php array conversion

You can do grouping easily if you use the grouper (client id) as the key of an array. The important thing is to check the existence of the key to decide between initializing or adding:

$nuevo = [];
foreach( $ventas as $i => &$x )
   if( !isset($nuevo[$x->id]) ){  // ¿Primera vez que se procesa el id?
      $x->codigos = [$x->codigo]; // Inicializa un arreglo de códigos
      unset( $x->codigo );        
      $x->primerId = $i;          // Temporalmente la llave final
      $nuevo[$x->id] = $x;
   }else                                      // Cliente ya procesado
      $nuevo[$x->id]->codigos[] = $x->codigo; // Sólo agregar código

Here the recommendation is to handle the codes as an array instead of a comma separated list. If required, at any time it is easy to do implode(',', $ventas[6]->codigos).

An example of the result for client 20 would be:

    [20] => stdClass Object
        (
            [id] => 20
            [cliente] => Juan Pérez
            [codigos] => Array
                (
                    [0] => 01/A2
                    [1] => 01/A3
                    [2] => 01/A1
                )
            [primerId] => 3
        )

If the original key ( primerId) is disposable, the following code is not needed to reassign:

// Cambiar por la llave inicial
foreach( $nuevo as $key => $val ){
   $id = $val->primerId;
   unset( $val->primerId, $nuevo[$key] );
   $nuevo[$id] = $val;
}

The end result should be something like this:

Array
(
    [3] => stdClass Object
        (
            [id] => 20
            [cliente] => Juan Pérez
            [codigos] => Array
                (
                    [0] => 01/A2
                    [1] => 01/A3
                    [2] => 01/A1
                )
        )

    [6] => stdClass Object
        (
            [id] => 19
            [cliente] => Autoescuelas Perez S.L
            [codigos] => Array
                (
                    [0] => 01/A3
                    [1] => 01/A1
                    [2] => 01/A2
                )
        )

    [9] => stdClass Object
        (
            [id] => 18
            [cliente] => bfsfdgsdfg
            [codigos] => Array
                (
                    [0] => 01/A2
                )
        )

I can think of 3 ways to solve it:

1.- Detect the odd ones through an if

2.- Create an if that saves the ids, and compare them in each round and validate that it is different from

3.- In your query order with an ORDER BY ID ASC/DESC and add a DISTINCT id to order the values ​​and remove your duplicates

This would be an example since I don't know what your complete structure is like, simulate the values ​​to obtain the odd ones

https://paiza.io/projects/LeuyO6y_dcGFLzP8JbDBlg

<?php
// Your code here!
    $array = [
       3 => [
            "id"=> 20,
            "cliente"=> "Juan Pérez",
            "codigo"=>"01/A2"
           ],
        4 => [
            "id"=> 20,
            "cliente"=> "Juan Pérez",
            "codigo"=>"01/A2"
           ],
        5 => [
            "id"=> 20,
            "cliente"=> "Juan Pérez",
            "codigo"=>"01/A2"
           ],
        6 => [
            "id"=> 19,
            "cliente"=> "Autoescuelas Perez S.L",
            "codigo"=>"01/A2"
           ],
        7 => [
            "id"=> 19,
            "cliente"=> "Autoescuelas Perez S.L",
            "codigo"=>"01/A3"
           ],
        8 => [
            "id"=> 19,
            "cliente"=> "Autoescuelas Perez S.L",
            "codigo"=>"01/A3"
           ],
        9 => [
            "id"=> 18,
            "cliente"=> "Juan Pérez",
            "codigo"=>"01/A2"
           ],
    ];
    
    
    $success = [];
    $index = 3;
    
    for($i = 0; $i < count($array); $i++)
    {
        if (($index % 2) != 0) 
        {
            $success[] = $array[$index];
        }
        $index++;
    }
    
    unset($success[0]);
    var_dump($success);
?>

The most recommended is to use the DISTINCT since it is easier, otherwise the next thing is to create the if and take the unique ids avoiding the duplicates, to maintain the consistency of the data, and the part of the odd ones can vary, you would have to analyze if in your odd ones, more duplicates may come, it is not highly recommended, but analyze and try to adapt any of these solutions