Equivalent to python's set.pop() for C++'s unordered sets

Question

Does C++ have an equivalent to python's set.pop()? I've been looking at the documentation for unordered_sets here, but there doesn't seem to be a way to 1. Access an arbitrary element, and/or 2. Access + remove an arbitrary element (popping).

Answer 1

You can either pop the first element

auto i = *set.begin();
set.erase(set.begin());

or if you're overly concerned about the implementation-defined internal ordering of the buckets (hint: you probably shouldn't be), you could remove a random element with something like

#include <unordered_set>
#include <iostream>
#include <random>

int main()
{
  std::unordered_set<int> set{0, 1, 2, 3, 4, 5};
  std::default_random_engine ran{std::random_device{}()};

  auto it = set.begin();
  std::advance(it, std::uniform_int_distribution<>{0, set.size() - 1}(ran));

  std::cout << *it << '\n';
  set.erase(it);
}

The above isn't particularly efficient however and you might be better off by filling a std::vector, removing the duplicates, randomizing the order and then just pop_backing the elements.

#include <algorithm>
#include <vector>
#include <iostream>
#include <random>

int main()
{
  std::vector<int> vec{0, 1, 2, 3, 3, 4, 5, 5};
  std::sort(vec.begin(), vec.end());
  vec.erase(std::unique(vec.begin(), vec.end()), vec.end());

  std::shuffle(
    vec.begin(), 
    vec.end(), 
    std::default_random_engine{std::random_device{}()}
  );

  while (!vec.empty()) {
    std::cout << vec.back() << '\n';
    vec.pop_back();
  }
}

(n.b. depending on your platform random_device may not be a very good seed).