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问题描述

有什么方法可以有效地将行扩展到列?

问题

我尝试分别用'where team = 1'和'where team = 2'过滤原始数据集,首先相应地获取数据集1和数据集2,然后在收入级别上加入两个数据集。但是,当income_level 具有太多不同的值时,会很不方便。有什么有效的方法可以得到我想要的结果吗?

标签: presto

解决方案


Prestodb 提供了一项map_agg功能,可以帮助您将长数据转换为您正在寻找的宽格式。不幸的是,似乎没有一种方法可以动态创建列名,但是这种方法应该比加入每个团队更有效(并且输入更少:))。

WITH raw_data AS (
  SELECT 1 AS team, 'a' AS income_level, 1 AS time, 11 AS ord
  UNION
  SELECT 1 AS team, 'b' AS income_level, 2 AS time, 12 AS ord
  UNION
  SELECT 1 AS team, 'c' AS income_level, 3 AS time, 13 AS ord
  UNION
  SELECT 2 AS team, 'a' AS income_level, 4 AS time, 14 AS ord
  UNION
  SELECT 2 AS team, 'b' AS income_level, 5 AS time, 15 AS ord
  UNION
  SELECT 2 AS team, 'c' AS income_level, 6 AS time, 16 AS ord
  UNION
  SELECT 3 AS team, 'a' AS income_level, 7 AS time, 17 AS ord
  UNION
  SELECT 3 AS team, 'b' AS income_level, 8 AS time, 18 AS ord
  UNION
  SELECT 3 AS team, 'c' AS income_level, 9 AS time, 19 AS ord
)

SELECT
  income_level,
  team_time[1] AS time_1,
  team_ord[1] AS ord_1,
  team_time[2] AS time_2,
  team_ord[2] AS ord_2,
  team_time[3] AS time_3,
  team_ord[3] AS ord_3
FROM (
  SELECT
    income_level,
    map_agg(team, time) AS team_time,
    map_agg(team, ord) AS team_ord
  FROM raw_data
  GROUP BY income_level
);

输出:

| income_level | time_1 | ord_1 | time_2 | ord_2 | time_3 | ord_3 |
|--------------|--------|-------|--------|-------|--------|-------|
| a            | 1      | 11    | 4      | 14    | 7      | 17    |
| b            | 2      | 12    | 5      | 15    | 8      | 18    |
| c            | 3      | 13    | 6      | 16    | 9      | 19    |

这个网站提供了另一个例子来说明如何做到这一点。


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