replacing a string with another and vice versa
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Hi, I would like to write two functions:
1- replace all occurrences of 'a(number)' and 'bb(number)' with respectively 'a_number' and 'bb_number'. More concretely, a(1) would become a_1 and bb(1285) would become bb_1285. the typical string would look like
string='log(a(2))+a(3)*cosh(exp(bb(5))'
and the result would be
string='log(a_2)+a_3*cos(exp(bb_5))'
2- do the inverse operation
Is there any efficient way of doing this? My sense is that it can be done with regular expressions but I am just a beginner on that front and I would not know how to go about this. Your help will be appreciated.
thanks,
Pat.
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per isakson
le 3 Oct 2012
Modifié(e) : per isakson
le 4 Oct 2012
A start of one way to use regular expression:
string = regexprep( string, '(?<=a)\((\d+)\)', '_$1' );
string = regexprep( string, '(?<=bb)\((\d+)\)', '_$1' );
.
-- in one line ---
Look for "(one or more digits)" that comes directly after "bb" or "a"
>> string = regexprep( string, '(?<=((bb)|a))\((\d+)\)', '_$1' )
string =
log(a_2)+a_3*cosh(exp(bb_5)
- (?<=((bb)|a)) "Look behind from current position and test if expr is found." Where expr evaluate to "a" or "bb".
--- another one-liner ---
>> str = log(a(2))+a(3)*cosh(exp(bb(5));
>> regexprep( str, '((bb)|a)\((\d+)\)', '$1_$2' )
ans =
log(a_2)+a_3*cosh(exp(bb_5)
- (bb)|a stands for "bb" or "a"
- (expr) stands for group regular expressions and capture tokens
- *\(* stands for "("
- \d+ stands for one ore more digits
3 commentaires
per isakson
le 3 Oct 2012
Modifié(e) : per isakson
le 3 Oct 2012
Yes it is, but why bother? Matt, has done it. With regular expressions one must not make it more complicated than one master.
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