Question

我有一个每天上午9点，上午10点，上午11点等的日内价格清单（数字格式如15011 15012等）。

我想仅保留最近5天的观察结果以及从't'开始的接下来的5个可用天数，并删除其他所有内容。有没有办法做到这一点？

我尝试过使用如果日期＆lt; ＆amp; t - 5或日期＆gt; ＆amp; t + 5然后删除; 但是，由于有周末/假期，我没有得到我想要的所有观察结果。

提前多多感谢！

Answer 1

没有太多信息可以继续，但这是一个可能的解决方案：

/* Invent some data */
data have;
  do date=15001 to 15020;
     do time='09:00't,'10:00't,'11:00't;
        price = ranuni(0) * 10;
        output;
        end;
     end;
run;

/* Your macro variable identifying the target "date"  */
%let t=15011;

/* Subset for current and following datae*/
proc sort data=have out=temp(where=(date >= &t));
   by date;
run;

/* Process to keep only current and following five days */
data current_and_next5;
   set temp;
      by date;
   if first.date then keep_days + 1;  /* Set counter for each day */
   if keep_days <= 6;                 /* Days to keep (target and next five) */
   drop keep_days;                    /* Drop this utility variable */
run;

/* Subset for previous and sort by date descending */
proc sort data=have out=temp(where=(date < &t));
   by descending date;
run;

/* Process to keep only five previous days */
data prev5;
   set temp;
      by descending date;
   if first.date then keep_days + 1;  /* Set counter for each day */
   if keep_days <= 5;                 /* Number of days to keep */
   drop keep_days;                    /* Drop this utility variable */
run;

/* Concatenate together and re-sort by date */

data want;
   set current_and_next5
       prev5;
run;

proc sort data=want;
   by date;
run;

当然，此解决方案表明您的起始数据包含所有有效“交易日”的观察结果，并返回所有内容而不进行日期算术运算。更好的解决方案是要求您创建一个包含所有有效日期的“交易日历”数据集。你可以很容易地处理周末，但假期和其他“非交易日”是非常具体的网站;因此，使用日历几乎总是首选。

更新：Joe的评论让我更仔细地重新阅读了这个问题。这应该总共返回十一（11）天的数据;五天前，五天后和目标日期。但是，更好的解决方案是使用日历参考表。

Answer 2

试试这个

/* Get distinct dates before and after &T */
proc freq data=mydata noprint ;
  table Date /out=before (where=(Date < &T)) ;
  table Date /out=after (where=(Date > &T)) ;
run ;

/* Take 5 days before and after */
proc sql outobs=5 ;
  create table before2 as
  select Date 
  from before 
  order by Date descending ;

  create table after2 as
  select Date 
  from after 
  order by Date ;
quit ;

/* Subset to 5 available days before & after */
proc sql ;
  create table final as
  select *
  from mydata 
  where Date >= (select min(date) from before2)
    and Date <= (select max(date) from after2)
  order by Date ;
quit ;

SAS返回上次可用5天的数据

2 个答案: