How to use '%' in if ... then statements

[haukewiegand]

Probably an easy task for a pro but I got stuck with this trivial problem for quite some time now:

I have this table as a start:

data inTable;
input ICD $ Drugs MDD ;
datalines;
F111 . .
F60 . .
F132 . .
F323 . .
F331 . .

;

 I want to get the following results table:

data outtableDesired;
input ICD $ Drugs MDD ;
datalines;
F111 1.
F60 . .
F132 1 .
F323 . 1
F331 . 1

;

I tried the following:

data outtable;
set inTable;
if ICD in ('F11%','F13%') then do;
Drugs=1;
end;
if ICD in ('F32%','F33%') then do;
MDD=1;
end;
run;

Apparently '%' does not work in 'if ... then' procedures ... or? Any alternatives? Of course, my real table is much larger with many more options for ICD, that's why I have to write just the beginning of the character string and then '%'.

Thanks in advance for your time to help me with a solution! 

[novinosrin]

Shouldn't the syntax be

data outtable;
set inTable;
if ICD like ('F11%') or ICD like ('F13%') then do;
Drugs=1;
end;
run;

?

[Cynthia_sas · Accepted solution]

Hi:
Are you trying to use the % as a wildcard? The % and the _ work with the LIKE operator in a WHERE statement or clause. You cannot use them in an IF statement.
But, if you are searching for F11 or F13 in the beginning of a string (so your full value is something like F11AB or F1134 or F13BC, then using the =: logical operator can work in an IF statement.

See this example.

Cynthia

[haukewiegand]

Thanks, that's working well!

Hauke

[Tom]

Note that the : modifier works with the IN operator also.

if code in: ('F11' 'F13') then .... ;

[Cynthia_sas]

Hi, Tom:
Good point!
Cynthia

[haukewiegand]

Thanks, that really makes it easier!

Hauke