我有一张看起来像这样的表:
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
可能是 ['biology', 'biology', 'geometry', 'geometry']
.在这种情况下,字段 l_0, l_1, l_2, l_3
的输出将是一个嵌套的重复字段,其中包含 l_0.name
中的所有元素以及 l_0.count
中的所有相应计数.
最佳答案
这应该有效,假设您想以每个数组为基础:
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;
为了避免如此多的重复,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数组中的多个不同名称,您可以让 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,如果要执行跨行计数,则需要取消嵌套数组并执行
GROUP BY
在外部层面,但看起来这不是您基于问题的意图。
关于google-bigquery - 如何计算bigquery数组字段中元素的频率,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48411331/