我的表有两个整数列:a
和b
。对于每一行,我想在b
值较小的行中选择a
的第n个最小值。这是一个输入/输出样本,n = 2。
输入:
a | b
-------
1 | 4
2 | 2
3 | 5
4 | 3
5 | 9
6 | 1
7 | 7
8 | 6
9 | 0
输出:
a | 2th min b
-------------
1 | null ← only 1 element in [4], no 2nd min
2 | 4 ← 2nd min between [4,2]
3 | 4 ← 2nd min between [4,2,5]
4 | 3 ← 2nd min between [4,2,5,3]
5 | 3 ← etc.
6 | 2
7 | 2
8 | 2
9 | 1
我在这里使用n = 2来保持简单,但在实践中,我想要第2000个最小值(或其他一些大的常数)。可以假定列a
包含不同的整数(甚至是1,2,3,......如果这样更容易)。
问题在于,如果我在我的窗口子句和ORDER BY b
中使用NTH_VALUE
,它只会在错误的值集上计算答案:
WITH data AS (
SELECT 1 AS a, 4 AS b
UNION ALL SELECT 2 AS a, 2 AS b
UNION ALL SELECT 3 AS a, 5 AS b
UNION ALL SELECT 4 AS a, 3 AS b
UNION ALL SELECT 5 AS a, 9 AS b
UNION ALL SELECT 6 AS a, 1 AS b
)
SELECT nth_value(b, 2) over (order by a)
from data
返回[null, 2, 2, 2, 2, 2]
:值按a
排序(因此顺序与它们出现的顺序相同),因此值b=2
始终是第二位的值。我想按a和然后订购b的第n个最小值。知道怎么用BigQuery(最好是标准SQL)写这个吗?
答案 0 :(得分:3)
下面是BigQuery Standard SQL,并为给定的示例生成正确的结果。
#standardSQL
WITH `project.dataset.table` AS (
SELECT 1 a, 4 b UNION ALL
SELECT 2, 2 UNION ALL
SELECT 3, 5 UNION ALL
SELECT 4, 3 UNION ALL
SELECT 5, 9 UNION ALL
SELECT 6, 1 UNION ALL
SELECT 7, 7 UNION ALL
SELECT 8, 6 UNION ALL
SELECT 9, 0
)
SELECT
a,
(SELECT b FROM
(SELECT b FROM UNNEST(c) b ORDER BY b LIMIT 2)
ORDER BY b DESC LIMIT 1
) b2
FROM (
SELECT a, IF(ARRAY_LENGTH(c) > 1, c, [NULL]) c
FROM (
SELECT a, ARRAY_AGG(b) OVER (ORDER BY a) c
FROM `project.dataset.table`
)
)
-- ORDER BY a
预期结果如下
Row a b2
1 1 null
2 2 4
3 3 4
4 4 3
5 5 3
6 6 2
7 7 2
8 8 2
9 9 1
注意:要使其适用于第2000个元素,您可以在LIMIT 2
快速更新
下面是一个看起来不那么丑陋的版本(当然是相同的输出)
#standardSQL
WITH `project.dataset.table` AS (
SELECT 1 a, 4 b UNION ALL
SELECT 2, 2 UNION ALL
SELECT 3, 5 UNION ALL
SELECT 4, 3 UNION ALL
SELECT 5, 9 UNION ALL
SELECT 6, 1 UNION ALL
SELECT 7, 7 UNION ALL
SELECT 8, 6 UNION ALL
SELECT 9, 0
)
SELECT a, c[SAFE_ORDINAL(2)] b2 FROM (
SELECT x.a, ARRAY_AGG(y.b ORDER BY y.b LIMIT 2) c
FROM `project.dataset.table` x
CROSS JOIN `project.dataset.table` y
WHERE y.a <= x.a
GROUP BY x.a
)
-- ORDER BY a
对于第2000个元素,将2
替换为2000
中的LIMIT 2
和SAFE_ORDINAL(2)
由于(现在)明确的CROSS JOIN