c++ - Best way to sort all columns of an armadillo matrix by an index vector -

i'm wondering whether there's better way achieve i'm doing here. have arma matrix , want reorder of it's columns indices stored in uvec vector. think i'm copying whole matrix.

#include <armadillo> using namespace arma;  int main(){              // discrete random matrix             // defined umat because want             // order given column of a. irrelevant now.     umat = randi<umat>(4,6,distr_param(0,3));     std::cout << "a " << std::endl;     std::cout << << std::endl;      // index vector row order     uvec b;     b << 3 << 2 << 1 << 0;      std::cout << "sort b:" << std::endl;     std::cout << b << std::endl;       // col indices     uvec cols = linspace<uvec>(0,a.n_cols-1,a.n_cols);      // order cols of b             // i'm afraid makes copy     = a.submat(b, cols );      std::cout << "reordered b" << std::endl;     std::cout << << std::endl;       return 0;  } 

you right in code creates new matrix , not exchange rows in place.

alternatively express permutation product of transpositions , swap rows of a one-by-one swap_rows. of course not trivial implement , go route if memory usage of concern or if need permute few of rows , leave rest are. otherwise rebuilding matrix faster due cache efficiency.

for example case, reverses row order, might of course want swap last , first row, last-1'th , 2nd , on.