Regex: All except leading global substitution...

[Peter Hill]

I think I am misunderstanding how /g works; the following (incorrect)
program does not remove all non-leading vowels, just the 1st non-leading
vowel. Corrections please?

#! /usr/bin/perl -w
use strict;
#remove all non-leading vowels
my $word = 'ABEDU';
$word =~ s/(^..*?)[AEIOU](.*)/$1$2/g;
print qq{$word\n};

tia
Peter Hill

 

[Xicheng]

I think I am misunderstanding how /g works; the following (incorrect)
program does not remove all non-leading vowels, just the 1st non-leading
vowel. Corrections please?

#! /usr/bin/perl -w
use strict;
#remove all non-leading vowels
my $word = 'ABEDU';
$word =~ s/(^..*?)[AEIOU](.*)/$1$2/g;
print qq{$word\n};

you need two archors, one is the beginning of your string, and the
other is \G which records the last matched position, so you may try
things like:

$word =~ s/(^.|\G)(.*?)([aeiou])/\1\2/gi;

also, you need to use \1 \2 instead of $1 $2 if you don't use /e
modifier..

Xicheng

 

[Uri Guttman]

I think I am misunderstanding how /g works; the following (incorrect)
program does not remove all non-leading vowels, just the 1st non-leading
vowel. Corrections please?

#! /usr/bin/perl -w
use strict;
#remove all non-leading vowels
my $word = 'ABEDU';
$word =~ s/(^..*?)[AEIOU](.*)/$1$2/g;
print qq{$word\n};

X> you need two archors, one is the beginning of your string, and the
X> other is \G which records the last matched position, so you may try
X> things like:

X> $word =~ s/(^.|\G)(.*?)([aeiou])/\1\2/gi;

X> also, you need to use \1 \2 instead of $1 $2 if you don't use /e
X> modifier..

you are all working too hard. this seems to work fine (according to what
i understand of the specs):

perl -pe 's/([^aeiou]+)[aeiou]+/$1/ig'
andhekjajd
andhkjjd
weh1ajajkrkew
wh1jjkrkw
aaammmmddeww
aaammmmddww

uri

 

[Xicheng]

I think I am misunderstanding how /g works; the following (incorrect)
program does not remove all non-leading vowels, just the 1st non-leading
vowel. Corrections please?

#! /usr/bin/perl -w
use strict;
#remove all non-leading vowels
my $word = 'ABEDU';
$word =~ s/(^..*?)[AEIOU](.*)/$1$2/g;
print qq{$word\n};

X> you need two archors, one is the beginning of your string, and the
X> other is \G which records the last matched position, so you may try
X> things like:

X> $word =~ s/(^.|\G)(.*?)([aeiou])/\1\2/gi;

X> also, you need to use \1 \2 instead of $1 $2 if you don't use /e
X> modifier..

you are all working too hard. this seems to work fine (according to what
i understand of the specs):

perl -pe 's/([^aeiou]+)[aeiou]+/$1/ig'
andhekjajd
andhkjjd
weh1ajajkrkew
wh1jjkrkw
aaammmmddeww
aaammmmddww

Good point, but we don't need to stick to one way, one tool, one
thought to do anything in Perl or in whatever, hehe..

Good day,
Xicheng

 

[Anno Siegel]

I think I am misunderstanding how /g works; the following (incorrect)
program does not remove all non-leading vowels, just the 1st non-leading
vowel. Corrections please?

#! /usr/bin/perl -w
use strict;
#remove all non-leading vowels
my $word = 'ABEDU';
$word =~ s/(^..*?)[AEIOU](.*)/$1$2/g;
print qq{$word\n};

you need two archors, one is the beginning of your string, and the
other is \G which records the last matched position, so you may try
things like:

$word =~ s/(^.|\G)(.*?)([aeiou])/\1\2/gi;

That only keeps the initial vowel if there is another vowel in the string.
"andhkjjd" is changed to "ndhkjjd". Also, as I understand the OP, any
initial *group* of vowels should be kept, not only the first one.

also, you need to use \1 \2 instead of $1 $2 if you don't use /e
modifier..

This is plain wrong. \1, \2, etc. should be used if you need to refer
to a capture in the regex itself. On the substitution side of s///,
$1, $2 etc. are correct. Run it under warnings and see.

Anno

 

[Matt Garrish]

I think I am misunderstanding how /g works; the following (incorrect)
program does not remove all non-leading vowels, just the 1st non-leading
vowel. Corrections please?

#! /usr/bin/perl -w
use strict;
#remove all non-leading vowels
my $word = 'ABEDU';
$word =~ s/(^..*?)[AEIOU](.*)/$1$2/g;
print qq{$word\n};

$word =~ s/(^.|\G)(.*?)([aeiou])/\1\2/gi;

also, you need to use \1 \2 instead of $1 $2 if you don't use /e
modifier..

Huh? Care to explain where you got that idea from? You use backreferences
inside the match and numbered variables on the substitution side. You *can*
do what you've done above, but see perlre for the pitfalls.

Matt

 

[Xicheng]

I think I am misunderstanding how /g works; the following (incorrect)
program does not remove all non-leading vowels, just the 1st non-leading
vowel. Corrections please?

#! /usr/bin/perl -w
use strict;
#remove all non-leading vowels
my $word = 'ABEDU';
$word =~ s/(^..*?)[AEIOU](.*)/$1$2/g;
print qq{$word\n};

$word =~ s/(^.|\G)(.*?)([aeiou])/\1\2/gi;

also, you need to use \1 \2 instead of $1 $2 if you don't use /e
modifier..

Huh? Care to explain where you got that idea from? You use backreferences
inside the match and numbered variables on the substitution side. You *can*
do what you've done above, but see perlre for the pitfalls.

Hi, Thanks Anno and Matt for correcting me out. these days, I am
busying in learning some other new stuff, so may mix up something
here.. But anyhow in the last several months, I've learnt a lot from
this group either from other ppl's posts or from my own mistakes. As
for me, it's not that bad to refresh myself everyday, and upon my
graduation, I hope I will be gathering more knowledge in Perl and feel
more confidence about it..

Have a good weekend,
Xicheng

 

[Xicheng]

you need two archors, one is the beginning of your string, and the
other is \G which records the last matched position, so you may try
things like:

$word =~ s/(^.|\G)(.*?)([aeiou])/\1\2/gi;

That only keeps the initial vowel if there is another vowel in the string.
"andhkjjd" is changed to "ndhkjjd". Also, as I understand the OP, any
initial *group* of vowels should be kept, not only the first one.

There is a bug in my regex as you said, but it might be fixed by
separating ^. part with the other part, the way to solve this problem
should not be wrong...,say:
change:
$word =~ s/(^.|\G)(.*?)([aeiou])/\1\2/gi;
to:
$word =~ s/(^.)|\G(.*?)[aeiou]/$1$2/gi;

this deletes the degenerate case when there is no match of
"^.(.*?)[aeiou]" and "\G" becomes the beginning of the string..

Xicheng

 

[robic0]

I think I am misunderstanding how /g works; the following (incorrect)
program does not remove all non-leading vowels, just the 1st non-leading
vowel. Corrections please?

#! /usr/bin/perl -w
use strict;
#remove all non-leading vowels
my $word = 'ABEDU';
$word =~ s/(^..*?)[AEIOU](.*)/$1$2/g;
print qq{$word\n};

tia
Peter Hill

I might be missing some hidden detail in yur explaination.
It might be a convolution of the sample 'ABEDU' with the
problem statement "does not remove all non-leading vowels,
just the 1st non-leading vowel."

This removes *ALL* of the character class defined.
You can't say all non leading vowels without defining
what "leading" is, and you certaintly don't do that here.

use strict;
#remove all non-leading vowels
my $word = 'ABEDU';
$word =~ s/[AEIOU]//g;
print qq{$word\n};

Back to class pigmy!!
-robic0-