SAS 数组 跳跃 10

Question

我想实现相同的输出，但不是使用 var1 - var10 之类的东西对每个数组元素进行编码，但这会跳动 10 个，就像几十年一样。

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;

我仍然是初学者，因此我不确定这是否可能或容易实现。谢谢!

Answer 1

不是自动的，但也不是那么困难。首先创建名称的数据集，然后转置并使用未执行的集引入名称，然后定义数组。请注意如何使用 [*] 和 name: 定义数组，就像您对 Population_array 所做的那样。

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;