在 BigQuery 中选择每天的最高计数

Question

我正在尝试编写一个查询，该查询将返回一系列 BigQuery 表中每天计数最高的条目。

我只写了以下查询，它返回每天的所有条目及其计数，按天排序，然后按条目最高的产品从高到低排序。

SELECT 
   STRFTIME_UTC_USEC(UTC_USEC_TO_day((ts-25200000)*1000),"%Y-%m-%d") AS day,
   products.id as product, 
   count(products.id) as num_entries
FROM 
   TABLE_DATE_RANGE([table_name_], timestamp('20170801'), timestamp(current_date()))
GROUP BY day, product
ORDER BY day, num_entries desc

例如

2017-08-01 . product A . 10
2017-08-01 . product B . 8
2017-08-01 . product C . 4
2017-08-01 . product D . 2
2017-08-02 . product X . 18
2017-08-02 . product Y . 15
2017-08-02 . product Z . 11
2017-08-03 . product N . 20
2017-08-03 . product M . 12
2017-08-03 . product N . 5
2017-08-03 . product O . 3
...

如何更改查询以仅返回每天的最高条目（最高 num_entries）？

例如

2017-08-01 . product A . 10
2017-08-02 . product X . 18
2017-08-03 . product N . 20
...

Answer 1

如果出于某种原因您仍然需要 BigQuery Legacy SQL - 请在下面使用
只需用很少的额外逻辑包装您的原始查询

#legacySQL
SELECT 
  day, 
  product, 
  num_entries
FROM (
  SELECT 
    day, 
    product, 
    num_entries, 
    ROW_NUMBER() OVER(PARTITION BY day ORDER BY num_entries DESC) AS win
  FROM (
  -- your original query START
    SELECT 
       STRFTIME_UTC_USEC(UTC_USEC_TO_day((ts-25200000)*1000),"%Y-%m-%d") AS day,
       products.id as product, 
       COUNT(products.id) as num_entries
    FROM 
       TABLE_DATE_RANGE([table_name_], TIMESTAMP('20170801'), TIMESTAMP(CURRENT_DATE()))
    GROUP BY day, product        )
  -- your original query END
)
WHERE win = 1  

同时，将 migrating 考虑为 BigQuery 标准 SQL
在您的情况下，查询将如下所示

#standardSQL
WITH days AS (
  SELECT
    PARSE_DATE('%Y%m%d', _TABLE_SUFFIX) AS day,
    products.id AS product,
    COUNT(*) AS num_entries
  FROM `table_name_*`
  WHERE _TABLE_SUFFIX >= '20170801'
  GROUP BY day, product
)
SELECT top.* FROM (
  SELECT ARRAY_AGG(days ORDER BY num_entries DESC LIMIT 1)[OFFSET(0)] AS top
  FROM days
  GROUP BY day
) 

您可以开始使用问题中的虚拟数据进行简化查询

#standardSQL
WITH days AS (
  SELECT '2017-08-01' AS day, 'product A' AS product, 10 AS num_entries UNION ALL
  SELECT '2017-08-01', 'product B', 8 UNION ALL
  SELECT '2017-08-01', 'product C', 4 UNION ALL
  SELECT '2017-08-01', 'product D', 2 UNION ALL
  SELECT '2017-08-02', 'product X', 18 UNION ALL
  SELECT '2017-08-02', 'product Y', 15 UNION ALL
  SELECT '2017-08-02', 'product Z', 11 UNION ALL
  SELECT '2017-08-03', 'product N', 20 UNION ALL
  SELECT '2017-08-03', 'product M', 12 UNION ALL
  SELECT '2017-08-03', 'product N', 5 UNION ALL
  SELECT '2017-08-03', 'product O', 3
)
SELECT top.* FROM (
  SELECT ARRAY_AGG(days ORDER BY num_entries DESC LIMIT 1)[OFFSET(0)] AS top
  FROM days
  GROUP BY day
)

结果如预期：

Row day         product     num_entries  
1   2017-08-01  product A   10   
2   2017-08-03  product N   20   
3   2017-08-02  product X   18   

Answer 2

这应该可以，但请注意，您应该使用standard SQL 进行查询：

#standardSQL
WITH ProductCounts AS (
  SELECT
    PARSE_DATE('%Y%m%d', _TABLE_SUFFIX) AS date,
    products.id AS product,
    COUNT(*) AS num_entries
  FROM `your_table_*`
  WHERE _TABLE_SUFFIX >= '20170801'
  GROUP BY date, product
)
SELECT
  date,
  ARRAY_AGG(product ORDER BY num_entries DESC LIMIT 1)[OFFSET(0)] AS top_product
FROM ProductCounts
GROUP BY date;