[英]Given a sequence of numbers how to identify the missing numbers
我想在数字序列中获取所有缺失的数字。
只是想知道是否有比下面更好的方法?
SELECT x
FROM
(
SELECT x,
LAG(x,1) OVER ( ORDER BY x ) prev_x
FROM
( SELECT * FROM
( SELECT 1 AS x ),
( SELECT 2 AS x ),
( SELECT 3 AS x ),
( SELECT 4 AS x ),
( SELECT 5 AS x ),
( SELECT 6 AS x ),
( SELECT 8 AS x ),
( SELECT 10 AS x ),
( SELECT 11 AS x )
)
)
WHERE x-prev_x > 1;
让我对你诚实!
任何其他可行的解决方案都会比提出问题更好 - 原因很简单 - 这是错误的! 它根本不返回丢失的数字! 它而是显示下一个间隙后的数字。 就是这样(希望你会明白我睁开了你的眼睛)
现在,关于更好的解决方案 - 有很多选择供您追求。
注意:以下选项仅适用于 BigQuery!
选项1
BigQuery 标准 SQL - 请参阅如何启用标准 SQL
WITH YourTable AS (
SELECT 1 AS x UNION ALL
SELECT 2 AS x UNION ALL
SELECT 3 AS x UNION ALL
SELECT 6 AS x UNION ALL
SELECT 8 AS x UNION ALL
SELECT 10 AS x UNION ALL
SELECT 11 AS x
),
nums AS (
SELECT num
FROM UNNEST(GENERATE_ARRAY((SELECT MIN(x) FROM YourTable), (SELECT MAX(x) FROM YourTable))) AS num
)
SELECT num FROM nums
LEFT JOIN YourTable ON num = x
WHERE x IS NULL
ORDER BY num
选项 2
BigQuery Legacy SQL您可以在下面尝试(这里您需要在 nums 表的选择表达式中设置开始/最小值和结束/最大值
SELECT num FROM (
SELECT num FROM (
SELECT ROW_NUMBER() OVER() AS num, *
FROM (FLATTEN((SELECT SPLIT(RPAD('', 11, '.'),'') AS h FROM (SELECT NULL)), h))
) WHERE num BETWEEN 1 AND 11
) AS nums
LEFT JOIN (
SELECT x FROM
(SELECT 1 AS x),
(SELECT 2 AS x),
(SELECT 3 AS x),
(SELECT 6 AS x),
(SELECT 8 AS x),
(SELECT 10 AS x),
(SELECT 11 AS x)
) AS YourTable
ON num = x
WHERE x IS NULL
选项 3
BigQuery Legacy SQL - 如果您不想依赖 min 和 max 并且需要设置这些值 - 您可以使用以下解决方案 - 它只需要设置足够高的 max 以适应您的预期增长(例如我放了 1000)
SELECT num FROM (
SELECT num FROM (
SELECT ROW_NUMBER() OVER() AS num, *
FROM (FLATTEN((SELECT SPLIT(RPAD('', 1000, '.'),'') AS h FROM (SELECT NULL)), h))
) WHERE num BETWEEN 1 AND 1000
) AS nums
LEFT JOIN YourTable
ON num = x
WHERE x IS NULL
AND num BETWEEN (SELECT MIN(x) FROM YourTable) AND (SELECT MAX(x) FROM YourTable)
选项 4(出于某种原因 - 到目前为止我最喜欢的)
BigQuery 标准 SQL - 没有显式连接
WITH YourTable AS (
SELECT 1 AS x UNION ALL
SELECT 2 AS x UNION ALL
SELECT 3 AS x UNION ALL
SELECT 6 AS x UNION ALL
SELECT 8 AS x UNION ALL
SELECT 10 AS x UNION ALL
SELECT 11 AS x
)
SELECT num
FROM (SELECT x, LEAD(x) OVER(ORDER BY x) AS next_x FROM YourTable),
UNNEST(GENERATE_ARRAY(x + 1,next_x - 1)) AS num
WHERE next_x - x > 1
ORDER BY x
您的查询可以写得更简洁,如下所示:
SELECT x
FROM (
SELECT x,
lag(x, 1) OVER ( ORDER BY x ) prev_x
FROM ( VALUES (1), (2), (3), (4), (5), (6), (8), (10), (11) ) v(x)
) sub
WHERE x-prev_x > 1;
这将返回缺失值 ( 8, 10
) 之后的下一个最高值,而不是缺失值本身 ( 7, 9
)。 但是,当然,您没有方便的值。
如果您知道序列中值的范围,则可以使用以下命令:
SELECT s.x
FROM generate_series(<<min>>, <<max>>) s(x)
LEFT JOIN my_table t ON s.x = t.x
WHERE t.x IS NULL;
这将返回实际的缺失值。
如果不知道取值范围,则需要添加子查询:
SELECT s.x
FROM ( SELECT min(x), max(x) FROM my_table ) r
JOIN generate_series(r.min, r.max) s(x) ON true
LEFT JOIN my_table t ON s.x = t.x
WHERE t.x IS NULL;
或者,而不是LEFT JOIN
:
SELECT x
FROM ( SELECT min(x), max(x) FROM my_table ) r,
generate_series(r.min, r.max) s(x)
WHERE NOT EXISTS (SELECT 1 FROM my_table t WHERE t.x = s.x);
Postgres 中最短的解决方案是使用标准 SQL EXCEPT
:
WITH tbl(x) AS (SELECT unnest ('{1,2,3,4,5,6,8,10,11}'::int[]))
-- the CTE provides a temp table - might be an actual table instead
SELECT generate_series(min(x), max(x)) FROM tbl
EXCEPT ALL
TABLE tbl;
集合返回函数unnest()
是 Postgres 特定的,并且是将您的一组数字作为表格提供的最短语法。
也适用于数据中的重复值或 NULL 值。
TABLE tbl
是(标准 SQL!) SELECT * FROM tbl
简短语法:
相关(有更多解释):
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