Question

我要转换此数据：

id  value
---------
1   v1=10
1   v2=20
1   v3=30
2   v1=40

看起来像这样：

id  v1     v2     v3
---------------------------
1   10     20     30
2   40     null   null

有什么解决方案吗？我想解决它bigquery的标准sql。

Answer 1

SELECT
  id,
  MAX(IF(name = 'v1', value, NULL)) AS v1,
  MAX(IF(name = 'v2', value, NULL)) AS v2,
  MAX(IF(name = 'v3', value, NULL)) AS v3
FROM (
  SELECT
    SPLIT(value, '=')[OFFSET(0)] AS name,
    SPLIT(value, '=')[OFFSET(1)] AS value
  FROM dataset.table
)
GROUP BY id

Answer 2

以下几个选项-全部用于BigQuery Standrad SQL

选项1-硬编码枢轴

假设您预先知道要创建的列的编号和名称
在这种情况下，您需要做什么

#standardSQL
SELECT 
  id, 
  MAX(IF(key = 'v1', val, NULL)) v1,
  MAX(IF(key = 'v2', val, NULL)) v2,
  MAX(IF(key = 'v3', val, NULL)) v3
FROM `project.dataset.table`, 
UNNEST([STRUCT<key STRING, val STRING>(SPLIT(value,"=")[OFFSET(0)], SPLIT(value, "=")[OFFSET(1)])])
GROUP BY id

您可以使用下面的问题中的虚拟数据进行测试，操作

#standardSQL
WITH `project.dataset.table` AS (
  SELECT 1 id, 'v1=10' value UNION ALL
  SELECT 1, 'v2=20' UNION ALL
  SELECT 1, 'v3=30' UNION ALL
  SELECT 2, 'v1=40' 
)
SELECT 
  id, 
  MAX(IF(key = 'v1', val, NULL)) v1,
  MAX(IF(key = 'v2', val, NULL)) v2,
  MAX(IF(key = 'v3', val, NULL)) v3
FROM `project.dataset.table`, 
UNNEST([STRUCT<key STRING, val STRING>(SPLIT(value,"=")[OFFSET(0)], SPLIT(value, "=")[OFFSET(1)])])
GROUP BY id
-- ORDER BY id

结果符合预期：

Row     id      v1      v2      v3   
1       1       10      20      30   
2       2       40      null    null

选项2-动态旋转

如果您不知道列的数目和名称-您将需要首先生成类似于上述选项＃1中的脚本

您应该在下面运行以获得它

#standardSQL
SELECT CONCAT('SELECT id, ', 
   STRING_AGG(
      CONCAT('MAX(IF(key = "', key, '", val, NULL)) as ', key)
   ) 
   ,' FROM `project.dataset.table`, UNNEST([STRUCT<key STRING, val STRING>(SPLIT(value, "=")[OFFSET(0)], SPLIT(value, "=")[OFFSET(1)])]) GROUP BY id')
FROM (
  SELECT SPLIT(value, '=')[OFFSET(0)] key 
  FROM `project.dataset.table`
  GROUP BY key
)

例如，如果您要对同一个虚拟人运行

#standardSQL
WITH `project.dataset.table` AS (
  SELECT 1 id, 'v1=10' value UNION ALL
  SELECT 1, 'v2=20' UNION ALL
  SELECT 1, 'v3=30' UNION ALL
  SELECT 2, 'v1=40' 
)
SELECT CONCAT('SELECT id, ', 
   STRING_AGG(
      CONCAT('MAX(IF(key = "', key, '", val, NULL)) as ', key)
   ) 
   ,' FROM `project.dataset.table`, UNNEST([STRUCT<key STRING, val STRING>(SPLIT(value, "=")[OFFSET(0)], SPLIT(value, "=")[OFFSET(1)])]) GROUP BY id')
FROM (
  SELECT SPLIT(value, '=')[OFFSET(0)] key 
  FROM `project.dataset.table`
  GROUP BY key
)

您将获得以下字符串

SELECT id, MAX(IF(key = "v1", val, NULL)) AS v1,MAX(IF(key = "v2", val, NULL)) AS v2,MAX(IF(key = "v3", val, NULL)) AS v3 FROM `project.dataset.table`, UNNEST([STRUCT<key STRING, val STRING>(SPLIT(value, "=")[OFFSET(0)], SPLIT(value, "=")[OFFSET(1)])]) GROUP BY id

现在，如果您针对虚拟数据运行此脚本

#standardSQL
WITH `project.dataset.table` AS (
  SELECT 1 id, 'v1=10' value UNION ALL
  SELECT 1, 'v2=20' UNION ALL
  SELECT 1, 'v3=30' UNION ALL
  SELECT 2, 'v1=40' 
)
SELECT id, MAX(IF(key = "v1", val, NULL)) AS v1,MAX(IF(key = "v2", val, NULL)) AS v2,MAX(IF(key = "v3", val, NULL)) AS v3 FROM `project.dataset.table`, UNNEST([STRUCT<key STRING, val STRING>(SPLIT(value, "=")[OFFSET(0)], SPLIT(value, "=")[OFFSET(1)])]) GROUP BY id

您将获得与选项1相同的结果-但不是-现在最终为您动态生成了最终查询

选项3-展平键-值与枢轴

透视很酷，但在许多实际情况下，发现以下简单方法非常有用，并且更适合处理

#standardSQL
WITH `project.dataset.table` AS (
  SELECT 1 id, 'v1=10' value UNION ALL
  SELECT 1, 'v2=20' UNION ALL
  SELECT 1, 'v3=30' UNION ALL
  SELECT 2, 'v1=40' 
)
SELECT 
  id, 
  SPLIT(value,"=")[OFFSET(0)] key, 
  SPLIT(value, "=")[OFFSET(1)] val
FROM `project.dataset.table`

给出普通的香草键值扁平化结构

Row     id      key     val  
1       1       v1      10   
2       1       v2      20   
3       1       v3      30   
4       2       v1      40

SQL：如何根据其值透视行

2 个答案: