Armadillo equivalent of Matlab permute?

问题

I have a arma::cube mycube(5,10,15); and I want to permute its dimensions, as one would do in matlab:

mycube = ones(5,10,15);
mycube = permute(mycube,[3 1 2]);
size(mycube) % returns (15 5 10)

Is there a way to do that?
Would it be too inefficient?

I actually want to do a 3D FFT, so I thought of permuting the first and third dimensions to be able to use arma::fft, and then permuting back.

回答1:

Armadillo ibrary does not contain such a function, but you can implement a simplified version. For example so:

#include <iostream>
#include <armadillo>
#include <tuple>
#include <algorithm>
#include <vector>

typedef std::tuple<std::size_t,std::size_t,std::size_t> D3tuple;

void printSize(const arma::cube &cube);
void simplePermute(arma::cube &cube, const D3tuple &order);
arma::uword getSize(const arma::cube &cube,
                    const std::size_t &n);
D3tuple get_coeff(arma::cube &cube, const D3tuple &order);

int main(int argc, char** argv)
  {

  arma::cube mycube = arma::randu<arma::cube>(2,2,2);
  std::cout<<mycube<<std::endl;
  printSize(mycube);

  simplePermute(mycube,D3tuple(3,1,2));
  printSize(mycube);
  std::cout<<mycube<<std::endl;
  return 0;
  }

void printSize(const arma::cube &cube)
    {
    std::cout<<cube.n_rows<<" "<<cube.n_cols<<" "<<cube.n_slices<<std::endl;
    }

void simplePermute(arma::cube &cube, const D3tuple &order)
    {
    auto first = std::get<0>(order),
        second = std::get<1>(order),
        third = std::get<2>(order);
    std::size_t cols = getSize(cube,first),
        rows = getSize(cube,second) ,
        slices = getSize(cube,third);

    arma::cube temp(cols,rows,slices);
    std::size_t c1,c2,c3;
    std::tie(c3,c2,c1) = get_coeff(cube,order);
    std::size_t index = 0;
    for(std::size_t i = 0;i<cols;i++)
        for(std::size_t j = 0;j<rows;j++)
            for(std::size_t k = 0;k<slices;k++)
                temp[index++] = cube[c1*i+c2*j+c3*k];

    cube = temp;
    }

arma::uword getSize(const arma::cube &cube,
                    const std::size_t &n)
    {
    switch (n)
        {
        case 1 : return cube.n_rows;
        case 2 : return cube.n_cols;
        case 3 : return cube.n_slices;
        }
    return 0;
    }

D3tuple get_coeff(arma::cube &cube, const D3tuple &order)
    {
    std::size_t c1,c2,c3;
    switch (std::get<0>(order))
        {
        case 1 : 
            c1 =  1;break;
        case 2 : 
            c1 =  cube.n_rows; break;
        case 3 : 
            c1 =  cube.n_rows*cube.n_cols; break;
        }
    switch (std::get<1>(order))
        {
        case 1 : 
            c2 =  1; break;
        case 2 : 
            c2 =  cube.n_rows; break;
        case 3 : 
            c2 =  cube.n_rows*cube.n_cols; break;
        }
    switch (std::get<2>(order))
        {
        case 1 : 
            c3 =  1; break;
        case 2 : 
            c3 =  cube.n_rows; break;
        case 3 : 
            c3 =  cube.n_rows*cube.n_cols; break;
        }
    return std::make_tuple(c1,c2,c3);
    }

回答2:

Another easy-way to make 3-dim array (arma::cube) permutation is the one below. This is not really elegant, but easy understandable.

Because the permutation of 3 unique numbers is 6 (to be exact 5 without the reference order), it's quick to avoid algorithmic method.

Permutation of dim 1, 2, 3 :

123 (base order) 132 213 231 312 321.

So a simple switch between the differents permutations :

template <typename T>
static Cube<T> permute (Cube<T>& cube, const std::tuple<uword,uword,uword>& order)
{
    uword idx1 = std::get<0>(order);
    uword idx2 = std::get<1>(order);
    uword idx3 = std::get<2>(order);

    u32_vec dimension = shape(cube);

    uword rows = dimension(idx1 - 1);
    uword cols = dimension(idx2 - 1);
    uword slis = dimension(idx3 - 1);

    Cube<T> output; 
    output.zeros(rows, cols, slis);

    uword perm = idx1*100 + idx2*10 + idx3;

    switch (perm)
    {
        case 123:
        {
            output = cube; // identity
        }
        break;
        case 132:
        {
            for (int c = 0; c < cube.n_cols; ++c)
                for (int r = 0; r < cube.n_rows; ++r)
                    for (int s = 0; s < cube.n_slices; ++s)
                        output(r, s, c) = cube(r, c, s);
        }
        break;
        case 213:
        {
            for (int c = 0; c < cube.n_cols; ++c)
                for (int r = 0; r < cube.n_rows; ++r)
                    for (int s = 0; s < cube.n_slices; ++s)
                        output(c, r, s) = cube(r, c, s);
        }
        break;
        case 231:
        {
            for (int c = 0; c < cube.n_cols; ++c)
                for (int r = 0; r < cube.n_rows; ++r)
                    for (int s = 0; s < cube.n_slices; ++s)
                        output(c, s, r) = cube(r, c, s);
        }
        break;
        case 312:
        {
            for (int c = 0; c < cube.n_cols; ++c)
                for (int r = 0; r < cube.n_rows; ++r)
                    for (int s = 0; s < cube.n_slices; ++s)
                        output(s, r, c) = cube(r, c, s);
        }
        break;
        case 321:
        {
            for (int c = 0; c < cube.n_cols; ++c)
                for (int r = 0; r < cube.n_rows; ++r)
                    for (int s = 0; s < cube.n_slices; ++s)
                        output(s, c, r) = cube(r, c, s);
        }
        break;
    }

    return output;
}

The order tuple is in matlab-style (1-based) while armadillo is zero-based array.

The shape(cube) function is just a little helper that return the equivalent of Size() in matlab, a N-dim array with each dimension size.

template <typename T> 
inline u32_vec shape (const Cube<T>& x) 
{ 
    return { x.n_rows, x.n_cols, x.n_slices };
}

The code need to be used with :

using namespace arma;

来源：https://stackoverflow.com/questions/35796248/armadillo-equivalent-of-matlab-permute

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matlab

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