Select text from brackets

[Christen]

Hi all! 
Pleas help me with my problem..i want select text from brackets, but i don't khow how to do it.
For example i have 3 values in one variable - var
What i have:
var:  1   I'm falling. (In all the good times I find) myself
        2   Longin (for change) and in the bad times (I fear myself) 

        3   Tell me something, boy (aren't you (tired) tryin' to fill that void?) 

What i want:

        1     In all the good times I find

        2     I fear myself

        3     aren't you (tired) tryin' to fill that void?

i wtite something like this, but for second value this dont work..
data New;
set Old;
str_pos = find(var, "(");
end_pos = find(var, ")", -length(var));
len = end_pos - str_pos;
want = substrn(var, str_pos+1, len-1);
run;

for second value my code prodused this: for change) and in the bad times (I fear myself  

Thank you!

[jimbarbour]

If there are two parenthetical expressions in one variable, do you consistently want the second one?

Jim

[Christen]

Hi Jim! Yes, i need the second one

[jimbarbour · Accepted solution]

Why don't you try the following program?

Jim

DATA	Lyrics;
	INFILE	Datalines	MISSOVER;
	LENGTH	Text		$1026;
	INPUT	Text	&	$;
DATALINES;
I'm falling. (In all the good times I find) myself
Longin (for change) and in the bad times (I fear myself) 
Tell me something, boy (aren't you (tired) tryin' to fill that void?) 
;
RUN;

DATA	Extracted_Text;
	DROP	_:;
	SET	Lyrics;

	_Left_Paren_Cnt						=	COUNTC(Text, '(');

	IF	_Left_Paren_Cnt					>	2					THEN
		DO;
			PUTLOG	'WARNING:  Encountered more left parentheses than expected.  Using whole record.';
		END;
	ELSE
	IF	_Left_Paren_Cnt					=	2					THEN
		DO;
			_First_Right_Paren			=	INDEX(Text, ')');
			_Next_Right_Paren			=	INDEX(SUBSTR(Text, _First_Right_Paren + 1), ')') + _First_Right_Paren;
			_First_Left_Paren			=	INDEX(Text, '(');
			_Next_Left_Paren			=	INDEX(SUBSTR(Text, _First_Left_Paren + 1), '(') + _First_Left_Paren;
			IF	_Next_Left_Paren		<	_First_Right_Paren	THEN
				DO;
					Text				=	SUBSTR(Text, (_First_Left_Paren + 1), (_Next_Right_Paren - _First_Left_Paren - 1));
				END;
			ELSE
				DO;
					Text				=	SUBSTR(Text, (_Next_Left_Paren + 1), (_Next_Right_Paren - _Next_Left_Paren - 1));
				END;
		END;
	ELSE
	IF	_Left_Paren_Cnt					=	1					THEN
		DO;
			Text						=	SUBSTR(Text, INDEX(Text, '(') + 1, INDEX(Text, ')') - INDEX(Text, '(') - 1);
		END;
	ELSE
		DO;
			PUTLOG	'WARNING:  No left parentheses are in the text.  Using whole record.';
		END;
RUN;

[Christen]

Thank you very much, Jim!!!
It is work. But, I saw that the data have errors, for example, an unclosed parenthesis, which looks like this: Longin (for change) and in the bad times (I (fear myself)

Сould you please help me to account for this problem?
In this case i need to get someting like that: (I (fear myself) )

[jimbarbour]

OK, that should not be too hard.  I have some things to do right now, but let me look at this in a little be later.

Jim

[Christen]

Ok, thank you! )

[jimbarbour]

Hi, @Christen,

Lot's of things to do today, but I finally got a chance to look at the code.  Please try the below revision.

@ChrisNZ has a good point that sometimes it's good to clean up the data instead of spending too much time writing overly complex code.  However, do please try the program below and let me know how it goes.

Jim

%LET	Max_Paren							=	3;
%LET	db									=	*;

DATA	Extracted_Text;
	DROP	_:;
	SET	Lyrics;
	ARRAY	_L_Paren	[*]		_L_Paren1	-	_L_Paren&Max_Paren;
	ARRAY	_R_Paren	[*]		_R_Paren1	-	_R_Paren&Max_Paren;

	LINK	Get_Parens;

	IF	_L_Paren_Cnt						>	1							THEN
		DO;
			IF	_L_Paren[_L_Paren_Cnt]		<	_R_Paren[_L_Paren_Cnt - 1]	THEN
				DO;
					Text					=	SUBSTR(Text, _L_Paren[_L_Paren_Cnt - 1] + 1, _R_Paren[_L_Paren_Cnt] - _L_Paren[_L_Paren_Cnt - 1] - 1);
&db					PUTLOG	"&Nte1  A. "  _L_Paren_Cnt=  _L_Paren[_L_Paren_Cnt - 1]=  _R_Paren[_L_Paren_Cnt]=  Text=;
				END;
			ELSE
				DO;
					Text					=	SUBSTR(Text, _L_Paren[_L_Paren_Cnt] + 1, _R_Paren[_L_Paren_Cnt] - _L_Paren[_L_Paren_Cnt] - 1);
&db					PUTLOG	"&Nte1  B. "  _L_Paren_Cnt=  _L_Paren[_L_Paren_Cnt]=  _R_Paren[_L_Paren_Cnt]=  Text=;
				END;		
		END;
	ELSE
		IF	_L_Paren_Cnt					=	1							THEN
			DO;
				Text						=	SUBSTR(Text, _L_Paren[1] + 1, _R_Paren[1] - _L_Paren[1] - 1);
&db				PUTLOG	"&Nte1  C. "  _L_Paren_Cnt=  _L_Paren[1]=  _R_Paren[1]=  Text=;
			END;
	ELSE
		DO;
			PUTLOG	'WARNING:  No left parentheses are in the text.  Using whole record.';
		END;

	******;
	RETURN;
	******;

	**********;
	Get_Parens:
	**********;
		_Temp								=	LENGTHN(Text);
&db		PUTLOG	"&Nte2  ";
&db		PUTLOG	"&Nte1  "  _N_=  _Temp=  ' (Text length)';

		_L_Paren_Cnt						=	COUNTC(Text, '(');
		_p									=	0;
		DO	_i								=	1	TO	_L_Paren_Cnt	
			UNTIL						(_p	=	0); 
			_p								=	FIND(Text, '(', (_p + 1));
			_L_Paren[_i]					=	_p;
		END;

		_R_Paren_Cnt						=	COUNTC(Text, ')');
		_p									=	0;
		DO	_i								=	1	TO	_R_Paren_Cnt	
			UNTIL						(_p	=	0); 
			_p								=	FIND(Text, ')', (_p + 1));
			_R_Paren[_i]					=	_p;
		END;

		IF	_L_Paren_Cnt					=	_R_Paren_Cnt	+	1	THEN
			DO;
				_R_Paren[_L_Paren_Cnt]		=	LENGTHN(Text)	+	1;
&db				PUTLOG	"&Nte1  D. "  _R_Paren[_L_Paren_Cnt]=  ' (Text length)';
			END;
	******;
	RETURN;
	******;
RUN;

[Christen]

Thank you! It works=)

[jimbarbour]

OK, great. 👍  Perhaps we can mark this topic as solved, then?

Jim

[ChrisNZ]

One way:

data HAVE;
  input STR :& $80.;
cards;
I'm falling. (In all the good times I find) myself
Longin (for change) and in the bad times (I fear myself) 
Tell me something, boy (aren't you (tired) tryin' to fill that void?) 
run;

data WANT;
  set HAVE;
  STR1=prxchange('s/ ( \( [^(]* ) \( ( [^)]* ) \) ( [^)]* \) ) /\1^\2#\3/x',-1,STR);
  STR2=prxchange('s/ .* \( ( .* ) \) [^)(]* \Z/\1/x',1,STR1);
  STR3=translate(STR2,'()','^#');
run;

STR	STR1	STR2	STR3
I'm falling. (In all the good times I find) myself	I'm falling. (In all the good times I find) myself	In all the good times I find	In all the good times I find
Longin (for change) and in the bad times (I fear myself)	Longin (for change) and in the bad times (I fear myself)	I fear myself	I fear myself
Tell me something, boy (aren't you (tired) tryin' to fill that void?)	Tell me something, boy (aren't you ^tired# tryin' to fill that void?)	aren't you ^tired# tryin' to fill that void?	aren't you (tired) tryin' to fill that void?

[Christen]

Thank you a lot!!!
This code cuts down the program significantly. Could you tell me please where I can read about regular expressions in sas?
Is it possible to get someting like this: (I (fear myself) ) if i have string with issue: Longin (for change) and in the bad times (I (fear myself) ?

[ChrisNZ]

SAS uses Perl regular expressions. There are countless tutorials on the web, that's how I learnt.
Rather than write crazy complicated code to cater for weird data, it's best to clean the data before processing. 

[Christen]

Thank you)