How do I programmatically render a menu's label?

Question

I'm trying to render out a menu with its label in Drupal 8. I am able to get the menu and its links inside of a preprocess function using a helper function, but I can't locate the menu's title.

Here is how I'm getting the menu. I'm able to add the return value as a twig variable $variable['menu']['menu-name'] and render out an <ul> anywhere I need to.

function get_menu($menu_name) {
  $menu_tree = \Drupal::menuTree();
  $parameters = $menu_tree->getCurrentRouteMenuTreeParameters($menu_name);
  $tree = $menu_tree->load($menu_name, $parameters);
  $manipulators = array(
    // Only show links that are accessible for the current user.
    array('callable' => 'menu.default_tree_manipulators:checkAccess'),
    // Use the default sorting of menu links.
    array('callable' => 'menu.default_tree_manipulators:generateIndexAndSort'),
  );
  $tree = $menu_tree->transform($tree, $manipulators);
  $menu = $menu_tree->build($tree);

  return array('#markup' => drupal_render($menu));
}

Answer 1

Here is the snippet to change the menu name:

 function example_preprocess_menu(&$variables) {
      $element = &$variables['items'];
      foreach ($element as &$item) {
        if ($item['original_link']->getMenuName() == 'menu-machine-name') {
        //For the menu label    
       $menu = entity_load('menu', 'menu-machine-name');
          $menu_label = $menu->label(); 
          if ($item['title'] == 'Example menu') {
            $item['title'] = 'New menu name';
          }
        }
      }
    }