问一个很小的问题

（手机发帖，格式乱，请见谅）

向各位老师请教一个小问题，例如有abc01ef这样的变量二十个（直到abc20ef）

现在我想在data步中描述这么一个问题：如果全部变量都为空，则为真。我想请老师用数组和宏变量的方式分别做一下，感激不尽。

附，和这个相似的问题我解决了：如果全部变量均非空，则为真，我的实现是这样的：

data a;
set a;
if missing(abc01ef + ... + abc20ef) then ;else 判断实现;
run;

谢谢。

data ;
retain s 1;
d1=.;
d2=.;
d3=.;
d4=.;
d5=.;
array var{5} d1-d5;
      do i=1 to 5;
         s=s * (var{i}=' ');
      end;
put s=;
/*if s=1 then do;  */
/*...statement...*/
/*end;*/
run;

data ex;
input abc01ef abc02ef abc03ef  ;
cards;
1 2 3
1 . 3
. . .
;
run;

/*数组运算*/
data ex1(drop=i j);
set ex;
count=0;
array  aa(*) abc01ef abc02ef abc03ef;
do i=1 to dim(aa);
if missing(aa(i)) then j=0;
else j=1 ;
count+j;
end;
if count=0 then put "true";
else put "false";
run;

/*用宏做个循环没多大意思，用n函数更方便直接*/
data ex1;
set ex;
if n(abc01ef,abc02ef,abc03ef)=0 then put "true";
else put "false";
run;

1. /*最简单方法*/
   
2. data b;
   
3. set a;
   
4. if n(x,y,z)=0 then hello="全为空";
   
5. run;
   
6. 
   
7. /*如果你非要用数组*/
   
8. data c;
   
9. set a;
   
10. array fuk(3) x--z;
    
11. point=0;
    
12. do i=1 to 3;
    
13. if missing(fuk(i))=0 then point+1;
    
14. end;
    
15. if point=0 then hello="全为空";
    
16. drop i point;
    
17. run;
    
18. 
    
19. /*至于宏，宏仅仅是文本替代，上述两个方法都可以改写成宏，只不过将简单的变麻烦了，不做赘述了*/

复制代码

不求别的，只为论坛币

ziyenano 发表于 2012-10-16 15:54
data ex;
input abc01ef abc02ef abc03ef  ;
cards;

我靠，你跟我师从的同一个祖师吧？

Imasasor 发表于 2012-10-16 16:01
我靠，你跟我师从的同一个祖师吧？

这种问题，着实想不出什么出彩的思路

Imasasor 发表于 2012-10-16 15:58
不求别的，只为论坛币

如你所愿了吧 O(∩_∩)O~

数据分析师3K 发表于 2012-10-16 16:22
如你所愿了吧 O(∩_∩)O~

谢谢啊！！！！

Using the sum function can bypass an looping of array index. It is not only simpler, but also it runs faster as well. See the comparison at bottom.

There is no macro necessary in this context.

data t1;
  array abc[*] abc01ef abc02ef abc03ef abc04ef;
  do i=1 to 5;
      
     do j=1 to dim(abc);
             if i in (1,4,5) then abc[j]=ranuni(123);
                 else abc[j]=.;
         end;
         output;
  end;
  keep abc:;
run;

data t2;
  set t1;
  array abc[*] abc01ef abc02ef abc03ef abc04ef;
  missflag=(sum(of abc[*])=.);
run;

proc print;run;

******************************
251  data t1;
252    array abc[*] abc01ef abc02ef abc03ef abc04ef;
253    do k=1 to 1e6;
254    do i=1 to 5;
255
256       do j=1 to dim(abc);
257           if i in (1,4,5) then abc[j]=ranuni(123);
258           else abc[j]=.;
259       end;
260       output;
261    end;
262    end;
263    keep abc:;
264  run;

NOTE: The data set WORK.T1 has 5000000 observations and 4 variables.
NOTE: DATA statement used (Total process time):
      real time           1.64 seconds
      cpu time            1.57 seconds


265
266  data t2;
267    set t1;
268    array abc[*] abc01ef abc02ef abc03ef abc04ef;
269    missflag=(sum(of abc[*])=.);
270  run;

NOTE: Missing values were generated as a result of performing an operation on missing values.
      Each place is given by: (Number of times) at (Line):(Column).
      2000000 at 269:13
NOTE: There were 5000000 observations read from the data set WORK.T1.
NOTE: The data set WORK.T2 has 5000000 observations and 5 variables.
NOTE: DATA statement used (Total process time):
      real time           1.57 seconds
      cpu time            1.35 seconds


271
272  data t2;
273
274    set t1;
275    array abc[*] abc01ef abc02ef abc03ef abc04ef;
276    s=1;
277    do i=1 to dim(abc);
278       s=s*missing(abc[i]);
279    end;
280    drop i;
281  run;

NOTE: There were 5000000 observations read from the data set WORK.T1.
NOTE: The data set WORK.T2 has 5000000 observations and 5 variables.
NOTE: DATA statement used (Total process time):
      real time           3.89 seconds
      cpu time            1.81 seconds