

问题描述

这个问题急匆匆地问了。我原始程序中的错误，不是此处显示的代码中的错字。错误是我的程序v由于某些条件而没有填充。

从该线程获取的更有用的收获是将std :: vector复制到本征矩阵的所有行或列，在接受的答案中。

---

我想将向量复制到a的列中矩阵，如下所示：

  #include< Eigen / Dense> 
 #include< vector> 
 #include< iostream> 
 
 int main（）{
 int m = 10; 
 
 std :: vector< Eigen :: VectorXd> v（m）； 
 Eigen :: MatrixXd S（m，m）; 
 
 for（int i = 0; i！= m; ++ i）{
 v [i] .resize（m）; 
 
 for（int j = 0; j！= m; ++ j）{
 v [i]（j）= rand（）％m; 
} 
 
 //S.cols(i）= v [i]; //需要这样的东西
} 
 
返回0; 
} 
  

S的类型为Eigen :: MatrixXd，尺寸为mxm。 v是Eigen :: VectorXd的std :: vector，其中每个Eigen :: VectorXd的大小为m，v中有m个。

解决方案

关于原始问题，您需要用 Eigen :: Map std :: vector $ c>。您还可以（也应该）使该操作成为单线操作。 S.cols（i）应该是 S.col（i）。

  int main（）
 {
 size_t sz = 6; 
 Eigen :: MatrixXd S（sz，sz）; 
 std :: vector< double> v（sz）; 
 std :: vector< Eigen :: VectorXd> vv（sz）; 
 for（int i = 0; i< sz; i ++）
 {
 v [i] = i * 2; 
 vv [i] = Eigen :: VectorXd :: LinSpaced（sz，（i + sz），（i + sz）* 2）; 
} 
 
 for（int i = 0; i！= sz; ++ i）
 S.col（i）= vv [i]; 
 std :: cout<< S<<  \n\n； 
 
 S.rowwise（）= Eigen :: Map< Eigen :: RowVectorXd>（v.data（），sz）; 
 std :: cout<< S<<  \n\n； 
 
 S.colwise（）= Eigen :: Map< Eigen :: VectorXd>（v.data（），sz）; 
 std :: cout<< S<<  \n\n； 
 
返回0; 
} 
  

将会输出

< blockquote>

6 7 8 9 10 11

7.2 8.4 9.6 10.8 12 13.2

8.4 9.8 11.2 12.6 14 15.4

9.6 11.2 12.8 14.4 16 17.6

10.8 12.6 14.4 16.2 18 19.8

12 14 16 18 20 22

0 2 4 6 8 10

0 2 4 6 8 10

0 2 4 6 8 10

0 2 4 6 8 10

0 2 4 6 8 10

0 2 4 6 8 10

0 0 0 0 0 0

2 2 2 2 2 2

4 4 4 4 4 4

6 6 6 6 6 6

8 8 8 8 8 8

10 10 10 10 10 10

This question was asked in haste. The error in my original program, was not the typo in the code that is displayed here. The error was that in my program v was not getting populated due to some conditions.

The more useful takeaway from this thread is the demonstration of copying a std::vector to all rows or columns of an Eigen Matrix, in the accepted answer.

---

I want to copy vectors into the columns of a matrix, like the following:

#include <Eigen/Dense>
#include <vector>
#include <iostream>

int main() {
 int m = 10;

 std::vector<Eigen::VectorXd> v(m);
 Eigen::MatrixXd S(m,m);

 for (int i = 0; i != m; ++i) {
  v[i].resize(m);

  for (int j = 0; j != m; ++j) {
   v[i](j) = rand() % m;
  }

  //S.cols(i) = v[i]; //needed something like this
 }

 return 0;
}

S is of type Eigen::MatrixXd and dimension mxm. v is a std::vector of Eigen::VectorXd, where each Eigen::VectorXd is of size m and there are m of them in v.

解决方案

Regarding the original question, you need to wrap the std::vector with an Eigen::Map. You could/should also make the operation a one-liner.

The reworded question is reduced to a typo. S.cols(i) should be S.col(i).

int main()
{
    size_t sz = 6;
    Eigen::MatrixXd S(sz, sz);
    std::vector<double> v(sz);
    std::vector<Eigen::VectorXd> vv(sz);
    for(int i = 0; i < sz; i++)
    {
        v[i] = i*2;
        vv[i] = Eigen::VectorXd::LinSpaced(sz, (i+sz), (i+sz)*2);
    }

    for (int i = 0; i != sz; ++i)
        S.col(i) = vv[i];
    std::cout << S << "\n\n";

    S.rowwise() = Eigen::Map<Eigen::RowVectorXd>(v.data(), sz);
    std::cout << S << "\n\n";

    S.colwise() = Eigen::Map<Eigen::VectorXd>(v.data(), sz);
    std::cout << S << "\n\n";

    return 0;
}

which would output

6 7 8 9 10 11
7.2 8.4 9.6 10.8 12 13.2
8.4 9.8 11.2 12.6 14 15.4
9.6 11.2 12.8 14.4 16 17.6
10.8 12.6 14.4 16.2 18 19.8
12 14 16 18 20 22

0 2 4 6 8 10
0 2 4 6 8 10
0 2 4 6 8 10
0 2 4 6 8 10
0 2 4 6 8 10
0 2 4 6 8 10

0 0 0 0 0 0
2 2 2 2 2 2
4 4 4 4 4 4
6 6 6 6 6 6
8 8 8 8 8 8
10 10 10 10 10 10

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