AboutPROC MODEL :有关样本外预测值(y_hat)递迭摆置的问题

I using insample file a by(group) to estimate a0 and a1 coef. and used those to predict threey (y_hate) under given outsample d, how to put those 3 y_hat  rooling outcom like below:

group	199111	199112	199201	199202
1	yhat(1,1)	yhat(1,2)	yhat(1,3)
2		yhat(2,1)	yhat(2,2)	yhat(2,3)

PROC MODEL DATA= a ;

        y=a0+a1*x;
FIT y / OUTEST=b1 ;
SOLVE y / DATA=d ESTDATA=b1 OUT=c1 ;
BY group;
RUN;
PROC PRINT DATA=c1 (DROP _MODE_  _ERRORS_) ;
RUN;
insample a.file

MON	group	y	x
199101	1	1	-3.45
199102	1	2	8.05
199103	1	2	-0.66
199104	1	3	0.19
199105	1	4	-0.54
199106	1	3	-5.59
199107	1	6	1.21
199108	1	7	-0.95
199109	1	2	6.54
199102	2	9	8.05
199103	2	8	-0.66
199104	2	7	0.19
199105	2	4	-0.54
199106	2	6	-5.59
199107	2	5	1.21
199108	2	4	-0.95
199109	2	3	6.54
199110	2	2	-2.33

out sample file d:

MON group       x 199110 1 2.33 199111 1 9 199112 1 7 199111 2 9 199112 2 7 199201 2 2
I hope results:	group 199111 199112 199201 199202 1 yhat(1,1) yhat(1,2) yhat(1,3) 2 yhat(2,1) yhat(2,2) yhat(2,3)

data a;
input mon group x;
cards;
199110 1 2.33
199111 1 9
199112 1 7
199111 2 9
199112 2 7
199201 2 2
;
run;

proc transpose data=a out=b(drop=_name_);
by group;
id mon;
run;

proc print data=b noobs;
run;


group _199110 _199111 _199112 _199201
1    2.33    9     7    .
2      .       9     7    2



是要这个结果么？

data a;
input mon group x;
cards;
199110 1 2.33
199111 1 9
199112 1 7
199111 2 9
199112 2 7
199201 2 2
;
run;

proc transpose data=a out=b(drop=_name_);
by group;
id mon;
run;

proc print data=b noobs;
run;


group _199110 _199111 _199112 _199201
1    2.33    9     7    .
2      .       9     7    2


是要这个结果么？

NO! NO!
y regress on x (using in sample to forecas outsample y_hat)
PROC MODEL DATA= a ;
        y=a0+a1*x;
FIT y / OUTEST=b1 ;
SOLVE y / DATA=d ESTDATA=b1 OUT=c1 ;
BY group;
RUN;

This is correct that i hope results :
group 1----->199110 to 199112
group 2----->199111 to 199201
-----------------------------------------------------
group        199110        199111        199112        199201
1        yhat(1,1)        yhat(1,2)        yhat(1,3)       
2                yhat(2,1)        yhat(2,2)        yhat(2,3)

那你只能预测199110，199111，199112的group1 的y
和199111，199112，199201的group2的y吧
1999202的怎么预测？

data a;
input mon group y x @@;
cards;
199101 1 1 -3.45
199102 1 2 8.05
199103 1 2 -0.66
199104 1 3 0.19
199105 1 4 -0.54
199106 1 3 -5.59
199107 1 6 1.21
199108 1 7 -0.95
199109 1 2 6.54
199102 2 9 8.05
199103 2 8 -0.66
199104 2 7 0.19
199105 2 4 -0.54
199106 2 6 -5.59
199107 2 5 1.21
199108 2 4 -0.95
199109 2 3 6.54
199110 2 2 -2.33  
;
run;

data d;
input month group x@@;
cards;
199110 1 2.33
199111 1 9
199112 1 7
199111 2 9
199112 2 7
199201 2 2
;
run;


PROC MODEL DATA= a ;
        y=a0+a1*x;
                by group;
FIT y / OUTEST=b1 ;
SOLVE y / DATA=d ESTDATA=b1 OUT=c1 ;
BY group;
RUN;

data c2(keep=month group y);
merge d c1;
by group;
run;


proc transpose data=c2 out=after(drop=_name_);
by group;
id month;
proc print noobs;
run;


是这样么？

Intheangel :Can you check the codes, I construct a macro program under your giving codes,
because I have lots of “a” file : totala1(start)---->totala135(end)
                      lots of   “d” file : totald1(start)---->totald135(end)
I hope all results of  “afters” file : after1(start)---->after135(end)
--------------------------------------------------
%macro country(cstart,cend)
%Do i=&cstart %to &cend;
procimportdatafile=”c:\temp\toatala.&i.csv”
out=a.&i
        dbms=csv
        replace;
getnames=no;
run;
procimportdatafile=”c:\temp\toatalb.&i.csv”
out=d.&i
        dbms=csv
        replace;
getnames=no;
run;
PROCMODEL DATA= a.&i ;
         y=a0+a1*x;
by group;
FIT y / OUTEST=b.&i ;
SOLVE y / DATA=d.&i ESTDATA=b.&i OUT=c.&i ;
BY group;
RUN;
Data cc.&i(keep=month group y);
Merge d.&i c.&i;
by group;
run;
proctranspose data=cc.&I out=after.&i(drop=_name_);
by group;
id month;
procexport data= after&i
outfile="c:\temp\after&i.csv"
dbms=csv
replace;
run;
%end;
%mend country;
Good luck today !

3qsir 发表于 2014-3-10 09:02
Intheangel :Can you check the codes, I construct a macro program under your giving codes,
because I ...

%macro country(cstart,cend)
%Do i=&cstart %to &cend;
proc import datafile=”c:\temp\toatala&i.csv”
out=a_&i
        dbms=csv
        replace;
getnames=yes;
run;
proc import datafile=”c:\temp\toatalb&i.csv”
out=d_&i
        dbms=csv
        replace;
getnames=no;
run;
PROC MODEL DATA= a_&i;
         y=a0+a1*x;
by group;
FIT y / OUTEST=b_&i ;
SOLVE y / DATA=d_&i ESTDATA=b_&i OUT=c_&i ;
BY group;
RUN;

Data cc_&i(keep=month group y);
Merge d_&i c_&i;
by group;
run;
proctranspose data=cc_&i out=after_&i(drop=_name_);
by group;
id month;
proc export data= after_&i
outfile="c:\temp\after&i.csv"
dbms=csv
replace;
run;
%end;
%mend country;




Without your resource file ，i can not sure about the code is right or not.

Still ERROR.try it !
Can your card  "data a" save as c:\temp\totala1.csv and your card "data d" ---->c: \temp\otald1.csv
then excute %macro(1,2)
--------------------------------------
ERROR: Expected semicolon not found.  The macro will not be compiled.
ERROR: A dummy macro will be compiled.
15         %macro country(cstart,cend)
16          %Do i=&cstart %to &cend;
17          proc import datafile=”c:\temp\toatala&i.csv”
18          out=a_&i
19                  dbms=csv
20                  replace;
21          getnames=yes;
22          run;
23          proc import datafile=”c:\temp\toatalb&i.csv”
24          out=d_&i
25                  dbms=csv
26                  replace;
27          getnames=no;
28          run;
29          PROC MODEL DATA= a_&i;
30                   y=a0+a1*x;
31          by group;
32          FIT y / OUTEST=b_&i ;
33          SOLVE y / DATA=d_&i ESTDATA=b_&i OUT=c_&i ;
34          BY group;
35          RUN;
36         
37         Data cc_&i(keep=month group y);
38          Merge d_&ic_&i;
39          by group;
40          run;
41          proctranspose data=cc_&i out=after_&i(drop=_name_);
42          by group;
43          id month;
44          proc export data= after_&i
45          outfile="c:\temp\after&i.csv"
46          dbms=csv
47          replace;
48          run;
49          %end;
ERROR: There is no matching %DO statement for the %END. This statement will be ignored.
50          %mend country;
51         % country(1,2)
           _
           180
ERROR 180-322: Statement is not valid or it is used out of proper order.

52         
53         GOPTIONS NOACCESSIBLE;

54         %LET _CLIENTTASKLABEL=;
55         %LET _CLIENTPROJECTPATH=;
56         %LET _CLIENTPROJECTNAME=;
57         %LET _SASPROGRAMFILE=;
58         
59         ;*';*";*/;quit;run;
60         ODS _ALL_ CLOSE;
61         
62         
63         QUIT; RUN;
---------------------------------------