Is the a better way to do an average of a unordered_multimap key?

Question

I'm searching for get all value of one key in my unordered_multimap, which is actually stacking an int and a value who represent an execution time in nanosecs. I need to get all value to replace the multi keys by one key with an average.

I tried some code, this one is the most valuable actually :

std::unordered_multimap<int, std::chrono::nanoseconds> data;
std::chrono::nanoseconds average;
// *filling my map with value*
for (auto & pair : data){
  auto range = data.equal_range(pair.first);

  for_each (
    range.first,
    range.second,
    [](std::unordered_multimap<int, std::chrono::nanoseconds>::value_type& x){
      average = average + x.second;
    }
  );
  average = average / data.count(pair.first);
  data.erase(pair.first);
  data.insert({pair.first, average});
  }

The error i get error: 'average' is not captured : average = average + x.second;

Answer 1

Rather than std::for_each, use std::accumulate.

Note that removing entries from a std::unordered_multimap within a ranged-for over it is undefined behaviour. It's safer to fill a different container.

std::unordered_multimap<int, std::chrono::nanoseconds> data;
// *filling my map with value*

std::unordered_multimap<int, std::chrono::nanoseconds> new_data;
for (auto it = data.begin(); it != data.end(); ++it){
  auto range = data.equal_range(it->first);

  auto average = std::accumulate (
    range.first,
    range.second,
    std::chrono::nanoseconds{0},
    [](auto sum, auto & x){
      return sum + x.second;
    }
  ) / std::distance(range.first, range.second);

  new_data.emplace(key, average);
}
data.swap(new_data);

Alternatively, if you have C++17, std::transform_reduce.

  std::transform_reduce(
    range.first,
    range.second,
    std::chrono::nanoseconds{0},
    std::plus<>{},
    [](auto & x) { return x.second; }
  )