5

我有一个看起来像这样的表：

在此处输入图像描述

我正在寻找一个表格，它给出了字段中元素的频率计数l_0, l_1, l_2, l_3。

例如，输出应如下所示：

| author_id  | year | l_o.name         | l_0.count| l1.name    | l1.count | l2.name             | l2.count| l3.name            | l3.count|
| 2164089123 | 1987 | biology          | 3        | botany     | 3        |                     |         |                    |         |
| 2595831531 | 1987 | computer science | 2        | simulation | 2        | computer simulation | 2       | mathematical model | 2       |

编辑：

在某些情况下，数组字段可能有不止一种类型的元素。例如l_0可以是 ['biology', 'biology', 'geometry', 'geometry']. 在这种情况下，字段的输出l_0, l_1, l_2, l_3将是一个嵌套的重复字段，其中包含所有元素l_0.name以及l_0.count.

4

1 回答 1

8

这应该可以工作，假设您想基于每个数组计算：

SELECT
  author_id,
  year,
  (SELECT AS STRUCT ANY_VALUE(l_0) AS name, COUNT(*) AS count
   FROM UNNEST(l_0) AS l_0) AS l_0,
  (SELECT AS STRUCT ANY_VALUE(l_1) AS name, COUNT(*) AS count
   FROM UNNEST(l_1) AS l_1) AS l_1,
  (SELECT AS STRUCT ANY_VALUE(l_2) AS name, COUNT(*) AS count
   FROM UNNEST(l_2) AS l_2) AS l_2,
  (SELECT AS STRUCT ANY_VALUE(l_3) AS name, COUNT(*) AS count
   FROM UNNEST(l_3) AS l_3) AS l_3
FROM YourTable;

为避免如此多的重复，您可以使用 SQL UDF：

CREATE TEMP FUNCTION GetNameAndCount(elements ARRAY<STRING>) AS (
  (SELECT AS STRUCT ANY_VALUE(elem) AS name, COUNT(*) AS count
   FROM UNNEST(elements) AS elem)
);

SELECT
  author_id,
  year,
  GetNameAndCount(l_0) AS l_0,
  GetNameAndCount(l_1) AS l_1,
  GetNameAndCount(l_2) AS l_2,
  GetNameAndCount(l_3) AS l_3
FROM YourTable;

如果您可能需要考虑一个数组中的多个不同名称，则可以让 UDF 返回一个包含相关计数的数组：

CREATE TEMP FUNCTION GetNamesAndCounts(elements ARRAY<STRING>) AS (
  ARRAY(
    SELECT AS STRUCT elem AS name, COUNT(*) AS count
    FROM UNNEST(elements) AS elem
    GROUP BY elem
    ORDER BY count
  )
);

SELECT
  author_id,
  year,
  GetNamesAndCounts(l_0) AS l_0,
  GetNamesAndCounts(l_1) AS l_1,
  GetNamesAndCounts(l_2) AS l_2,
  GetNamesAndCounts(l_3) AS l_3
FROM YourTable;

请注意，如果您想跨行执行计数，则需要取消嵌套数组并GROUP BY在外部级别执行，但根据问题，这看起来并不是您的意图。

于 2018-01-24T00:42:13.463 回答