Arrays, Indicator, missing array

[smend]

I have the following data set.  I would like to create another separate array c1_1-c1_10 and c2_1-c2_1 where they are assigned 1 from the start a{1, j} >0 till b{i, j}=1.  For example, a1_3 through a1_7 is equal to 1.  And, b1_5 is equal to 1.  Therefore, the array c1_1 - c1_10 would have c1_3 through c1_5 equal to 1 and rest equal to ".".   Let me know if you need further clarification.  Thank you.

data a;
input a1_1-a1_10 a2_1-a2_10 b1_1-b1_10 b2_1-b2_10;
datalines;
. . 1 1 1 1 1 . . . . . . . 1 1 1 1 1 . . . . . . 1 . . . . . . . . 1 . . . . .
run;

Would like to the result to look something like following data set.

data c;
input c1_1-c1_10 c2_1-c2_10;
datalines;
. . 1 1 1 1 . . . . . . . . 1 . . . . .
run;

c1_1  c1_2  c1_3 c1_4 c1_5 c1_6 c1_7 c1_8 c1_9 c1_10   c2_1 c2_2 c2_3 c2_4 c2_5 c2_6 c2_7 c2_8 c2_9 c2_10

.            .       1       1       1      1         .       .           .    .             .       .        .         .        1      .          .      .       .           .  

[ballardw]

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@smend wrote:

I have the following data set.  I would like to create another separate array c1_1-c1_10 and c2_1-c2_1 where they are assigned 1 from the start a{1, j} >0 till b{i, j}=1. 

For example, a1_3 through a1_7 is equal to 1. 

And, b1_5 is equal to 1. 

Therefore, the array c1_1 - c1_10 would have c1_3 through c1_5 equal to 1 and rest equal to ".". 

I am not seeing the impact role of the B variable, and what happens with there are gaps in a1_3 through a1_7 (such as a1_5 is missing)

Hint: cut back on the scale so you can provide more concrete examples, such as a1_1 to a1_5  instead of a1_10 and provide a few more worked examples.

Your statement:

Therefore, the array c1_1 - c1_10 would have c1_3 through c1_5 equal to 1 and rest equal to ".". 

looks a lot like c1_1-c1_10 is a copy of a1_1 to a1_10. If not provide why not.

Since you haven't shown any definitions for arrays that might help as to what you think an array in this instance would be. I see potential for at least 6 arrays in the solution.

[ChrisNZ · Accepted solution]

Like this?

data HAVE;
  input A1_1-A1_10 A2_1-A2_10 B1_1-B1_10 B2_1-B2_10;
datalines;
. . 1 1 1 1 1 . . . . . . . 1 1 1 1 1 . . . . . . 1 . . . . . . . . 1 . . . . .
run;

data WANT; 
  array A1_[10]; 
  array A2_[10]; 
  array B1_[10]; 
  array B2_[10]; 
  array C1_[10]; 
  array C2_[10]; 
  set HAVE;
  do I=whichn(1,of A1_[*]) to whichn(1,of B1_[*]);
    C1_[I]=1;
  end;
  do I=whichn(1,of A2_[*]) to whichn(1,of B2_[*]);
    C2_[I]=1;
  end;
run;

proc print noobs;
  var C:;
run;
  

C1_1	C1_2	C1_3	C1_4	C1_5	C1_6	C1_7	C1_8	C1_9	C1_10	C2_1	C2_2	C2_3	C2_4	C2_5	C2_6	C2_7	C2_8	C2_9	C2_10
.	.	1	1	1	1	.	.	.	.	.	.	.	.	1	.	.	.	.	.