Thanks for reply 虽然我没有彻底理解。
I will continue to raise a qustion. what happens when set 'have_2' only has one observation?
答案是 set 'have' 只有前四个观测。
what about set 'have_1' having first 4 observations while keeping 'have_2' unchanged, that is, 2 observations?
答案是 set 'have' 只有四个观测。
似乎这个code也可以套入set a; set b;的模式,即截取最少的观测数。从另一点来讲,set have_2 如何读取数据很令人惊奇。
举例问题1
[code:3bm1qh39]have_1 meet condition? have_2
obs1 Y obs1 -->output
obs2 N -->output
obs3 N -->output
obs4 N -->output
obs5 Y end of have_2 -->stop[/code:3bm1qh39]
推荐的文章也是sas的精髓文章之一。而猪头也颇有Ian之风呢!