postgres LAG() 使用错误的先前值
postgres LAG() using wrong previous value
获取以下数据和查询:
create table if not exists my_example(a_group varchar(1)
,the_date date
,metric numeric(4,3)
);
INSERT INTO my_example
VALUES ('1','2018-12-14',0.514)
,('1','2018-12-15',0.532)
,('2','2018-12-15',0.252)
,('3','2018-12-14',0.562)
,('3','2018-12-15',0.361);
select
t1.the_date
,t1.a_group
,t1.metric AS current_metric
,lag(t1.metric, 1) OVER (ORDER BY t1.a_group, t1.the_date) AS previous_metric
from
my_example t1;
产生以下结果:
+------------+---------+----------------+-----------------+
| the_date | a_group | current_metric | previous_metric |
+------------+---------+----------------+-----------------+
| 2018-12-14 | 1 | 0.514 | NULL |
| 2018-12-15 | 1 | 0.532 | 0.514 |
| 2018-12-15 | 2 | 0.252 | 0.532 |
| 2018-12-14 | 3 | 0.562 | 0.252 |
| 2018-12-15 | 3 | 0.361 | 0.562 |
+------------+---------+----------------+-----------------+
我预计单独的 a_group==2
行的 previous_metric 的值为 NULL
。但是,如您所见,该值显示为 0.532
,这是从上一行中提取的。我如何修改此查询以产生我预期的 NULL
值?
您需要将 LAG
与 a_group
上的分区一起使用,因为您需要来自特定帧的滞后值:
SELECT
t1.the_date,
t1.a_group,
t1.metric AS current_metric,
LAG(t1.metric, 1) OVER (PARTITION BY t1.a_group ORDER BY t1.the_date)
AS previous_metric
FROM my_example t1;
获取以下数据和查询:
create table if not exists my_example(a_group varchar(1)
,the_date date
,metric numeric(4,3)
);
INSERT INTO my_example
VALUES ('1','2018-12-14',0.514)
,('1','2018-12-15',0.532)
,('2','2018-12-15',0.252)
,('3','2018-12-14',0.562)
,('3','2018-12-15',0.361);
select
t1.the_date
,t1.a_group
,t1.metric AS current_metric
,lag(t1.metric, 1) OVER (ORDER BY t1.a_group, t1.the_date) AS previous_metric
from
my_example t1;
产生以下结果:
+------------+---------+----------------+-----------------+
| the_date | a_group | current_metric | previous_metric |
+------------+---------+----------------+-----------------+
| 2018-12-14 | 1 | 0.514 | NULL |
| 2018-12-15 | 1 | 0.532 | 0.514 |
| 2018-12-15 | 2 | 0.252 | 0.532 |
| 2018-12-14 | 3 | 0.562 | 0.252 |
| 2018-12-15 | 3 | 0.361 | 0.562 |
+------------+---------+----------------+-----------------+
我预计单独的 a_group==2
行的 previous_metric 的值为 NULL
。但是,如您所见,该值显示为 0.532
,这是从上一行中提取的。我如何修改此查询以产生我预期的 NULL
值?
您需要将 LAG
与 a_group
上的分区一起使用,因为您需要来自特定帧的滞后值:
SELECT
t1.the_date,
t1.a_group,
t1.metric AS current_metric,
LAG(t1.metric, 1) OVER (PARTITION BY t1.a_group ORDER BY t1.the_date)
AS previous_metric
FROM my_example t1;