根据与上一行值的差异跳过 bigquery 中的行

Question

假设下面的 table 是按值排序的 (DESC)，我如何才能 return 只有当前值与上一行中的值之差小于某个数字 x (例如 2)，并在第一次满足此条件后丢弃下一行

即return下面只有第1行和第2行，因为第3行和第2行的值相差（9.0-4.0=5.0）>2，所以我们跳过第3行和第4行

with table as (
    select 1 as id, "a" as name, 10.0 as value UNION ALL
    select 2, "b", 9.0 UNION ALL 
    select 3, "c", 4.0 UNION ALL 
    select 4, "d", 1.0 UNION ALL 
)

输出

id, name, value
1,   a,   10.0
2,   b,   9.0

Answer 1

我们可以使用 lag() 求差，结合 id<=id 和 max(difference)<=2 来过滤结果。

with t1 as (
    select 1 as id, 'a' as name, 10.0 as value UNION ALL
    select 2, 'b', 9.0 UNION ALL 
    select 3, 'c', 4.0 UNION ALL 
    select 4, 'd', 1.0  
)

select 
  a.id, a.name, a.value,
  max(b.value_diff) max_diff
from t1 a
join (select id, abs(coalesce(value - lag(value) over (order by id),0)) as value_diff from t1 )b
on  a.id >= b.id
group by a.id, a.name, a.value
having max(b.value_diff) <= 2;

根据与上一行值的差异跳过 bigquery 中的行

Skip rows in bigquery based on difference from value in previous row

sql

google-bigquery