SQLite 窗口函数：当没有足够的窗口大小时给出 NULL 结果答案

【问题标题】：SQLite Window functions: give NULL result when sufficient window size is not availableSQLite 窗口函数：当没有足够的窗口大小时给出 NULL 结果
【发布时间】：2021-11-26 19:19:40
【问题描述】：

create table boll as select
     *, (avg(close) over win)-2*sqrt((avg(close*close) over win)-pow((avg(close) over win),2)) as BOLD,
     (avg(close) over win)+2*sqrt((avg(close*close) over win)-pow((avg(close) over win),2)) as BOLU
     from bhav
     window win as (partition by isin order by timestamp rows 19 preceding);

使用上述查询，如果当前行之前的行数少于 19 行，是否可以通过窗口 win 的计算返回 NULL？

【问题讨论】：

标签： sql sqlite count average window-functions

【解决方案1】：

只需使用LAG 回顾 19 个时间戳：

create table boll as 
  select *,
  case when lag(timestamp, 19) over (order by timestamp) is not null then
    avg(close) over win) - 2 * sqrt((avg(close*close) over win) - pow((avg(close) over win), 2)
  end as BOLD,
  case when lag(timestamp, 19) over (order by timestamp) is not null then
    avg(close) over win) + 2 * sqrt((avg(close*close) over win) - pow((avg(close) over win), 2) 
  end as BOLU
from bhav
window win as (partition by isin order by timestamp rows 19 preceding);

【讨论】：

【解决方案2】：

您可以在您定义的窗口上使用COUNT()窗口函数来检查有多少行：

create table boll as 
select *, 
       CASE WHEN COUNT(*) OVER win >= 19 THEN (avg(close) over win) - 2 * sqrt((avg(close*close) over win)-pow((avg(close) over win),2)) END as BOLD,
       CASE WHEN COUNT(*) OVER win >= 19 THEN (avg(close) over win)+2*sqrt((avg(close*close) over win)-pow((avg(close) over win),2)) END as BOLU
from bhav
window win as (partition by isin order by timestamp rows 19 preceding);

【讨论】：