Proc sql：基于回顾期的新客户和继续客户

作者: 关于贤的记忆
发布时间: 2024-12-09 05:02:49 (12天前)
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仙风道骨刘憨憨 | 2019-08-31 10-32

<div class =“post-text”itemprop =“text”> <P> 您似乎需要两个不同的“状态”变量，一个用于连续性的<strong> 过度 </强> 前一年和一年的连续性的<strong> 内 </强> 月。 </p> <P> 在SQL中，存在性反身相关的子查询结果可以是满足该行的行的案例测试的<strong> 过度 </强> 和的<strong> 内 </强> 标准。日期算术用于计算相隔天数和 <code> INTCK </code> 用于计算相隔数月： </p> <pre> <code> data have; input customer $ date& date9. item& $; format date date9.; datalines; wei 01feb2018 car wei 02feb2018 car wei 02mar2019 bike carlin 01feb2018 car carlin 05feb2018 bike carlin 07mar2018 bike carlin 01mar2019 car run; proc sql; create table want as select *, case when exists ( select * from have as inner where inner.customer=outer.customer and (outer.date - inner.date) between 1 and 365 ) then 'cont.' else 'new' end as status_year, case when exists ( select * from have as inner where inner.customer=outer.customer and outer.date > inner.date and intck ('month', outer.date, inner.date) = 0 ) then 'cont.' else 'new' end as status_month from have as outer ; quit; </code> </pre> </DIV>

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無口君 | 2019-08-31 10-32

<div class =“post-text”itemprop =“text”> <P> 您可以使用 <code> retain </code> ： </p> <pre> <code> proc sort data=test out=test2; by name type date; run; data test2 ; set test2; retain retain 'new'; by name type date; if first.type then retain='new'; else retain='con'; run; proc sort data=test2 out=test2; by name date; run; </code> </pre> <P> 输出： </p> <pre> <code> +--------+-----------+------+--------+ | name | date | type | retain | +--------+-----------+------+--------+ | carlin | 01FEB2018 | car | new | | carlin | 05FEB2018 | bike | new | | carlin | 01MAR2019 | car | con | | wei | 01FEB2018 | car | new | | wei | 02FEB2018 | car | con | | wei | 02MAR2019 | bike | new | +--------+-----------+------+--------+ </code> </pre> </DIV>

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