如何根据表中的某些条件为每个记录分配不同的单元格值？

Question

给定表中的某些条件，我必须为每条记录指定不同的单元格值。

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在图像中，column1= elig, column2=status, column3= type, column4 =metro, column5=quartile, column6=urb, column7=cell。我刚才写的前三个条件是一个if，然后是else (它们来自另一个表)。

但是，一旦我开始使用变量“四分位数”、“urb”和“类型”，我需要比我所写的代码更好的代码。

• 地铁将从1-4增加。
• 四分位数与每一条其他记录一起递增。
• 类型更改为一个组。

对于第一个块type = 1，对于第二个块type maybe be 4 or 5 (不一定是增量的)。下一个类型组可能是type 7,8,9,10。

只有单元格值对每条记录都有更改。我知道我可以替代宏var来缩短名称和保存类型，但是如何使这个代码更紧凑、更高效。

在此之前，非常感谢您。

 If elig=0 then cell=0;

 else if elig =1 then 
do;
   if status in ('2','3') then cell=1;
   else if ( status = ' ' and typec=25 )  then cell =2;
    else if (status ='1','4','') and (quartile = . )  then cell=2;
end;

     else if  elig= '1' and type =1 and metro eq='1' then 
do;
       if quartile = 1 and urb in ('1','2') then cell =1111;
else if quartile = 1 and urb = '3'  then cell =1112; 
else if quartile = 2 and urb in ('1','2') then cell =1121;
else if quartile = 2 and urb = '3'  then cell =1122; 
else if quartile =3 and urb in ('1','2') then cell =1141;
else if quartile = 3 and urb = '3'  then cell =1142; 
else if quartile = 4 and urb in ('1','2') then cell =1121;
else if quartile = 4 and urb = '3'  then cell =1172; 
end;
/*here will be 3 more blocks of code for metro =2,3,4* /
/*note type changes value after metro cycles through 4 iterations*/

    else if elig='1' and  type =('4','5') and metro eq='1 then 
do;
if quartile = 1 and urb in ('1','2') then cell =1211;
else if quartile = 1 and urb = '3'  then cell =1212; 
else if quartile = 2 and urb in ('1','2') then cell =1221;
else if quartile = 2 and urb = '3'  then cell =1222; 
else  if quartile =3 and urb in ('1','2') then cell =1241;
else if quartile = 3 and urb = '3'  then cell =1242; 
else if quartile = 4 and urb in ('1','2') then cell =1271;
else if quartile = 4 and urb = '3'  then cell =1272; 
end;
/*3 more blocks of code for metro =2,3,4* /
/*then type changes and metro=1 and so on*/
else if elig='1' and type  type =('7','8') and metro eq=1 then 
    do;  
/*more code until my groups end* /
   end;

Answer 1

从您发布的数据的外观来看，您只需重新编码原始变量并连接结果即可。

例如：

data out_data;
    set in_data;
length class_12 $2 class_3 class_4 $1 class $4.;

select (type);
    when (1) 
     do;
        select (metro);
            when (1) class_12 = '11';
            when (2) class_12 = '12';
            when (3) class_12 = '14';
            otherwise;
        end;
     end;
    when (4, 5) 
      do;
        select (metro);
            when (1) class_12 = '21';
            when (2) class_12 = '22';
            when (3) class_12 = '24';
            otherwise;
        end;
    end;
    otherwise;
end;
select (quartile);
    when (1) class_3 = '1';
    when (2) class_3 = '2';
    when (3) class_3 = '4';
    when (4) class_3 = '7';
    otherwise;
end;
select (urb);
    when (1, 2) class_4 = '1';
    when (3) class_4 = '2';
    otherwise;
end;
class = class_12||class_3||class_4;
run; 

添加附加条件，并在需要时引入进一步的变量。