Finding the id whose average sum is maximum compared to other id - mysql

Question

I have a MySQL table named "Rating" which contains (id, rateable_id, rating, user_id) columns. rateable_id is the id of a record which is rated and rated score is stored in rating (star rating). All I want to do is to find particular rateable_id (grouped by) whose sum average is maximum among other rateable_id.
In the above sample table , i should get rateable_id = 26 because its rating is max ((3+3)/2) = 3 as compared to other rateable_id.
raw sql or eloquent any preferred.
Edit: Sorry to mention but i have done roughly in not standard way Anyway it returns the answer but i am looking using nested select answer. averageRating is willvincent package for counting avg sum of rating. $popular_post returns the id whose average rating sum maximum.
$posts = "App\Opportunity"::all(); //where(createdate < 1month)
$i=11;
$cr=0;
$pr=0;
$mi = 0; //max index
$maxR=0; //max rating value
for($i=0; $i< count($posts); $i++)
{
$cr = $posts[$i]->averageRating; //current rating
if($cr)<br>
{
if($cr > $maxR)
{
$maxR = $cr;
$mi = $i;
}
}
else
{
// echo "skip<br>";
}
}
$popular_post = ($posts[$mi]->id);

Answer 1

You are going to need 2 selects:
SELECT MAX(rating) rating, rateable_id FROM (
SELECT AVG(rating) rating, rateable_id FROM table GROUP BY reateble_id
) GROUP BY rateable_id ORDER BY rating LIMIT 1

Answer 2

I think this will help you little bit.
$ratings = Rating::all()->groupBy('rateable_id');
$topList = [];
foreach($ratings as $key => $value){
$topList[$key] = ($value->sum('rating')) / count($value);
}
arsort($topList); //sort the array by top values
$max_rateable_id = key($topList); //key of first item of the array
$max_rating = reset($topList); //value of first item of the array
$both = [$max_rateable_id => $max_rating]; //key and value of the first item of the array