我要选择最后一行（按 mod_date )where列 new_status 与给定的上一个条目不同 object_id .
一开始我尝试了行数，但没有成功，后来我提出了超前/滞后函数，我认为我更接近解决方案，但仍然不是理想的结果。
下面是代码和小提琴：https://www.db-fiddle.com/f/ks8sai2wsajfflomd7t2it/0

CREATE TABLE changes
(object_id integer,
 new_status smallint,
 comment text,
 mod_date timestamp);

INSERT INTO changes
 VALUES
 (1001, 0, null, '2020-06-01 12:01'),
  (1001, 1, 'XYZ', '2020-06-01 12:05'),
   (1001, 1, 'YZX', '2020-06-01 12:11'),
    (1002, 1, 'XYZ', '2020-06-01 13:21'),
     (1002, 1, 'AAA', '2020-06-01 13:25'),
      (1002, 0, 'BCA', '2020-06-01 14:11'),
       (1003, 1, 'AXX', '2020-06-01 14:12'),
        (1003, 0, 'YZX', '2020-06-01 14:13'),
         (1003, 0, 'YYY', '2020-06-01 14:17');

SELECT object_id, min(mod_date), new_status FROM (
    SELECT 
      object_id
    , mod_date 
    , new_status
    --, row_number() over (partition BY object_id ORDER BY mod_date desc) rn
    , lag(new_status) OVER (partition by object_id ORDER BY mod_date desc) as next_status

    FROM changes
    ORDER BY 1)x
WHERE new_status = next_status 
OR next_status is null
GROUP BY 1,3

1001和1003的输出正常，1002的输出应该是状态为0的行。
感谢您的帮助和建议！