C++ sorting and keeping track of indexes

Question

Using C++, and hopefully the standard library, I want to sort a sequence of samples in ascending order, but I also want to remember the original indexes of the new samples.

For example, I have a set, or vector, or matrix of samples A : [5, 2, 1, 4, 3]. I want to sort these to be B : [1,2,3,4,5], but I also want to remember the original indexes of the values, so I can get another set which would be: C : [2, 1, 4, 3, 0 ] - which corresponds to the index of each element in 'B', in the original 'A'.

For example, in Matlab you can do:

 [a,b]=sort([5, 8, 7])
 a = 5 7 8
 b = 1 3 2

Can anyone see a good way to do this?

Answer 1

Using C++ 11 lambdas:

#include <iostream>
#include <vector>
#include <numeric>      // std::iota
#include <algorithm>    // std::sort, std::stable_sort

using namespace std;

template <typename T>
vector<size_t> sort_indexes(const vector<T> &v) {

  // initialize original index locations
  vector<size_t> idx(v.size());
  iota(idx.begin(), idx.end(), 0);

  // sort indexes based on comparing values in v
  // using std::stable_sort instead of std::sort
  // to avoid unnecessary index re-orderings
  // when v contains elements of equal values 
  stable_sort(idx.begin(), idx.end(),
       [&v](size_t i1, size_t i2) {return v[i1] < v[i2];});

  return idx;
}

Now you can use the returned index vector in iterations such as

for (auto i: sort_indexes(v)) {
  cout << v[i] << endl;
}

You can also choose to supply your original index vector, sort function, comparator, or automatically reorder v in the sort_indexes function using an extra vector.

Answer 2

You could sort std::pair instead of just ints - first int is original data, second int is original index. Then supply a comparator that only sorts on the first int. Example:

Your problem instance: v = [5 7 8]
New problem instance: v_prime = [<5,0>, <8,1>, <7,2>]

Sort the new problem instance using a comparator like:

typedef std::pair<int,int> mypair;
bool comparator ( const mypair& l, const mypair& r)
   { return l.first < r.first; }
// forgetting the syntax here but intent is clear enough

The result of std::sort on v_prime, using that comparator, should be:

v_prime = [<5,0>, <7,2>, <8,1>]

You can peel out the indices by walking the vector, grabbing .second from each std::pair.

Answer 3

Suppose Given vector is

A=[2,4,3]

Create a new vector

V=[0,1,2] // indicating positions

Sort V and while sorting instead of comparing elements of V , compare corresponding elements of A

 //Assume A is a given vector with N elements
 vector<int> V(N);
 std::iota(V.begin(),V.end(),0); //Initializing
 sort( V.begin(),V.end(), [&](int i,int j){return A[i]<A[j];} );

Answer 4

vector<pair<int,int> >a;

for (i = 0 ;i < n ; i++) {
    // filling the original array
    cin >> k;
    a.push_back (make_pair (k,i)); // k = value, i = original index
}

sort (a.begin(),a.end());

for (i = 0 ; i < n ; i++){
    cout << a[i].first << " " << a[i].second << "\n";
}

Now a contains both both our values and their respective indices in the sorted.

a[i].first = value at i'th.

a[i].second = idx in initial array.

Answer 5

I wrote generic version of index sort.

template <class RAIter, class Compare>
void argsort(RAIter iterBegin, RAIter iterEnd, Compare comp, 
    std::vector<size_t>& indexes) {

    std::vector< std::pair<size_t,RAIter> > pv ;
    pv.reserve(iterEnd - iterBegin) ;

    RAIter iter ;
    size_t k ;
    for (iter = iterBegin, k = 0 ; iter != iterEnd ; iter++, k++) {
        pv.push_back( std::pair<int,RAIter>(k,iter) ) ;
    }

    std::sort(pv.begin(), pv.end(), 
        [&comp](const std::pair<size_t,RAIter>& a, const std::pair<size_t,RAIter>& b) -> bool 
        { return comp(*a.second, *b.second) ; }) ;

    indexes.resize(pv.size()) ;
    std::transform(pv.begin(), pv.end(), indexes.begin(), 
        [](const std::pair<size_t,RAIter>& a) -> size_t { return a.first ; }) ;
}

Usage is the same as that of std::sort except for an index container to receive sorted indexes. testing:

int a[] = { 3, 1, 0, 4 } ;
std::vector<size_t> indexes ;
argsort(a, a + sizeof(a) / sizeof(a[0]), std::less<int>(), indexes) ;
for (size_t i : indexes) printf("%d\n", int(i)) ;

you should get 2 1 0 3. for the compilers without c++0x support, replace the lamba expression as a class template:

template <class RAIter, class Compare> 
class PairComp {
public:
  Compare comp ;
  PairComp(Compare comp_) : comp(comp_) {}
  bool operator() (const std::pair<size_t,RAIter>& a, 
    const std::pair<size_t,RAIter>& b) const { return comp(*a.second, *b.second) ; }        
} ;

and rewrite std::sort as

std::sort(pv.begin(), pv.end(), PairComp(comp)()) ;

Answer 6

I came across this question, and figured out sorting the iterators directly would be a way to sort the values and keep track of indices; There is no need to define an extra container of pairs of ( value, index ) which is helpful when the values are large objects; The iterators provides the access to both the value and the index:

/*
 * a function object that allows to compare
 * the iterators by the value they point to
 */
template < class RAIter, class Compare >
class IterSortComp
{
    public:
        IterSortComp ( Compare comp ): m_comp ( comp ) { }
        inline bool operator( ) ( const RAIter & i, const RAIter & j ) const
        {
            return m_comp ( * i, * j );
        }
    private:
        const Compare m_comp;
};

template <class INIter, class RAIter, class Compare>
void itersort ( INIter first, INIter last, std::vector < RAIter > & idx, Compare comp )
{ 
    idx.resize ( std::distance ( first, last ) );
    for ( typename std::vector < RAIter >::iterator j = idx.begin( ); first != last; ++ j, ++ first )
        * j = first;

    std::sort ( idx.begin( ), idx.end( ), IterSortComp< RAIter, Compare > ( comp ) );
}

as for the usage example:

std::vector < int > A ( n );

// populate A with some random values
std::generate ( A.begin( ), A.end( ), rand );

std::vector < std::vector < int >::const_iterator > idx;
itersort ( A.begin( ), A.end( ), idx, std::less < int > ( ) );

now, for example, the 5th smallest element in the sorted vector would have value **idx[ 5 ] and its index in the original vector would be distance( A.begin( ), *idx[ 5 ] ) or simply *idx[ 5 ] - A.begin( ).

Answer 7

Consider using std::multimap as suggested by @Ulrich Eckhardt. Just that the code could be made even simpler.

Given

std::vector<int> a = {5, 2, 1, 4, 3};  // a: 5 2 1 4 3

To sort in the mean time of insertion

std::multimap<int, std::size_t> mm;
for (std::size_t i = 0; i != a.size(); ++i)
    mm.insert({a[i], i});

To retrieve values and original indices

std::vector<int> b;
std::vector<std::size_t> c;
for (const auto & kv : mm) {
    b.push_back(kv.first);             // b: 1 2 3 4 5
    c.push_back(kv.second);            // c: 2 1 4 3 0
}

The reason to prefer a std::multimap to a std::map is to allow equal values in original vectors. Also please note that, unlike for std::map, operator[] is not defined for std::multimap.

Answer 8

There is another way to solve this, using a map:

vector<double> v = {...}; // input data
map<double, unsigned> m; // mapping from value to its index
for (auto it = v.begin(); it != v.end(); ++it)
    m[*it] = it - v.begin();

This will eradicate non-unique elements though. If that's not acceptable, use a multimap:

vector<double> v = {...}; // input data
multimap<double, unsigned> m; // mapping from value to its index
for (auto it = v.begin(); it != v.end(); ++it)
    m.insert(make_pair(*it, it - v.begin()));

In order to output the indices, iterate over the map or multimap:

for (auto it = m.begin(); it != m.end(); ++it)
    cout << it->second << endl;

Answer 9

Beautiful solution by @Lukasz Wiklendt! Although in my case I needed something more generic so I modified it a bit:

template <class RAIter, class Compare>
vector<size_t> argSort(RAIter first, RAIter last, Compare comp) {

  vector<size_t> idx(last-first);
  iota(idx.begin(), idx.end(), 0);

  auto idxComp = [&first,comp](size_t i1, size_t i2) {
      return comp(first[i1], first[i2]);
  };

  sort(idx.begin(), idx.end(), idxComp);

  return idx;
}

Example: Find indices sorting a vector of strings by length, except for the first element which is a dummy.

vector<string> test = {"dummy", "a", "abc", "ab"};

auto comp = [](const string &a, const string& b) {
    return a.length() > b.length();
};

const auto& beginIt = test.begin() + 1;
vector<size_t> ind = argSort(beginIt, test.end(), comp);

for(auto i : ind)
    cout << beginIt[i] << endl;

prints:

abc
ab
a

Answer 10

Make a std::pair in function then sort pair :

generic version :

template< class RandomAccessIterator,class Compare >
auto sort2(RandomAccessIterator begin,RandomAccessIterator end,Compare cmp) ->
   std::vector<std::pair<std::uint32_t,RandomAccessIterator>>
{
    using valueType=typename std::iterator_traits<RandomAccessIterator>::value_type;
    using Pair=std::pair<std::uint32_t,RandomAccessIterator>;

    std::vector<Pair> index_pair;
    index_pair.reserve(std::distance(begin,end));

    for(uint32_t idx=0;begin!=end;++begin,++idx){
        index_pair.push_back(Pair(idx,begin));
    }

    std::sort( index_pair.begin(),index_pair.end(),[&](const Pair& lhs,const Pair& rhs){
          return cmp(*lhs.second,*rhs.second);
    });

    return index_pair;
}

ideone

Answer 11

If it's possible, you can build the position array using find function, and then sort the array.

Or maybe you can use a map where the key would be the element, and the values a list of its position in the upcoming arrays (A, B and C)

It depends on later uses of those arrays.

Answer 12

Well, my solution uses residue technique. We can place the values under sorting in the upper 2 bytes and the indices of the elements - in the lower 2 bytes:

int myints[] = {32,71,12,45,26,80,53,33};

for (int i = 0; i < 8; i++)
   myints[i] = myints[i]*(1 << 16) + i;

Then sort the array myints as usual:

std::vector<int> myvector(myints, myints+8);
sort(myvector.begin(), myvector.begin()+8, std::less<int>());

After that you can access the elements' indices via residuum. The following code prints the indices of the values sorted in the ascending order:

for (std::vector<int>::iterator it = myvector.begin(); it != myvector.end(); ++it)
   std::cout << ' ' << (*it)%(1 << 16);

Of course, this technique works only for the relatively small values in the original array myints (i.e. those which can fit into upper 2 bytes of int). But it has additional benefit of distinguishing identical values of myints: their indices will be printed in the right order.

Answer 13

I recently stepped upon the elegant projection feature of C++20 <ranges> and it allows to write shorter/clearer code:

std::vector<std::size_t> B(std::size(A));
std::iota(begin(B), end(B), 0);
std::ranges::sort(B, {}, [&](std::size_t i){ return A[i]; }); 

{} refers to the usual std::less<std::size_t>. So as you can see we define a function to call on each element before any comparaison. This projection feature is actually quite powerful since this function can be, as here, a lambda or it can even be a method, or a member value. For instance:

struct Item {
    float price;
    float weight;
    float efficiency() const { return price / weight; }
};

int main() {
    std::vector<Item> items{{7, 9}, {3, 4}, {5, 3}, {9, 7}};
    std::ranges::sort(items, std::greater<>(), &Item::efficiency);
    // now items are sorted by their efficiency in decreasing order:
    // items = {{5, 3}, {9, 7}, {7, 9}, {3, 4}}
}

If we wanted to sort by increasing price:

std::ranges::sort(items, {}, &Item::price);

Don't define operator< or use lambdas, use a projection!

Answer 14

Are the items in the vector unique? If so, copy the vector, sort one of the copies with STL Sort then you can find which index each item had in the original vector.

If the vector is supposed to handle duplicate items, I think youre better of implementing your own sort routine.

Answer 15

For this type of question Store the orignal array data into a new data and then binary search the first element of the sorted array into the duplicated array and that indice should be stored into a vector or array.

input array=>a
duplicate array=>b
vector=>c(Stores the indices(position) of the orignal array
Syntax:
for(i=0;i<n;i++)
c.push_back(binarysearch(b,n,a[i]));`

Here binarysearch is a function which takes the array,size of array,searching item and would return the position of the searched item

Answer 16

One solution is to use a 2D vector.

#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

int main() {
 vector<vector<double>> val_and_id;
 val_and_id.resize(5);
 for (int i = 0; i < 5; i++) {
   val_and_id[i].resize(2); // one to store value, the other for index.
 }
 // Store value in dimension 1, and index in the other:
 // say values are 5,4,7,1,3.
 val_and_id[0][0] = 5.0;
 val_and_id[1][0] = 4.0;
 val_and_id[2][0] = 7.0;
 val_and_id[3][0] = 1.0;
 val_and_id[4][0] = 3.0;

 val_and_id[0][1] = 0.0;
 val_and_id[1][1] = 1.0;
 val_and_id[2][1] = 2.0;
 val_and_id[3][1] = 3.0;
 val_and_id[4][1] = 4.0;

 sort(val_and_id.begin(), val_and_id.end());
 // display them:
 cout << "Index \t" << "Value \n";
 for (int i = 0; i < 5; i++) {
  cout << val_and_id[i][1] << "\t" << val_and_id[i][0] << "\n";
 }
 return 0;
}

Here is the output:

   Index   Value
   3       1
   4       3
   1       4
   0       5
   2       7