keesturam has asked for the wisdom of the Perl Monks concerning the following question:

Hello Monks, I have started learning perl and was trying to do some exercise in regular expressions. One of the exercise was this.. "Construct regular expressions which can pick out words from a file with the Words containing an exclamation mark" I can write a pgm to find if there is an exclamation in a line/file. But how do I extract the word containing a reg ex (exclamation as in this case)? I tried googling, but couldn't find a suitable answer. --Karthick

Replies are listed 'Best First'.
Re: pick a word contaning reg-ex
by 2teez (Vicar) on Dec 04, 2012 at 13:37 UTC

    You mean something like this: 

    use warnings;
use strict;


while(<DATA>){
  chomp;
    print grep{/.+?\!/} split
}


__DATA__
 Mary had a little lamb! ,
 whose fleece! was white! as snow.

 And everywhere! that Mary went,
 the lamb was sure! to go.

    [download]

    If you tell me, I'll forget.
    If you show me, I'll remember.
    if you involve me, I'll understand.
    --- Author unknown to me
[reply]
[d/l]
Re: pick a word contaning reg-ex
by Kenosis (Priest) on Dec 04, 2012 at 16:19 UTC

    And another option:

    use Modern::Perl;

while (<DATA>) {
    say $1 while /(\S+!)/g;
}

__DATA__
Mary had a little lamb! ,
whose fleece! was white! as snow.

And everywhere! that Mary went,
the lamb was sure! to go.

    [download]

    Output:

    lamb!
fleece!
white!
everywhere!
sure!

    [download]
[reply]
[d/l]
[select]
Re: pick a word contaning reg-ex
by Wiggins (Hermit) on Dec 04, 2012 at 17:16 UTC
    The problem says "Words containing an exclamation mark". This is not the same as "ending with". 
      /(\w*!\w*)/

    [download]
    This is flawed because it matches "!".
    I don't know how to specify at least 1 sub-part must have a length > 0.

    It is always better to have seen your target for yourself, rather than depend upon someone else's description.

[reply]
[d/l]
      #!/perl
use warnings;
use strict;

while(<DATA>){
    print map {"$_\n"} grep{/!/ and /[^!]/} split;
}

__DATA__
 Mary had a little lamb! ,
 whose fleece! was white! as snow.

 And everywhere! that Mary went,
 the lamb was sure! to go.

oops!eedaisy
! ! !! HI!!IH

      [download]
[reply]
[d/l]
Re: pick a word contaning reg-ex
by keesturam (Initiate) on Dec 04, 2012 at 13:31 UTC

    Oh.! Reply from previous post answers my question as well!!

    perl -ne '/(\w*!\w*)/g && print "$1\n";' code_file(s)

    Lesson learnt. search for existing threads before posting questions :-)

[reply]
[d/l]
[reply]

        I just tried the code above and it only prints the first word (update: on each line) that contains an exclamation point.

        #!/perl
use warnings;
use strict;

while(<DATA>){
    #print map {"$_\n"} grep{/!/} split;
    /(\w*!\w*)/g && print "$1\n";
}

__DATA__
 Mary had a little lamb! ,
 whose fleece! was white! as snow.

 And everywhere! that Mary went,
 the lamb was sure! to go.

oops!eedaisy

        [download]

        Produces this:

        lamb!
fleece!
everywhere!
sure!
oops!eedaisy

        [download]
[reply]
[d/l]
[select]
Re: pick a word contaning reg-ex
by locked_user sundialsvc4 (Abbot) on Dec 04, 2012 at 16:57 UTC

    There are several things that you should, yourself, search for answers to.   First, how to write a regular expression that searches for “one-or-more Word characters followed by an exclamation mark.”   Then, the meaning of the idea “greedy,” and how to specify it.   Finally, how to perform more than one regular-expression match (in a loop ...) within a single string.

    The information is there e.g. in perldoc perlre.   If patiently read from stem to stern.

      Thanks..!! This was really helpful.
[reply]

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