1 year later, but I have come across a solution that works on 2D arrays. so instead of 2 vectors, you can make it a Nx2 array, and sort the 2nd column based on the 1st column - no need for index list. Furthermore, if there are duplicates in the 1st column, you can then sort the duplicates based on the 2nd column - e.g.: [7,15],[7,10],[1,2],[3,4],[7,2] => [1,2],[3,4],[7,2],[7,10],[7,15] this can even be extended to Nx3 as done in the example below... most of the code is actually the cout to show the results - the code is a single line of qsort, and then the qsort comparison routine (which is the engine). Made a variation for int by changing the arr definition, and the qsort type - works like a charm...Since I am new to this forum, I am restricted in submitting code.... hope it comes out.

include <sstream>

include <string>

include <iostream>

include <stdlib.h>

using namespace std;

static int compfloat(const void* p1, const void* p2) { float* arr1 = (float)p1; float arr2 = (float*)p2; float diff1 = arr1[0] - arr2[0]; if (diff1) return diff1; return arr1[1] - arr2[1]; //only compares 2nd column if the 1st is the same - can remove }

define arraysize 5

int main() { float arr[arraysize][3] = {{5,10,1},{2,2,1}, {1,5,2}, {5,4,3}, {5,20,4}}; //example data for (int i=0; i<arraysize; i++)="" {="" for="" (int="" j="0;j&lt;3;j++)" 3="number" of="" columns...="" change="" as="" required="" cout="" &lt;&lt;="" arr[i][j]="" &lt;&lt;="" "="" ";="" cout="" &lt;&lt;="" endl;="" }="" cout="" &lt;&lt;="" endl;<="" p="">

qsort(arr, arraysize, 3*sizeof(float), compfloat); //3=number of columns... change as required

for (int i=0; i<arraysize; i++)="" {="" for="" (int="" j="0;j&lt;3;j++)" 3="number" of="" columns...="" change="" as="" required="" cout="" &lt;&lt;="" arr[i][j]="" &lt;&lt;="" "="" ";="" cout="" &lt;&lt;="" endl;="" }="" return="" 0;="" }<="" p="">

1 year later, but I have come across a solution that works on 2D arrays. so instead of 2 vectors, you can make it a Nx2 array, and sort the 2nd column based on the 1st column - no need for index list. Furthermore, if there are duplicates in the 1st column, you can then sort the duplicates based on the 2nd column - e.g.: e.g.:

[7,15],[7,10],[1,2],[3,4],[7,2] => [1,2],[3,4],[7,2],[7,10],[7,15] [1,2],[3,4],[7,2],[7,10],[7,15]

this can even be extended to Nx3 as done in the example below... most of the code is actually the cout to show the results - the code is a single line of qsort, and then the qsort comparison routine (which is the engine). Made a variation for int by changing the arr definition, and the qsort type - works like a charm...Since I am new to this forum, I am restricted in submitting code.... hope it comes out.

#include <sstream>
#include <string>
#include <iostream>
#include <stdlib.h>

using namespace std;
 
std;
static int compfloat(const void* p1, const void* p2) {
float* arr1 = (float)p1;
float (float*)p1;
float* arr2 = (float*)p2;
float diff1 = arr1[0] - arr2[0];
if (diff1) return diff1;
return arr1[1] - arr2[1]; //only compares 2nd column if the 1st is the same - can remove
}
 
define }
#define arraysize 5
 
5
int main()
{ float arr[arraysize][3] = {{5,10,1},{2,2,1}, {1,5,2}, {5,4,3}, {5,20,4}}; //example data
for (int i=0; i<arraysize; i++)="" {="" for="" (int="" j="0;j&lt;3;j++)" 3="number" of="" columns...="" change="" as="" required="" cout="" &lt;&lt;="" arr[i][j]="" &lt;&lt;="" "="" ";="" cout="" &lt;&lt;="" endl;="" }="" cout="" &lt;&lt;="" endl;<="" p="">
i++)
{ for (int j=0;j<3;j++) //3=number of columns... change as required
cout << arr[i][j] << " ";
cout << endl;
}
cout << endl;
qsort(arr, arraysize, 3*sizeof(float), compfloat); //3=number of columns... change as required
 
required
for (int i=0; i<arraysize; i++)="" {="" for="" (int="" j="0;j&lt;3;j++)" 3="number" of="" columns...="" change="" as="" required="" cout="" &lt;&lt;="" arr[i][j]="" &lt;&lt;="" "="" ";="" cout="" &lt;&lt;="" endl;="" }="" return="" 0;="" }<="" p="">
i++)
{ for (int j=0;j<3;j++) //3=number of columns... change as required
cout << arr[i][j] << " ";
cout << endl;
}
return 0;
}