Question

我的数据看起来像

名称|来自| To_City |请求日期

安迪|巴黎|伦敦| 08/21/2014 12:00
     莉娜|科隆|柏林| 08/22/2014 18:00

安迪|巴黎|伦敦| 08/22/2014 06:00

丽莎|罗马| Neapel | 08/25/2014 18:00
     莉娜|罗马|伦敦| 08/21/2014 20:00

丽莎|罗马| Neapel | 08/24/2014 18:00

安迪|巴黎|伦敦| 08/25/2014 12:00

我想知道一个人在+/-一天内有多少相同的驱动器请求。我很想收到一张桌子说：

名称|来自| To_City | avg请求日期| #requests

安迪|巴黎|伦敦| 08/21/2014 21:00 | 2
     莉娜|科隆|柏林| 08/22/2014 18:00 | 1

丽莎|罗马| Neapel | 08/25/2014 06:00 | 2
     莉娜|罗马|伦敦| 08/21/2014 20:00 | 1

安迪|巴黎|伦敦| 08/25/2014 12:00 | 1

这是分组子句的结果。但是，通常可以编写这样一个条件来检查在初始请求的24小时内是否有多少相同的请求？到目前为止，我在Excel中下载数据并在那里进行，但是有很多数据，因此效率不高......

示例数据：

让我们先构建一个样本数据集：

select * from (select 'Andy' as name,'Paris' as f,'London' as to, '2014-08-21 12:00' as date),
(select 'Lena' as name,'Koln' as f,'Berlin' as to, '2014-08-22 18:00' as date),
(select 'Andy' as name,'Paris' as f,'London' as to, '2014-08-22 06:00' as date),
(select 'Lisa' as name,'Rome' as f,'Neapel' as to, '2014-08-25 18:00' as date),
(select 'Lena' as name,'Rome' as f,'London' as to, '2014-08-21 20:00' as date),
(select 'Lisa' as name,'Rome' as f,'Neapel' as to, '2014-08-24 18:00' as date),
(select 'Andy' as name,'Paris' as f,'London' as to, '2014-08-25 12:00' as date)

Answer 1

一种方法是使用RANGE窗口的窗口函数。为此，首先需要将日期转换为天数，因为RANGE要求排序列为连续数字。 PARTITION BY子句类似于GROUP BY - 它列出了定义＆＃34;相同＆＃34;的列。驱动请求（在您的情况下 - 名称，从和到）。然后，您只需使用COUNT（*）来计算此窗口内的天数。

select name, f, to, date, count(*) 
  over(partition by name, f, to
       order by day
       range between 1 preceding and 1 following) from (
select name, f, to, date, integer(timestamp(date)/1000000/60/60/24) day from
(select 'Andy' as name,'Paris' as f,'London' as to, '2014-08-21 12:00' as date),
(select 'Lena' as name,'Koln' as f,'Berlin' as to, '2014-08-22 18:00' as date),
(select 'Andy' as name,'Paris' as f,'London' as to, '2014-08-22 06:00' as date),
(select 'Lisa' as name,'Rome' as f,'Neapel' as to, '2014-08-25 18:00' as date),
(select 'Lena' as name,'Rome' as f,'London' as to, '2014-08-21 20:00' as date),
(select 'Lisa' as name,'Rome' as f,'Neapel' as to, '2014-08-24 18:00' as date),
(select 'Andy' as name,'Paris' as f,'London' as to, '2014-08-25 12:00' as date))

Answer 2

您可以截断日期以排除小时，分钟和秒。然后按该列分组

SELECT SUBSTR(STRING(date-of-request), 0, 9) AS day
FROM t1
GROUP BY day

BigQuery在一个时间间隔内选择数据

2 个答案: