Zip several std::list iterators together

Question

Using the boost library it is possible to zip together a known number of iterators using a zip iterator, but what about when the number of iterators to be zipped is not known until runtime?

To expand a little bit, I have a list of lists that are all the same size, and I need to group together all the values at each each index and feed them into another operation. Right now this is all manual, and I feel like there should be a better way.

Example:

Say I have 3 lists:

• [1, 2, 3, 4, 5]
• [11, 12, 13, 14, 15]
• [21, 22, 23, 24, 25]

I need to transform these lists into:

• [1, 11, 12]
• [2, 12, 22]
• [3, 13, 23]
• [4, 14, 24]
• ... etc

I do not know how many lists are in the input until runtime.

Answer 1

Alright after spending almost 1/2 hour, I came up with this dynamic_zip_iterator class which can be further improved, to make it look like STL-like iterators. As of now, it's very specific, as I've hardcoded std::list in it which you can replace with std::vector or can make even more generic:

Anyway, have a look at it:

template<typename T>
struct dynamic_zip_iterator
{
   typedef typename std::list<T>::iterator list_iterator;
   std::list<list_iterator> iterators;
   std::list<std::list<T>> * plists;
   dynamic_zip_iterator(std::list<std::list<T>> & lists, bool isbegin) : plists(&lists) 
   {
         auto it = plists->begin();
         for( ; it != plists->end(); ++it)
         {
           if ( isbegin )
                iterators.push_back(it->begin()); 
           else
                iterators.push_back(it->end()); 
         }
   }
   dynamic_zip_iterator(const dynamic_zip_iterator & zip) : 
          plists(zip.plists),iterators(zip.iterators) {}

   dynamic_zip_iterator operator++()
   { 
     auto it = iterators.begin();
     for( ; it != iterators.end(); ++it)
          ++(*it);
     return *this;
   }
   std::list<T> operator*() 
   { 
     std::list<T> lst;
     auto it = iterators.begin();
     for( ; it != iterators.end(); ++it)
          lst.push_back(*(*it));     
     return lst;
   }
   bool operator!=(dynamic_zip_iterator &zip)
   { 
     auto it1 = iterators.begin();
     auto it2 = zip.iterators.begin();
     return (*it1) != (*it2);
   }
   static dynamic_zip_iterator begin(std::list<std::list<T>> & lists)
   {
      return dynamic_zip_iterator<T>(lists, true);
   }
   static dynamic_zip_iterator end(std::list<std::list<T>> & lists)
   {
      return dynamic_zip_iterator<T>(lists, false);
   }
};

Using it your problem reduces to this function:

std::list<std::list<int>> create_lists(std::list<std::list<int>>& lists)
{
  std::list<std::list<int>> results;
  auto begin = dynamic_zip_iterator<int>::begin(lists);
  auto end = dynamic_zip_iterator<int>::end(lists);
  for( ; begin != end ; ++begin)
  {
     results.push_back(*begin);
  }
  return results;    
}

Test code:

int main() {
        int a[] = {1, 2, 3, 4, 5}, b[] = {11, 12, 13, 14, 15}, c[] = {21, 22, 23, 24, 25};
        std::list<int> l1(a,a+5), l2(b,b+5), l3(c,c+5);
        std::list<std::list<int>> lists;
        lists.push_back(l1); 
        lists.push_back(l2);
        lists.push_back(l3);
        std::list<std::list<int>> newlists = create_lists(lists);
        for(auto lst = newlists.begin(); lst != newlists.end(); ++lst)
        {
                std::cout << "[";
                std::copy(lst->begin(), lst->end(), std::ostream_iterator<int>(std::cout, " "));
                std::cout << "]" << std::endl;
        }
        return 0;
}

Output:

[1 11 21 ]
[2 12 22 ]
[3 13 23 ]
[4 14 24 ]
[5 15 25 ]

Online demo : http://ideone.com/3FJu1

Answer 2

I am pretty sure nothing exists today for that. But why not create a very simple list of iterator elements ? That would do the trick, i am sure !

Create a function for all 3 parts of the for statement --> begin, end, increment

And that should be enough ! A bit more detail below.

• begin : (const ref to list of lists, reference to empty list of iterators) --> construct the iterator list using begin() for each sublist

• end : (const ref to list of iterators, const ref to list of lists) --> true if one iterator is at the end of its list

• increment : (ref to list of iterators) --> increment every iterator in the list