CODEWITHSUNDEEP
regexstringperlreplace
Tony Xu
Tony Xu

Reputation: 3091

How to replace consecutive and identical characters in Perl?

I have a string like XXXXYYYYZZZYYZZZYYYY which needs to be converted to XXXXAAAYZZZAYZZZAAAY

$s =~ s/Y{2}+/AY/g;

this has 2 problems, {2}+ will get YYYY to AYAY; and AY is not the same length as YYYY (expecting AAAY)

How to get this done in perl?

Upvotes: 3

Views: 207

Answers (3)

user7712945
user7712945

Reputation:

$s =~ s/Y{2}+/AY/g
RHS Pattern is ambiguously obscure pattern: Y{2}+, that's very rarely used regex pattern except if {}+ very rarely is available in few advanced regex engine, including perl maybe, as a regex feature called 'atomic grouping'.
You might have meant (Y{2})+ which is (YY)+ or Y{2,} which is YY+
in perl it's no brainer simple and easy as it supports lookaround feature

perl -e '$s=XXXXYYYYZZZYYZZZYYYY ;$s =~ s/Y(?=Y)/A/g;print $s'

actually lower regex engine such sed still can do it albeit in cumbersome, uneasy way

echo XXXXYYYYZZZYYZZZYYYY |sed -E 's/YY+/&\n/g;s/Y/A/g;s/A\n/Y/g'

Upvotes: 0

Tanktalus
Tanktalus

Reputation: 22254

There's always more than one way to do it. My suggestion is to grab all the Ys except the last one, and then use that to create a string of As of the same length. The e modifier tells perl to execute the code in the replacement side instead of using it directly, and the r modifier tells =~ to return the result of the substitution instead of modifying the input text directly (useful for these one-liner tests, among other places).

$ perl -E 'say shift =~ s/(Y+)(?=Y)/"A"x length$1/gre' XXXXYYYYZZZYYZZZYYYY
XXXXAAAYZZZAYZZZAAAY

Upvotes: 4

mob
mob

Reputation: 118605

Use a "look-ahead":

$s =~ s/Y(?=Y+)/A/g;

(?=Y+) means "followed by one or more Y characters", so any Y character that is followed by another Y character will be replaced with an A.

More info from perlretut

Upvotes: 6

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