Sql 对BigQuery数组类型执行循环和计算

我的原始数据B是INT64的数组：

我想计算B[n+1]-B[n]之间的差值，从而得到一个新的表，如下所示：

我发现我可以通过使用循环和IF条件以某种方式实现这一点：

DECLARE x INT64 DEFAULT 0;
LOOP
 SET x = x + 1
 IF(x < array_length(table.B))
 THEN INSERT INTO newTable (SELECT A, B[OFFSET(x+1)] - B[OFFSET(x)]) from table
 END IF;
END LOOP;

有人能告诉我怎么做吗？或者有更好的方法来解决这个问题

谢谢。

为此，您可以使用带有偏移量的

unnest（）

：

select id, a,
       array_agg(b_el - prev_b_el order by n) as b_diffs
from (select t.*, b_el, lag(b_el) over (partition by t.id order by n) as prev_b_el
      from t cross join
           unnest(b) b_el with offset n
     ) t
where prev_b_el is not null
group by t.id, t.a

为此，可以将

unnest（）与偏移量一起使用

：

select id, a,
       array_agg(b_el - prev_b_el order by n) as b_diffs
from (select t.*, b_el, lag(b_el) over (partition by t.id order by n) as prev_b_el
      from t cross join
           unnest(b) b_el with offset n
     ) t
where prev_b_el is not null
group by t.id, t.a

---

下面是实际工作-BigQuery

select * replace(
  array(select diff from (
      select offset, lead(el) over(order by offset) - el as diff
      from unnest(B) el with offset
    ) where not diff is null
    order by offset
  ) as B
)
from `project.dataset.table` t

如果要应用于问题中的样本数据，则输出为

---

以下内容实际有效-BigQuery

select * replace(
  array(select diff from (
      select offset, lead(el) over(order by offset) - el as diff
      from unnest(B) el with offset
    ) where not diff is null
    order by offset
  ) as B
)
from `project.dataset.table` t

如果要应用于问题中的样本数据，则输出为