topic SCAN(SUBSTRING) in SAS Enterprise Guide

SCAN(SUBSTRING)

Dsrountree — Tue, 15 Apr 2014 17:12:32 GMT

How do I write a function that will search for a character within a string then pull the 6 digits after the character including the character?

Ex:

0000M123450000

11M12345898989

222222m1234500

0M12345547

Each data line is different in length, but the output (work) should be

M12345

Keep in mind the character can be lowercase or uppercase.

In this situation the value is "M" and I'm looking for 6 digits including the M.

Thanks in advance.....

Re: SCAN(SUBSTRING)

Peter_C — Tue, 15 Apr 2014 17:38:07 GMT

Substr( inputstring, verify( inputstring, '012345679''), 6)

That VERIFY() function returns the position of the first character of the first parameter that is not in the second parameter. So in the example above it points to any non-numeric which means that M.

it is just about the most efficient of these string hunting functiins

(I am told)

Re: SCAN(SUBSTRING)

Reeza — Tue, 15 Apr 2014 17:38:30 GMT

Use a combination of find and substring.

You may need to play with the -1 and or 6 to get what you want, but this should be close (untested).

There's also Perl Expressions, but I avoid those

WANT = substr(word,findc(word, "M", 'TI')-1, 6);

Re: SCAN(SUBSTRING)

DBailey — Tue, 15 Apr 2014 17:39:03 GMT

newcol=upper(substring(oldcol,find(oldcol,'m',i),6));

Re: SCAN(SUBSTRING)

Dsrountree — Tue, 15 Apr 2014 18:24:24 GMT

Adjusting the value to -0 will account for the first position.

substr(t1.'PACKAGE ID'n,findc(t1.'PACKAGE ID'n, "M", 'TI')-0, 9)

Thanks Everyone - very Helpful

Re: SCAN(SUBSTRING)

Tom — Tue, 15 Apr 2014 18:31:58 GMT

You might need to use IF/THEN logic to handle cases that do not meet the pattern.

For this simple case of only one letter you could use the INDEXC() function to handle matching case.

loc = indexc(STRING,'mM');

if 0 < loc <= length(string)-5 then NEW = substr(STRING,loc,6) ;

Re: SCAN(SUBSTRING)

Peter_C — Tue, 15 Apr 2014 22:25:32 GMT

It shouldn't need conditional logic like IF or CASE when you are using FINDc or VERIFY.

Just extend the strings like

substr( inputstring !! " ", verify( inputstring !! "M", '1234567890'), 6 )

Or,

Substr( t1.'PACKAGE ID'n !! ' ', findc( t1.'PACKAGE ID'n !! 'M', 'ti' ), 9 )

by appending enough blanks and an M to the input string you can guarantee therr enough characters to extract from and an M will always be found in the dearch even if it is after the original 'package ID'

Nor should you need that -0 .

Was it 9 or 6 characters you wanted to extract?

Re: SCAN(SUBSTRING)

slchen — Wed, 16 Apr 2014 00:21:20 GMT

It seems that 'M/m' is only alpha character in string, so:

substr(string,anyalpha(string),6)

Re: SCAN(SUBSTRING)

Astounding — Wed, 16 Apr 2014 12:49:00 GMT

I hate to add to answers that are already plentiful and working, but this approach seemed interesting:

length newvar $ 6;

newvar = 'M' || scan(word, 2, 'Mm');

It will, however, return "M" with no digits following if the incoming word doesn't contain any M/M.

Re: SCAN(SUBSTRING)

Ksharp — Wed, 16 Apr 2014 13:05:48 GMT

But I like Perl Regular Expression.

For every pure programmer like Java , C++ ,they all use it to process string type variables . I recommend you to learn it, you will get a powerful tool .

data have;
input x : $40.;
cards;
0000M123450000
11M12345898989
222222m1234500
0M12345547
;
run;
data want;
 set have;
 xx=prxchange('s/.*([Mm]\d{5}).*/$1/',-1,x);
run;

Xia Keshan

Re: SCAN(SUBSTRING)

Dsrountree — Wed, 16 Apr 2014 14:17:39 GMT

Thanks again Gentlemen - i love learning new methods to parse data...

Keshan - I'll take a look at it, but the last time I touched Java was back in my college days.

Keep in mind - I'm new to this SAS thing, but love to learn..............

Re: SCAN(SUBSTRING)

Ksharp — Wed, 16 Apr 2014 14:52:37 GMT

You are going to love Perl Regular Expression.