如何计算 BigQuery 中每个 ID 的列值的出现次数

Question

如果我在 BigQuery 中有这样的table

id    success
-------------
01    true
01    true
02    false
02    true

我想这样结束：

id    true    false
--------------------
01    2       0  
02    1       1

我知道了，但我想知道是否有更优雅的方法来做到这一点？

SELECT 
    t1.id,
    (
        SELECT 
            count(*)
        FROM `my_table` t2
        WHERE t2.success = true and t2.id = t1.id
    ) as trueCount,
    (
        SELECT 
            count(*)
        FROM `my_table` t3
        WHERE t3.success = false and t3.id = t1.id
    ) as falseCount
FROM `my_table` t1
GROUP BY id

Answer 1

您可以尝试使用条件聚合函数代替子查询。

SELECT id,
       COUNT(CASE WHEN success = true THEN 1 END) trueCount,
       COUNT(CASE WHEN success = false THEN 1 END) falseCount
FROM T
GROUP BY id

Answer 2

考虑应该适用于几乎所有数据库的条件聚合：

SELECT 
    t.id,
    SUM(CASE WHEN t.success = true THEN 1 ELSE 0 END) AS trueCount,
    SUM(CASE WHEN t.success = false THEN 1 ELSE 0 END) AS falseCount
FROM `my_table` t
GROUP BY t.id

Answer 3

考虑以下 BigQuery 方法

select * from your_table
pivot (count(*) for success in (true, false))

如果应用于您问题中的示例数据 - 输出为

如何计算 BigQuery 中每个 ID 的列值的出现次数

How to count occurrences of a column value per ID in BigQuery

sql

google-bigquery