我目前正在使用两个SAS数据集(样本和主数据)进行SAS编程。以下是为说明目的而创建的假设数据或虚拟数据,以通过SAS编程解决我的问题。我想从主数据集(测试)中提取样本数据集中id的数据。我给出了一个例子,其中很少有id作为样本数据集,我需要根据年度信息(主要表格中的所需输出)从主表(测试)中提取下一个12个月的信息。第三输出)。
以下是提取前12个月数据的代码,但我不知道如前几个月提取下一个12个月的记录,任何人都可以帮助我使用优化方式的SAS编程来解决这个问题。
proc sort data=test;
by id yearmonth;
run;
data result;
set test;
array prev_month {13} PREV_MONTH_0-PREV_MONTH_12;
by id;
if first.id then do;
do i =1 to 13;
prev_month(i)=0;
end;
end;
do i = 13 to 2 by -1;
prev_month(i)=prev_month(i-1);
end;
prev_month(1)=no_of_cust;
drop i prev_month_0;
retain prev_month:;
run;
data sample1;
set sample(drop=no_of_cust);
run;
proc sort data=sample1;
by id yearmonth;
run;
data all;
merge sample1(in=a) result(in=b);
by id yearmonth;
if a;
run;
一个样本数据集(数据集名称 - 样本)。
ID YEARMONTH NO_OF_CUST
1 200909 50
1 201005 65
1 201008 78
1 201106 95
2 200901 65
2 200902 45
2 200903 69
2 201005 14
2 201006 26
2 201007 98
一个主数据集 - 数据集名称(测试)(从帐户开始到日期的每个ID的年份庞大数据集。)
ID YEARMONTH NO_OF_CUST
1 200808 125
1 200809 125
1 200810 111
1 200811 174
1 200812 98
1 200901 45
1 200902 74
1 200903 73
1 200904 101
1 200905 164
1 200906 104
1 200907 22
1 200908 35
1 200909 50
1 200910 77
1 200911 86
1 200912 95
1 201001 95
1 201002 87
1 201003 79
1 201004 71
1 201005 65
1 201006 66
1 201007 66
1 201008 78
1 201009 88
1 201010 54
1 201011 45
1 201012 100
1 201101 136
1 201102 111
1 201103 17
1 201104 77
1 201105 111
1 201106 95
1 201107 79
1 201108 777
1 201109 758
1 201110 32
1 201111 15
1 201112 22
2 200711 150
2 200712 150
2 200801 44
2 200802 385
2 200803 65
2 200804 66
2 200805 200
2 200806 333
2 200807 285
2 200808 265
2 200809 222
2 200810 220
2 200811 205
2 200812 185
2 200901 65
2 200902 45
2 200903 69
2 200904 546
2 200905 21
2 200906 256
2 200907 214
2 200908 14
2 200909 44
2 200910 65
2 200911 88
2 200912 79
2 201001 65
2 201002 45
2 201003 69
2 201004 54
2 201005 14
2 201006 26
2 201007 98
所需的输出应如下所示,
ID YEARMONTH NO_OF_CUST AFTER_MONTH_1 AFTER_MONTH_2 AFTER_MONTH_3 AFTER_MONTH_4 AFTER_MONTH_5 AFTER_MONTH_6 AFTER_MONTH_7 AFTER_MONTH_8 AFTER_MONTH_9 AFTER_MONTH_10 AFTER_MONTH_11 AFTER_MONTH_12
1 200909 50 77 86 95 95 87 79 71 65 66 66 78 88
答案 0 :(得分:0)
Step1: 加入您的样本表与主(测试)表,并使用intnx获取未来12个月的所有值。
第2步:在“月之后”制作列名称
第3步: 转置以获得最终输出
proc sql;
create table abc as
select a.id,a.yearmonth,b.yearmonth as yearmonth1, b.no_of_cust
from
sample a
left join
test b
on a.id = b.id and a.yearmonth <= b.yearmonth <= intnx("month",a.yearmonth,12)
order by a.id,a.yearmonth,b.yearmonth;
quit;
data abc1(drop=col yearmonth1);
set abc;
by id yearmonth;
if first.yearmonth then col=-1;
col+1;
columns = compress("after_month_"||col);
run;
proc transpose data=abc1 out=abc2(rename=(after_month_0 = no_of_cust) drop=_name_);
by id yearmonth;
id columns;
var no_of_cust;
run;
或强>
如果您想在查询中进行更改,则可以使用以下代码。
proc sort data=test;
by id descending yearmonth;
run;
data result;
set test;
array after_month {13} after_MONTH_0-after_MONTH_12;
by id;
if first.id then do;
do i = 1 to 13;
after_month(i) = 0;
end;
end;
do i = 13 to 2 by -1;
after_month(i) = after_month(i-1);
end;
after_month(1) = NO_OF_CUST;
drop i after_MONTH_0;
retain after_MONTH:;
run;
data sample1;
set sample(drop=no_of_cust);
run;
proc sort data=result;
by id yearmonth;
run;
proc sort data=sample1;
by id yearmonth;
run;
data all;
merge sample1(in=a) result(in=b);
by id yearmonth;
if a;
run;
如有任何疑问,请与我联系。