在 BigQuery 中取消嵌套多个列答案

【问题标题】：Unnesting Multiple Columns in BigQuery在 BigQuery 中取消嵌套多个列
【发布时间】：2020-05-13 18:49:02
【问题描述】：

我在处理一个 UNNEST 查询时遇到问题。下面是一个查询示例，我当前得到的结果以及我希望从中得到的结果。

一点上下文，我目前正在上传的内容是将ID's 与A 隔开，而不是, 将其强制为字符串而不是数字，因为其中有多个ID同一个单元格。 Price 也被 , 隔开。正在上传的数据示例：

    Name    |    Date    |   Item_ID   |  Price
    John    |  4/17/2020 | 123A456A678 | 19.99,21.99,30.00
    Joe     |  4/17/2020 | 555A777A888 | 8.99,10.00,15.99
    Jake    |  4/18/2020 |   444A333   | 15.99,9.00
    John    |  4/18/2020 |     432     | 75.99
    Megan   |  4/18/2020 | 12A890A23A99| 5.99,6.99,9.99,10.00

这是尝试 UNNEST 之前表中数据的示例。下面是当前 UNNEST 查询的示例以及输出示例。

With data AS(
  SELECT
    Name,
    Date,
    SPLIT(Item_ID, 'A') AS Item_ID_Split,
    SPLIT(Price, ',') AS Price_Split
FROM
  Example.Table
SELECT
  Name,
  Date,
  Item_ID_Split,
  Price_Split
FROM data,
UNNEST(Item_ID_Split) Item_ID_Split WITH OFFSET pos1
UNNEST(Price_Split) Price_Split WITH OFFSET pos2

当前输出如下所示：

    Name   |    Date   |  Item_ID_Split | Price_Split
    John   | 4/17/2020 |      123       |   19.99
    John   | 4/17/2020 |      456       |   19.99
    John   | 4/17/2020 |      678       |   19.99
    John   | 4/17/2020 |      123       |   21.99
    John   | 4/17/2020 |      456       |   21.99
    John   | 4/17/2020 |      678       |   21.99
    John   | 4/17/2020 |      123       |   30.00
    John   | 4/17/2020 |      456       |   30.00
    John   | 4/17/2020 |      678       |   30.00
    Joe    | 4/17/2020 |      555       |   8.99
    Joe    | 4/17/2020 |      777       |   8.99
    Joe    | 4/17/2020 |      888       |   8.99
    Joe    | 4/17/2020 |      555       |   10.00
    Joe    | 4/17/2020 |      777       |   10.00
    Joe    | 4/17/2020 |      888       |   10.00
    Joe    | 4/17/2020 |      555       |   15.99
    Joe    | 4/17/2020 |      777       |   15.99
    Joe    | 4/17/2020 |      888       |   15.99
    Jake   | 4/18/2020 |      444       |   15.99
    Jake   | 4/18/2020 |      333       |   15.99
    Jake   | 4/18/2020 |      444       |   9.00
    Jake   | 4/18/2020 |      333       |   9.00
    John   | 4/18/2020 |      432       |   75.99
    Megan  | 4/18/2020 |      12        |   5.99
    Megan  | 4/18/2020 |      890       |   5.99
    Megan  | 4/18/2020 |      23        |   5.99
    Megan  | 4/18/2020 |      99        |   5.99
    Megan  | 4/18/2020 |      12        |   6.99
    Megan  | 4/18/2020 |      890       |   6.99
    Megan  | 4/18/2020 |      23        |   6.99
    Megan  | 4/18/2020 |      99        |   6.99
    Megan  | 4/18/2020 |      12        |   9.99
    Megan  | 4/18/2020 |      890       |   9.99
    Megan  | 4/18/2020 |      23        |   9.99
    Megan  | 4/18/2020 |      99        |   9.99
    Megan  | 4/18/2020 |      12        |   10.00
    Megan  | 4/18/2020 |      890       |   10.00
    Megan  | 4/18/2020 |      23        |   10.00
    Megan  | 4/18/2020 |      99        |   10.00

这是上面查询的当前输出。如您所见，有重复的 Item_IDs/Prices，我想要的结果如下：

    Name   |    Date   |  Item_ID_Split | Price_Split
    John   | 4/17/2020 |      123       |   19.99
    John   | 4/17/2020 |      456       |   21.99
    John   | 4/17/2020 |      678       |   30.00
    Joe    | 4/17/2020 |      555       |   8.99
    Joe    | 4/17/2020 |      777       |   10.00
    Joe    | 4/17/2020 |      888       |   15.99
    Jake   | 4/18/2020 |      444       |   15.99
    Jake   | 4/18/2020 |      333       |   9.00
    John   | 4/18/2020 |      432       |   75.99
    Megan  | 4/18/2020 |      12        |   5.99
    Megan  | 4/18/2020 |      890       |   6.99
    Megan  | 4/18/2020 |      23        |   9.99
    Megan  | 4/18/2020 |      99        |   10.00

这是我正在寻找在Item_ID_Split 和Price_Split 之间完全没有重复的结果。我试图将SPLIT 函数放在UNNEST 中，但我得到了相同的输出。我不完全确定如何做到这一点，所以任何帮助将不胜感激！

提前谢谢你！

【问题讨论】：

标签： sql google-bigquery

【解决方案1】：

你可以使用with offset:

SELECT Name, Date, Item_ID_Split, Price_Split
FROM data LEFT JOIN
     UNNEST(Item_ID_Split) Item_ID_Split WITH OFFSET pos1
     ON 1=1 LEFT JOIN
     UNNEST(Price_Split) Price_Split WITH OFFSET pos2
     ON pos1 = po2;

【讨论】：

这成功了！！！我在第二个UNNEST 之前的ON 1=1 下添加了LEFT JOIN，因为我假设这就是您的意思。但这完美解决了问题，非常感谢！我真的很感激！

【解决方案2】：

以下是 BigQuery 标准 SQL

#standardSQL
SELECT Name, Day, Splits.*
FROM (
  SELECT Name, Day, 
    ARRAY(
      SELECT AS STRUCT Item_ID_Split, Price_Split
      FROM UNNEST(SPLIT(Item_ID, 'A')) AS Item_ID_Split WITH OFFSET
      JOIN UNNEST(SPLIT(Price, ',')) AS Price_Split WITH OFFSET
      USING(OFFSET)
    ) AS arr
  FROM `project.dataset.table`
), UNNEST(arr) Splits

如果要应用到您的问题中的示例数据，如下例所示

#standardSQL
WITH `project.dataset.table` AS (
  SELECT 'John' Name, '4/17/2020' Day, '123A456A678' Item_ID,'19.99,21.99,30.00' Price UNION ALL
  SELECT 'Joe', '4/17/2020', '555A777A888','8.99,10.00,15.99' UNION ALL
  SELECT 'Jake', '4/18/2020', '444A333','15.99,9.00' UNION ALL
  SELECT 'John', '4/18/2020', '432','75.99' UNION ALL
  SELECT 'Megan', '4/18/2020', '12A890A23A99','5.99,6.99,9.99,10.00' 
)
SELECT Name, Day, Splits.*
FROM (
  SELECT Name, Day, 
    ARRAY(
      SELECT AS STRUCT Item_ID_Split, Price_Split
      FROM UNNEST(SPLIT(Item_ID, 'A')) AS Item_ID_Split WITH OFFSET
      JOIN UNNEST(SPLIT(Price, ',')) AS Price_Split WITH OFFSET
      USING(OFFSET)
    ) AS arr
  FROM `project.dataset.table`
), UNNEST(arr) Splits

输出是

Row Name    Day         Item_ID_Split   Price_Split  
1   John    4/17/2020   123             19.99    
2   John    4/17/2020   456             21.99    
3   John    4/17/2020   678             30.00    
4   Joe     4/17/2020   555             8.99     
5   Joe     4/17/2020   777             10.00    
6   Joe     4/17/2020   888             15.99    
7   Jake    4/18/2020   444             15.99    
8   Jake    4/18/2020   333             9.00     
9   John    4/18/2020   432             75.99    
10  Megan   4/18/2020   12              5.99     
11  Megan   4/18/2020   890             6.99     
12  Megan   4/18/2020   23              9.99     
13  Megan   4/18/2020   99              10.00

【讨论】：