Question

我有一个如下所示的数据集：

    respondent_id   day_session daydiff
nmo876  11/19/2017  0
nmo876  11/20/2017  1
nmo876  11/21/2017  1
nmo876  11/23/2017  2
nmo876  11/24/2017  1
nmo876  11/25/2017  1
nmo876  11/26/2017  1
nmo876  11/27/2017  1
nmo876  11/28/2017  1
nmo876  11/29/2017  1
nmo876  11/30/2017  1
nmo876  12/1/2017   1
nmo876  12/2/2017   1
nmo876  12/3/2017   1
nmo876  12/4/2017   1
nmo876  12/5/2017   1
nmo876  12/6/2017   1
nmo876  12/7/2017   1
nmo876  12/8/2017   1
nmo876  12/9/2017   1
nmo876  12/10/2017  1
nmo876  12/11/2017  1
nmo876  12/12/2017  1
nmo876  12/13/2017  1
nmo876  12/14/2017  1
nmo876  12/15/2017  1
nmo876  12/16/2017  1
nmo876  12/17/2017  1
nmo876  12/18/2017  1
nmo876  12/19/2017  1
nmo876  12/20/2017  1
nmo876  12/23/2017  3
nmo876  12/24/2017  1
nmo876  12/26/2017  2
nmo876  12/27/2017  1
nmo876  12/28/2017  1
nmo876  12/29/2017  1
nmo876  12/30/2017  1
nmo876  12/31/2017  1
nmo876  1/2/2018    2
nmo876  1/3/2018    1
nmo876  1/4/2018    1
nmo876  1/5/2018    1

我想编写一个脚本，从用户可能有多个连续day_sessions的数据集中选择最大的连续day_sessions块，即daydiff = 1.对于nmo876，输出为27。

以下是更多数据，代码应该计算连续每日会话的最大块数。对于用户jkl567，输出将为37：

jkl567  11/19/2017  1
jkl567  11/20/2017  1
jkl567  11/21/2017  1
jkl567  11/22/2017  1
jkl567  11/23/2017  1
jkl567  11/24/2017  1
jkl567  11/25/2017  1
jkl567  11/26/2017  1
jkl567  11/27/2017  1
jkl567  11/28/2017  1
jkl567  11/29/2017  1
jkl567  11/30/2017  1
jkl567  12/1/2017   1
jkl567  12/2/2017   1
jkl567  12/3/2017   1
jkl567  12/4/2017   1
jkl567  12/5/2017   1
jkl567  12/6/2017   1
jkl567  12/7/2017   1
jkl567  12/8/2017   1
jkl567  12/9/2017   1
jkl567  12/10/2017  1
jkl567  12/11/2017  1
jkl567  12/12/2017  1
jkl567  12/13/2017  1
jkl567  12/14/2017  1
jkl567  12/15/2017  1
jkl567  12/16/2017  1
jkl567  12/17/2017  1
jkl567  12/18/2017  1
jkl567  12/19/2017  1
jkl567  12/20/2017  1
jkl567  12/21/2017  1
jkl567  12/22/2017  1
jkl567  12/23/2017  1
jkl567  12/24/2017  1
jkl567  12/25/2017  1
jkl567  12/26/2017  2
jkl567  12/28/2017  1
jkl567  12/29/2017  3
jkl567  1/1/2018    1
jkl567  1/2/2018    1
jkl567  1/3/2018    1
jkl567  1/4/2018    1

Answer 1

您可以减去row_number()以获取定义该组的常量值。获得每组的长度：

select respondent_id, (day_session - seqnum * interval '1 day') as grp, count(*) as days_in_row
from (select t.*,
             row_number() over (partition by respondent_id order by day_session) as seqnum
      from t
     ) t
group by respondent_id, (day_session - seqnum * interval '1 day');

然后，您可以使用distinct on为每位受访者获取最大值。我使用子查询：

select distinct on (respondent_id) t.*
from (select respondent_id, (day_session - seqnum * interval '1 day') as grp, count(*) as days_in_row
      from (select t.*,
                   row_number() over (partition by respondent_id order by day_session) as seqnum
            from t
           ) t
      group by respondent_id, (day_session - seqnum * interval '1 day')
     ) t
order by respondent_id, days_in_row desc;

严格地说，子查询不是必需的。我发现以这种方式突破逻辑更容易。

返回PostgreSQL中每个用户连续几天的最大块数

1 个答案: