data a;
 	   input id  date total cod flounder haddock perch tuna;
 	   cards;
 	  1  1983	75	3.00	8.20	1.30	 .90	10.00
	  1  1984	80	4.60	8.80	2.60	1.40	11.20
	  1  1985	90	5.00	9.30	4.10	2.20	10.00
	  2  1983	100	6.50	12.20	7.90	4.00	15.70
	  2  1984	125	8.00	9.50	8.90	4.10	15.80
	  2  1985	134	11.20	15.20	22.70	4.90	26.20
	  2  1986	152	24.20	34.90	32.80	10.30	33.30
	  2  1997	177	22.60	41.70	31.90	15.90	58.70
	 ;
	run;
A Simple Transposition:
proc sort data=a;by id date;run; 
proc transpose prefix=total data=a out=b; 
by id; 
var total; 
id date; 
run; 
Result:
 

Example :

Arrays are a convenient way of grouping variables,and can be used for simple repetitive tasks, reshaping data sets, and \remembering" values from observation-to-observation. Arrays can be used to allow some traditional matrix-style programming techniques to be used in the data step.

Array Statement: Syntax:

ARRAY name<fnelemg> <$> <<elements <(initial-values)>>;

Examples:
array x x1-x3;
array check{5} _temporary_;
array miss{4} _temporary_ (9 9 99 9);
array dept $ dept1-dept4 ('Sales','Research','Training');
array value{3}; * generates value1, value2 and value3;

• All variables in an array must have the same type (numeric or character).
• An array name can't have the same name as a variable .
• You must explicitly state the number of elements when using
  _temporary_; in other cases SAS figures it out from context,generating new variables if necessary.
  
  Advanced Features of Arrays :

data new;
set old;
array x x1-x10;
array mval _temporary_ (9 9 9 9 9 99 99 99 99 99);
do i=1 to dim(x);
if x{i} = mval{i} then x{i} = .;
end;
run;

array class class1-class5;
total = 0;
do i = 1 to 5 until(total >= 10);
total = total + class{i};
end;
year = i;
if total lt 10 then year = .;

Consider a data set with 4 variables (x1-x4) stored as follows:
ID X1 X2 X3 X4
1 17 19 22 24
2 18 14 33 16
3 19 28 31 42

data new;
set old;
array xx x1-x4;
do time=1 to 4;
x = xx{time};
output;
end;
drop x1-x4;

data next;
set new;
by id;
array xx x1-x4;
retain xx;
if first.id then do i=1 to 4;xx{i} = .;end;
xx{time} = x;
if last.id then output;
drop x i;
run;

data  PRODUCTS;
set    PRODUCTS;
array dich[15]  prod1-prod15;

array name{15} $ _temporary_ ('p1',....,'p2' );

do i=1 to 15;
category=dich [i];
prod=name[i];
num_product=i;
output;
end;
keep sd10_numerator category num_product ;
run;

data BASKET_ANALYSES;
set zevet1.for_model(where=(tzvt_activ_form='300'));
array PR prod1-prod15;
m=0;
do i=1 to 15;
do j=i+1 to 15;
do k=j+1 to 15;

comb=PR[i]+PR[j]+PR[k];
name=compress(vname(aa[i])||'_'||vname(aa[j])||'_'||vname(aa[k]));
m+1;
if comb<3 then continue;
else output;

end;
end;
end;
run;

Now we wante to fix for each kind of treatment the most frequent combinations:

proc freq data=basket_analyses;
tables name*treatment/out=basket_result;
run;

proc sort data=basket_result;
by treatment descending count;
run;

data basket_result;
set basket_result;
by  treatment ;
retain counter ;
if first.treatment  then counter=1;
if treatment =lag(treatment) then counter=counter+1;
output;
run;

data frequent_resultes;
set basket_result;
if counter  <=4;
run;

data product_transpose;
set product_transpose_200110_200201;
array product

ms_cards
salary
l_miuazim
miuazim
num_hk
sum_hk 
loan
misgeret_ashrai
osher
pasiv;

array name{10} $ 16;
do i=1 to dim(product);
name[i]=vname(product[i]);
amount=product [i];
varname=name[i];
month_lag=month_lag6;
output;
end;
keep amount varname month_lag;
run;

proc sql;
select distinct
year into: period
separated by '+'
from example;
quit;

%macro trans;
%do i=1 %to 2;
DATA example (DROP=I year x y year1 year2 z);
;
ARRAY years {2} year1- year2;
ARRAY xs {2} x1-x2;
ARRAY ys {2} y1-y2;
DO I=1 TO 2 UNTIL (LAST.b);
SET dug;
BY b;
years {I}=year;
x_%scan(&period,&i,+)=xs{I};

xs{I}=x;
ys{I}=y;
END;

run;
%end;
%mend;
%trans;

Sometimes We want to create the same distribution .

DATA  SAME_DISTR;
SET SAME_DISTR;
IF TARGET =0 THEN DO;
ttt=ranuni(31311115)*100;
if ttt<=5 then kod=200502;
else if ttt<=28 then kod=200503;
else if ttt<=40 then kod=200504;
else if ttt<=52 then kod=200505;
else if ttt<=62 then kod=200506;
else if ttt<=71 then kod=200507;
else if ttt<=80 then kod=200508;
else if ttt<=88 then kod=200509;
else if ttt<=93 then kod=200510;
else if ttt<=100 then kod=200511;
end;
if target=1 then kod=100* year( Loan_Value_Date)+month(Loan_Value_Date);
run;

This is more wise way to do the same:

proc freq DATA = required_destribution;
tables Loan_Value_Date /out = outkod outcum noprint;
run; data _null_ ; length kod_str pct_str $5000;
set outkod end=eof;
retain kod_str pct_str;
kod_str=compress(kod_str||','||Loan_Value_Date);
pct_str=compress(pct_str||','||cum_pct);
if eof then do;
call symput('a1',substr(pct_str,2));
call symput('a2',substr(kod_str,2));
call symput('nn',_n_);
end;
run;

DATA SAME_DISTR;
SET SAME_DISTR;
array a1 {&nn} _temporary_ (&a1);
array a2 {&nn} _temporary_ (&a2);
IF TARGET =0 THEN DO;
ttt=ranuni(31311115)*100;
do i=1 to dim(a1);
if i=1 then do;
if ttt<=a1[i] then Loan_Value_Date=a2[i];
end;
else do;
if a1[i-1]<ttt<=a1[i] then Loan_Value_Date=a2[i];
end;
end;
end;

kod=100* year( Loan_Value_Date)+month(Loan_Value_Date);
run;