SAS Array <array-elements> to jump by 10 - sas

Question

I want to achieve the same output but instead of harcoding each of the array-element use something like var1 - var10 but that would jump by 10 like decades.
data work.test(keep= statename pop_diff:);
set sashelp.us_data(keep=STATENAME POPULATION:);
array population_array {*} POPULATION_1910 -- POPULATION_2010;
dimp = dim(population_array);
/* here and below something like:
array pop_diff_amount {10} pop_diff_amount_1920 -- pop_diff_amount_2010;*/
array pop_diff_amount {10} pop_diff_amount_1920 pop_diff_amount_1930
pop_diff_amount_1940 pop_diff_amount_1950
pop_diff_amount_1960 pop_diff_amount_1970
pop_diff_amount_1980 pop_diff_amount_1990
pop_diff_amount_2000 pop_diff_amount_2010;
array pop_diff_prcnt {10} pop_diff_prcnt_1920 pop_diff_prcnt_1930
pop_diff_prcnt_1940 pop_diff_prcnt_1950
pop_diff_prcnt_1960 pop_diff_prcnt_1970
pop_diff_prcnt_1980 pop_diff_prcnt_1990
pop_diff_prcnt_2000 pop_diff_prcnt_2010;
do i=1 to dim(population_array) - 1;
pop_diff_amount{i} = population_array{i+1} - population_array{i};
pop_diff_prcnt{i} = (population_array{i+1} / population_array{i} -1) * 100;
end;
RUN;
I am still beginner in it therefore I am not sure is this possible or easy to achieve.
Thanks!

Answer 1

Not automatic but not all that difficult either. First create a data set of the names then transpose and use an unexecuted set to bring in the names and then define arrays. Note how arrays are define using [*] and name: as you did with population_array.
data names;
do type = 'Amount','Prcnt';
do year=1920 to 2010 by 10;
length _name_ $32;
_name_ = catx('_','pop_diff',type,year);
output;
end;
end;
run;
proc print;
run;
proc transpose data=names out=pop_diff(drop=_name_);
var;
run;
proc contents varnum;
run;
data pop;
set sashelp.us_data(keep=STATENAME POPULATION:);
array population_array {*} POPULATION_1910 -- POPULATION_2010;
if 0 then set pop_diff;
array pop_diff_amount[*] pop_diff_amount:;
array pop_diff_prcnt[*] pop_diff_prcnt:;
do i=1 to dim(population_array) - 1;
pop_diff_amount{i} = population_array{i+1} - population_array{i};
pop_diff_prcnt{i} = (population_array{i+1} / population_array{i} -1) * 100;
end;
run;
proc print data=pop;
run;

Answer 2

SAS is automatically going to increment the array elements by 1. Here is an alternative solution that creates the variables using one extra step to create a set of macro variables that hold the desired variable names. Since you are basing them off of the variable POPULATION_<year>, we will simply grab the years from those variable names, create the variable names for the arrays that we want, and store them into a few macro variables.
proc sql noprint;
select cats('pop_diff_amount_', scan(name, -1, '_') )
, cats('pop_diff_prcnt_', scan(name, -1, '_') )
into :pop_diff_amount_vars separated by ' '
, :pop_diff_prcnt_vars separated by ' '
from dictionary.columns
where libname = 'SASHELP'
AND memname = 'US_DATA'
AND upcase(name) LIKE 'POPULATION_%'
;
quit;
data work.test(keep= statename pop_diff:);
set sashelp.us_data(keep=STATENAME POPULATION:);
array population_array {*} POPULATION_1910 -- POPULATION_2010;
dimp = dim(population_array);
array pop_diff_amount {*} &pop_diff_amount_vars.;
array pop_diff_prcnt {*} &pop_diff_prcnt_vars.;
do i=1 to dim(population_array) - 1;
pop_diff_amount{i} = population_array{i+1} - population_array{i};
pop_diff_prcnt{i} = (population_array{i+1} / population_array{i} -1) * 100;
end;
RUN;

Answer 3

Getting the data out of the meta data (create variable year) would make coding life easier.
proc transpose data=sashelp.us_data out=us_pop(rename=(col1=Population));
by statename;
var population_:;
run;
data us_pop;
set us_pop;
by statename;
year = input(scan(_name_,-1,'_'),4.);
pop_diff_amount=dif(population);
pop_diff_prcnt =(population/lag(population))-1;
format pop_diff_prcnt percent10.2;
if first.statename then call missing(of pop_diff_amount pop_diff_prcnt);
drop _:;
run;
proc print data=us_pop(obs=10);
run;