I want to reindex a Eigen:Matrix by rolling N∈ℤ rows like this (here N=+1):
1 4 7 -> 3 6 9
2 5 8 1 4 7
3 6 9 2 5 8
Is there a simple way, or do I have to create a new matrix and copy over the data?
I suggest setting up a new matrix and copying the data. Eigen's block operations allow doing this in an efficient way. Here is how a shift by n rows can be done for the example above.
MatrixXi A(3,3);
A << 1, 2, 3, 4, 5, 6, 7, 8, 9;
A.transposeInPlace();
int n = 1; // number of shifts
n = n % A.rows();
MatrixXi B(A.rows(), A.cols());
B.bottomRows(A.rows() - n) = A.topRows(A.rows() - n);
B.topRows(n) = A.bottomRows(n);
std::cout << "B = " << B << std::endl;
If you are interested in a matlab-like syntax you can also use
MatrixXd A;
//... fill A
VectorXi indices = {{2,0,1}};
A(indices, Eigen::all);
I don't know, whether this internally makes a copy.
Note: This does not work for Sparse matrices, see Subset columns of sparse eigen matrix
Related
I have an object of std::vector<std::array<double, 16>>
vector entry Data
[0] - 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
[1] - 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
[2] - 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
[...] - 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
This is intended to represent a 4x4 matrix in ravel format.
To not duplicate information I would like to create a pointer to extract a 3x3 from the above structure:
I have mathematical operations for the 3x3 structure (std::array<double, 9>)
someStructure: pointing to data elements [0, 1, 2, 4, 5, 6, 8, 9 10]
The end goal is do: std::array<double, 9> tst = someStructure[0] + someStructure[1];
Is this doable?
Best Regards
The 3x3 part is not contiguous, hence a pointer alone wont help here.
You can write a view_as_3x3 that allows you to access elements of the submatrix of the 4x4 as if it was contiguous:
struct view_as_3x3 {
double& operator[](size_t index) {
static const size_t mapping[] = {0, 1, 2, 4, 5, 6, 8, 9, 10};
return parent[mapping[index]];
}
std::array<double, 16>& parent;
};
Such that for example
for (size_t = 0; i< 9; ++i) std::cout << " " << view_as_3x3{orignal_matrix}[i];
is printing the 9 elements of the 3x3 sub-matrix of the original 4x4 original_matrix.
Then you could more easily apply your 3x3 algorithms to the 3x3 submatrix of a 4x4 matrix. You just need to replace the std::array<double, 9> with some generic T. For example change
double sum_of_elements(const std::array<double, 9>& arr) {
double res = 0;
for (int i=0;i <9; ++i) res += arr[i];
return res;
}
To:
template <typename T>
double sum_of_elements(const T& arr) {
double res = 0;
for (int i=0;i <9; ++i) res += arr[i];
return res;
}
The calls are then
std::array<double, 16> matrix4x4;
sum_of_elements(view_as_3x3{matrix4x4});
// or
std::array<double, 9> matrix3x3;
sum_of_elements(matrix3x3);
It would be nicer to use iterators instead of indices, however, writing the view with custom iterators requires considerable amount of boilerplate. On the other hand, I would not suggest to use naked std::arrays in the first place, but rather some my_4x4matrix that holds the array as member and provides iterators and more convenience methods.
vector<vector<double>> weights
{
{1},
{1}
};
Above is my code to make a 2x1 vector each holding 1.
I would like to make a matrix of 2xN that I could use to multiply with that vector.
I have seen other stackoverflow questions that talk about creating matrices, and most of the ones I've seen are with fixed values, or user input.
But what I would like to do, is initialize the entire first column of N length with 1s, and the initialize the entire second column with a second vector I already have.
I am unsure how in C++ I could accomplish this. I'm way more familiar with R, and in R this is a pretty simple task. Any thoughts or guidance?
You mean like this?
std::vector<int> vinner {
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
};
std::vector<std::vector<int>> v {
std::vector<int>(10, 1),
vinner
};
int main(int argc, char **argv)
{
for (auto i : v) {
for (auto j : i) {
std::cout << j << " ";
}
std::cout << "\n";
}
return 0;
}
Output:
$ clang++ -o vect vect.cpp -std=c++17
$ ./vect
1 1 1 1 1 1 1 1 1 1
1 2 3 4 5 6 7 8 9 10
I need to multiply each matrix column by each vector element using Eigen C++ library. I tried colwise without success.
Sample data:
Eigen::Matrix3Xf A(3,2); //3x2
A << 1 2,
2 2,
3 5;
Eigen::Vector3f V = Eigen::Vector3f(2, 3);
//Expected result
C = A.colwise()*V;
//C
//2 6,
//4 6,
//6 15
//this means C 1st col by V first element and C 2nd col by V 2nd element.
Matrix A can have 3xN and V Nx1. Meaning (cols x rowls).
This is what I would do:
Code
Eigen::Matrix3Xf A(3, 2); // 3x2
A << 1, 2, 2, 2, 3, 5;
Eigen::Vector3f V = Eigen::Vector3f(1, 2, 3);
const Eigen::Matrix3Xf C = A.array().colwise() * V.array();
std::cout << C << std::endl;
Example output:
1 2
4 4
9 15
Explanation
You were close, the trick is to use .array() to do broadcasting multiplications.
colwiseReturnType doesn't have a .array() method, so we have to do our colwise shenanigans on the array view of A.
If you want to compute the element-wise product of two vectors (The coolest of cool cats call this the Hadamard Product), you can do
Eigen::Vector3f a = ...;
Eigen::Vector3f b = ...;
Eigen::Vector3f elementwise_product = a.array() * b.array();
Which is what the above code is doing, in a columnwise fashion.
Edit:
To address the row case, you can use .rowwise(), and you'll need an extra transpose() to make things fit
Eigen::Matrix<float, 3, 2> A; // 3x2
A << 1, 2, 2, 2, 3, 5;
Eigen::Vector2f V = Eigen::Vector2f(2, 3);
// Expected result
Eigen::Matrix<float, 3, 2> C = A.array().rowwise() * V.transpose().array();
std::cout << C << std::endl;
Example output:
2 6
4 6
6 15
In other words, you want to scale each column by a different factor, that is, apply a non uniform scaling. Scaling are best represented as a diagonal matrix, thus:
C = A * V.asDiagonal();
Since Eigen is based on expression template, this does not create any temporary and amount to a code similar to Jacob's answer:
C = A.array().rowwise() * V.transpose().array();
How to use eigen library to compute lower triangular of input matrix without changing columns order?
for example for matrix:
A=[1 2 3;4 5 6 ;7 8 9]
I want the result to be:
1 0 0
4 0 0
7 0 0
Your text and your example don't match. I'll go through the three possible ways I understood your question. First, we'll set up the matrix:
Matrix3d mat;
mat << 1, 2, 3, 4, 5, 6, 7, 8, 9;
If you wanted the actual lower triangular matrix, you would use:
std::cout << Matrix3d(mat.triangularView<Lower>()) << "\n\n";
or similar. The result is:
1 0 0
4 5 0
7 8 9
Note the 5,8,9 which are missing from your example. If you just wanted the left-most column, you would use:
std::cout << mat.col(0) << "\n\n";
which gives
1
4
7
If (as the second part of your example shows) you want mat * [1, 0, 0] then you could either do the matrix multiplication (not recommended) or just construct the result:
Matrix3d z = Matrix3d::Zero();
z.col(0) = mat.col(0);
std::cout << z << "\n\n";
which gives the same result as your example:
1 0 0
4 0 0
7 0 0
I have a matrix in SAS/IML:
x = {7 6 3 3 8,
2 3 5 2 5,
2 6 4 3 8,
7 4 8 1 3,
8 8 6 8 7,
3 2 6 1 5 };
I want to create a new matrix that contains the highest k values of each column in x. For example, if k=3, I want the result matrix to contain:
8 8 8 8 8
7 6 6 3 8
7 6 6 3 7
because, for instance, the largest 3 numbers in the first column of x are 8, 7, and 7.
I've unsuccessfully tried to figure out how to do this using the rank function.
Your code looks fine. Here's a minor revision:
do c=1 to ncol(x);
r = rank(x[,c]);
y = x[loc(r>=nrow(x)-k+1), c];
call sort(y);
tops[,c] = y;
end;
As to avoiding the loop to make it faster, it's not necessary. Even with 10,000 columns, this code runs in a fraction of a second. Try running the following timing code:
x = j(500, 10000);
call randgen(x,"normal");
k = 3;
t0=time();
tops = j(k,ncol(x),0);
do c=1 to ncol(x);
r = rank(x[,c]);
y = x[loc(r>=nrow(x)-k+1), c];
call sort(y);
tops[,c] = y;
end;
t=time()-t0;
print t;
Here's a partial answer I've come up with:
k = 3;
tops = j(k,ncol(x),0);
do c=1 to ncol(x);
r = rank(x[,c]);
h=loc(r>=nrow(x)-k+1);
tops[,c] = x[,c][h];
end;
This approach uses a loop, which I'd like to avoid, so please post improvements if possible!