自上一个空值以来滞后获得第一个非空值

Question

下面是我试图在 Redshift 数据库中实现的示例。

我有一个变量 current_value，我想创建一个新列 value_desired 即：

如果前一行为空，则与 current_value 相同
如果前一行为非空值，则等于前一个非空值

这听起来很容易，但我还没有找到实现它的方法。

row_numb     current_value   value_desired
1
2
3            47              47
4
5            45              45
6
7
8            42              42
9            41              42
10           40              42
11           39              42
12           38              42
13
14           36              36
15
16
17           33              33
18           32              33

我试过使用 LAG() 函数，但我只能得到以前的值（不是 "non-null" 块中的第一个值），这是我的看法：

SELECT *
    , CASE WHEN current_value is not null and LAG(current_value) is null THEN current_value
           WHEN current_value is not null and LAG(current_value) is not null 
            THEN LAG(current_value)
      ELSE NULL END AS value_desired
  FROM test1

非常感谢任何帮助，谢谢。

Answer 1

使用 FIRST_VALUE() 而不是 LAG()

参考：FIRST_VALUE and LAST_VALUE Window Functions

Answer 2

这是正确的答案，给出了正确的结果。这里有一些巧妙的技巧，我建议你仔细看看，让我知道需要澄清的地方

根据您的问题创建测试数据。

drop table if exists test_table ;
create table test_table (row_num int,current_value int);
insert into test_table(row_num, current_value)
values
  (1,null),
(2,null),
(3,47),
(4,null),
(5,45),
(6,null),
(7,null),
(8 ,42),
(9 ,41),
(10,40  ),
(11,39 ),
(12,38 ),
(13,null),
(14,36),
(15,null),
(16,null),
(17 ,33),
(18,32  )
;

然后运行这个代码

SELECT DISTINCT
  j1.row_num,
  CASE WHEN j1.current_value IS NULL
    THEN NULL
  ELSE
    last_value(j2.current_value)
    OVER (
      PARTITION BY j1.row_num
      ORDER BY j2.row_num
      ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING ) END AS value_desired
FROM test_table AS j1
  LEFT JOIN (SELECT
               row_num,
               current_value,
               lag(current_value, 1)
               OVER (
                 ORDER BY row_num ) AS prev_cval
             FROM test_table) AS j2
    ON j1.row_num >= j2.row_num AND j2.current_value IS NOT NULL
       AND j2.prev_cval IS NULL
ORDER BY j1.row_num;

Answer 3

我们的测试数据

DROP TABLE IF EXISTS test_table;
CREATE TABLE test_table (
  row_num       INT,
  current_value INT
);

INSERT INTO test_table (row_num, current_value)
VALUES
  (1, NULL),
  (2, NULL),
  (3, 47),
  (4, NULL),
  (5, 45),
  (6, NULL),
  (7, NULL),
  (8, 42),
  (9, 41),
  (10, 40),
  (11, 39),
  (12, 38),
  (13, NULL),
  (14, 36),
  (15, NULL),
  (16, NULL),
  (17, 33),
  (18, 32);

我们知道的：

当current_value不为空且前面的current_value为空时，desired_value等于current_value
- 让我们将其称为 first_desired_value，根据下面的 q_first_desired_value 子查询
- 当那个first_desired_value不为null时，就是我们的desired_value
- 当 current_value 不为 null
当 current_value 不为空时，期望值是所有前面行中的 the last first_desired_value （在当前行旁边）并排除可能已添加到先前帧中的 NULL 值

综上所述，查询如下

WITH q_first_desired_value AS
(
    SELECT
      row_num,
      current_value,
      CASE WHEN LAG(current_value, 1)
                OVER (
                  ORDER BY row_num ) IS NULL
        THEN current_value
      ELSE NULL END AS first_desired_value
    FROM test_table
    ORDER BY row_num
)
SELECT
  row_num,
  current_value,
  CASE WHEN first_desired_value IS NOT NULL
    THEN first_desired_value
  WHEN current_value IS NOT NULL
    THEN LAST_VALUE(first_desired_value) IGNORE NULLS
    OVER (
    ORDER BY row_num ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING ) END AS desired_value
FROM q_first_desired_value;

自上一个空值以来滞后获得第一个非空值

lag to get first non null value since the previous null value

sql

lag

window-functions

amazon-redshift