如何计算同一列中不同类别的日期之间的差异？

我相信有人会指出这是一个基本的重塑问题，但我猜我在Google上很穷，所以我提出了我的问题，直到Stack Overflow的神。如何计算同一列中不同类别的日期之间的差异？

我的问题是，我想计算来自同一列但在另一列（Rank）中具有不同值的两个日期（等级1和等级2）之间的天数。这里的数据集架构的一个例子：

User Date  Rank 
Bob 2016-12-01 3 
Bob 2016-12-07 2 
Bob 2016-12-10 1

我想是这样的：

User Date1  Date2  DaysBetween 
Bob 2016-12-07 2016-12-10  3

理想情况下，我想这样做的画面，但SQL/BigQuery是OK了。

来源

2016-12-17 dnaeye

难道一个用户曾经有超过一排的多个具有相同的等级？你完全忽略了第3名？或者你还想要3至2级的日子吗？在你的例子中，排名似乎只是日期的相反顺序。情况总是如此吗？ –

对于Tableau解决方案，我可能会使用表格计算。取决于上述问题的答案。 –

试试下面

WITH YourTable AS (
    SELECT 'Bob' AS User, DATE '2016-12-01' AS Date, 3 AS Rank UNION ALL 
    SELECT 'Bob' AS User, DATE '2016-12-07' AS Date, 2 AS Rank UNION ALL 
    SELECT 'Bob' AS User, DATE '2016-12-10' AS Date, 1 AS Rank 
) 
SELECT 
    User, 
    MAX(CASE WHEN Rank = 2 THEN Date END) AS Date1, 
    MAX(CASE WHEN Rank = 1 THEN Date END) AS Date2, 
    DATE_DIFF(MAX(CASE WHEN Rank = 1 THEN Date END), 
      MAX(CASE WHEN Rank = 2 THEN Date END), DAY) AS DaysBetween 
FROM YourTable 
GROUP BY User

注意 - 这是在BigQuery中Standard SQL

来源

2016-12-17 00:18:00

这个伎俩！谢谢，米哈伊尔！ – dnaeye

另一种选择

WITH YourTable AS (
    SELECT 'Bob' AS User, DATE '2016-12-01' AS Date, 3 AS Rank UNION ALL 
    SELECT 'Bob' AS User, DATE '2016-12-07' AS Date, 2 AS Rank UNION ALL 
    SELECT 'Bob' AS User, DATE '2016-12-10' AS Date, 1 AS Rank 
) 
SELECT 
    User, Date1, Date2, 
    DATE_DIFF(Date2, Date1, DAY) AS DaysBetween 
FROM (
    SELECT 
    User, Rank, Date as Date2, 
    LEAD(Date) OVER(PARTITION BY User ORDER BY Date DESC) AS Date1 
    FROM YourTable 
) 
WHERE Rank = 1

来源

2016-12-17 00:42:43

如何计算同一列中不同类别的日期之间的差异？

回答

相关问题