给定一个数字序列如何识别缺失的数字

Question

我想得到一个数字序列中所有缺失的数字。
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;

Answer 1

您的查询可以像这样写得更简洁：

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 return 是缺失值 (8, 10) 之后的下一个最高值，而不是缺失值本身 (7, 9)。但是当然你没有方便的值。

如果你知道序列中值的范围，那么你可以使用这个：

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 return 是实际缺失值。

如果不知道取值范围，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，代替 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);

Answer 2

老实说！
任何其他可行的解决方案都会比有问题的解决方案更好——出于一个简单的原因——这是错误的！它根本 return 没有遗漏号码！它宁愿显示下一个差距之后的数字。仅此而已（希望您会感激我让您大开眼界）

现在，关于更好的解决方案 - 有很多选择供您选择。
注意：以下选项仅适用于 BigQuery！

Option 1

BigQuery 标准 SQL - 请参阅 How to Enable Standard 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

Option 2

BigQuery Legacy SQL 你可以在下面尝试（这里你需要在 select 中设置 start/min 和 end/max 值nums table

的表达式

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

Option 3

BigQuery Legacy SQL - 如果您不想依赖最小值和最大值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最大值以适应您的预期增长（例如我放 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)

Option 4 (for some reason - my favorite so far)

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

Answer 3

最短 Postgres 中的解决方案是标准 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 特有的，它只是提供数字集合的最短语法 table。

也适用于数据中的重复值或 NULL 值。

TABLE tbl 是（标准 SQL!）SELECT * FROM tbl:

的简短语法

给定一个数字序列如何识别缺失的数字

Given a sequence of numbers how to identify the missing numbers

sql

postgresql

gaps-and-islands

google-bigquery

gaps-in-data