在分组时间内按病例计数

Question

我正在尝试按版本类型计算每周出现在我的数据中的不同 ID，但我不确定如何正确构建查询。

我希望按照以下方式制作一个表格：

      1.1     1.2     1.3    1.4
wk1     1       5       4      8
wk2     4       3       9      8
wk3     1       8       0      6

我尝试制作下面的查询，但它不会运行，因为它需要 group by 中的 Case 语句，然后它不会接受 count()。

  SELECT
  Case  when version like "1.1%" then Count(distinct ID)
     when version like "1.2%" then Count(distinct ID)
     when version like "1.3%" then Count(distinct ID)
     when version like "1.4%" then Count(distinct ID) end,
  CAST(((datediff(timestamp_pst,'2016-01-03') / 7)+1) as INT) as week_of_the_year
  FROM db.table
  where timestamp_pst >=  "2016-01-28"
  group by CAST(((datediff(timestamp_pst,'2016-01-03') / 7)+1) as INT)  
        order by week_of_the_year

Answer 1

  SELECT
    COUNT(DISTINCT (CASE WHEN version like '1.1%' THEN ID END)) as '1.1'
    ,COUNT(DISTINCT (CASE WHEN version like '1.2%' THEN ID END)) as '1.2'
    ,COUNT(DISTINCT (CASE WHEN version like '1.3%' THEN ID END)) as '1.3'
    ,COUNT(DISTINCT (CASE WHEN version like '1.4%' THEN ID END)) as '1.4'
  CAST(((datediff(timestamp_pst,'2016-01-03') / 7)+1) as INT) as week_of_the_year
  FROM aws_d3.iaanalytics_detail
  where timestamp_pst >=  "2016-01-28"
  group by CAST(((datediff(timestamp_pst,'2016-01-03') / 7)+1) as INT)  
        order by week_of_the_year

您想要使用“条件聚合”。这样做，case 语句实际上进入了聚合函数内部。因为您想要COUNT(DISTINCT)，您实际上需要通过在聚合中使用DISTINCT 关键字或通过制作派生表来做到这一点，因此只有不同的值出现，正如另一个答案所暗示的那样，但作为唯一的词它会避免重复是DISTINCT 我认为没有必要使用派生表使事情复杂化。

请注意，SUM(CASE WHEN blah THEN 1 ELSE 0 END) 将不为您工作，因为它会将所有出现的次数相加而不计算不同的值。此外，聚合函数会忽略空值，并且当您不包含 ELSE 语句时，如果不匹配，则 case 表达式 的值将为 NULL。

Answer 2

您可以将COUNT() 聚合函数与条件CASE 语句一起使用。

SELECT
    week_of_the_year
  , COUNT(CASE WHEN version LIKE '1.1%' THEN id END) AS v1_1
  , COUNT(CASE WHEN version LIKE '1.2%' THEN id END) AS v1_2
  , COUNT(CASE WHEN version LIKE '1.3%' THEN id END) AS v1_3
  , COUNT(CASE WHEN version LIKE '1.4%' THEN id END) AS v1_4
FROM (
  SELECT
    DISTINCT
      id
    , version
    , CAST(((datediff(timestamp_pst,'2016-01-03') / 7)+1) as INT) as week_of_the_year
  FROM aws_d3.iaanalytics_detail
  where timestamp_pst >= '2016-01-28'
  ) t
GROUP BY week_of_the_year
ORDER BY week_of_the_year

请注意，查询的DISTINCT 部分发生在派生表t 中。实际上不需要派生表，但我发现它是一个更简洁的解决方案，因为GROUP BY 子句不会重复相同的代码并使其更具可读性。这也引入了不能在聚合中完成的不同部分。

Answer 3

试试这个

SELECT
  SUM(Case  when version like "1.1%" then 1 ELSE 0 END) as '1.1',
  SUM(Case  when version like "1.2%" then 1 ELSE 0 END) as '1.2',
  SUM(Case  when version like "1.3%" then 1 ELSE 0 END) as '1.3', 
  SUM(Case  when version like "1.4%" then 1 ELSE 0 END) as '1.4',
  CAST(((datediff(timestamp_pst,'2016-01-03') / 7)+1) as INT) as week_of_the_year
  FROM aws_d3.iaanalytics_detail
  where timestamp_pst >=  "2016-01-28"
  group by CAST(((datediff(timestamp_pst,'2016-01-03') / 7)+1) as INT)  
        order by week_of_the_year