Yes! I just wasn't sure if I'd get undefs in my array or not (I don't).
Thanks!
On 8/20/06, Tom Phoenix <[EMAIL PROTECTED]> wrote:
On 8/20/06, Gerald Host <[EMAIL PROTECTED]> wrote:
> No. I want the most efficient way to do this:
> my @array0=(0..1000);
> my @array1;
> for (0..100) {
>
On 8/20/06, Gerald Host <[EMAIL PROTECTED]> wrote:
No. I want the most efficient way to do this:
my @array0=(0..1000);
my @array1;
for (0..100) {
if (defined $array0[0]) { #accept 0 as a valid item
push(@array1, shift @array0);
}
}
Do you want splice?
@array1 = splice @array0
No. I want the most efficient way to do this:
my @array0=(0..1000);
my @array1;
for (0..100) {
if (defined $array0[0]) { #accept 0 as a valid item
push(@array1, shift @array0);
}
}
I want to stop at the end of the @array0 too, so if it only had 20 items in
it I would break the loop (
A Diumenge 20 Agost 2006 22:49, Gerald Host va escriure:
> What is the most efficient way to shift the first 100 items off an array?
>
> Thanks!
if you mean that first element be the last one and so on, use reverse.
--
Xavier Mas
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For additional com
John W. Krahn <[EMAIL PROTECTED]> had this to say,
> Ok, here it is using the code you posted elsethread.
Much more elegant than mine, but by looking at the output, it's still not
quite right. Note that there really is no -m option.
> After running it I get this output:
>
> direct
>
Deb wrote:
>
> John wrote:
> > Did you try the code I posted Friday? (Message-ID: <[EMAIL PROTECTED]>)
>
> Thanks, yes, I did. But, the syntax was new to me, and I've been reading
> up on it. I couldn't really get it to do what I want (see my previous
> post to this one). But, that's probably
John wrote:
> Did you try the code I posted Friday? (Message-ID: <[EMAIL PROTECTED]>)
John,
Thanks, yes, I did. But, the syntax was new to me, and I've been reading
up on it. I couldn't really get it to do what I want (see my previous
post to this one). But, that's probably b/c I didn't expl
Deb wrote:
> This (code below) makes sense to me, but I was talking this over with a
> co-worker on Friday, and then I tried putting together some 2-dimensional
> hashes - which hurts my head at the moment.
Hi Deb
Actually, the hash hee--so far, anyway--is not two-dimensional. More later
> So
Hi Deb -
> -Original Message-
> From: Deb [mailto:[EMAIL PROTECTED]
> Sent: Saturday, March 01, 2003 7:17 PM
> To: R. Joseph Newton; Perl List
> Subject: Re: shifting through arrays of line data
>
>
> This (code below) makes sense to me, but I was talking this ov
This (code below) makes sense to me, but I was talking this over with a
co-worker on Friday, and then I tried putting together some 2-dimensional
hashes - which hurts my head at the moment. So I went to
perl.plover.com/FAQs to read (again) his article on references, and I still
have a mental block
Deb wrote:
> Hi Guys,
>
> I have an array in which each element is a line commandline data. It looks
> something like this -
>
> @Array contains lines:
>
> post1: -r [EMAIL PROTECTED] -x cat-100 -h post1
> post2: -x tel -h post2
> post3: -h post3 -x hifi
The getRelationships sub here has a few l
"John W. Krahn" wrote:
> "R. Joseph Newton" wrote:
> >
> > Deb wrote:
> > >
> > > I have an array in which each element is a line commandline data. It looks
> > > something like this -
> > >...
> > > post1: -r [EMAIL PROTECTED] -x cat-100 -h post1
> > > post2: -x tel -h post2
> > > post3: -h post
"R. Joseph Newton" wrote:
>
> Deb wrote:
> >
> > I have an array in which each element is a line commandline data. It looks
> > something like this -
> >
> > @Array contains lines:
> >
> > post1: -r [EMAIL PROTECTED] -x cat-100 -h post1
> > post2: -x tel -h post2
> > post3: -h post3 -x hifi
> >
>
Deb wrote:
> Hi Guys,
>
> I have an array in which each element is a line commandline data. It looks
> something like this -
>
> @Array contains lines:
>
> post1: -r [EMAIL PROTECTED] -x cat-100 -h post1
> post2: -x tel -h post2
> post3: -h post3 -x hifi
>
> What I really need to do is build a re
Deb wrote:
>
> Hi Guys,
>
> I have an array in which each element is a line commandline data. It looks
> something like this -
>
> @Array contains lines:
>
> post1: -r [EMAIL PROTECTED] -x cat-100 -h post1
> post2: -x tel -h post2
> post3: -h post3 -x hifi
>
> And so on. The order of the opt
>
> Thanks, I know how to use split (I think). Since the data
> comes in any order, and I have to corellate it, I can't think
> of a way that split will fix me up - Maybe I'm missing
> something. Can you give me an example?
>
Ok I'll give her a go
my %results;
my $cnt = 1;
foreach $line(
Thanks, I know how to use split (I think). Since the data comes in
any order, and I have to corellate it, I can't think of a way that split
will fix me up - Maybe I'm missing something. Can you give me an example?
deb
Dan Muey <[EMAIL PROTECTED]> had this to say,
> perldoc -f split
>
> Will
perldoc -f split
Will fic you up!
Dmuey
> Hi Guys,
>
> I have an array in which each element is a line commandline
> data. It looks something like this -
>
> @Array contains lines:
>
> post1: -r [EMAIL PROTECTED] -x cat-100 -h post1
> post2: -x tel -h post2
> post3: -h post3 -x hifi
>
> An
splice is your friend :)
splice(@data, 0, 9);
will solve all your problems ;)
On Tue, 25 Feb 2003, dan wrote:
> Hi,
>
> I've come across a dilemma, I can solve it the long way round, I'm enquiring
> if there's a shorter way to do this.
>
> I have an array, @data, which contains a lot of data w
There are probably oodles of ways of doing this.
Here are two:
# An array slice
@[EMAIL PROTECTED];
or
# start at position 0, remove 9 elements
splice @data, 0, 9;
Pete
On Tue, 2003-02-25 at 14:39, dan wrote:
> onwards. My way of doing this was:
> shift(@data); x 9
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Timothy, et al --
...and then Timothy Johnson said...
%
% David->"Hmmm... OK, so that explains it, but I still don't get it... So
% the match is going to spit out a scalar but in order to use it you have to
% capture it in a list context?"
%
% No, actually the opposite. The match returns a l
as a scalar. Thus:
($fullpath =~ m:/mp3/(.+):)[0]
is element 0 of the list created by evaluating the match. That way what
split sees is a scalar (the element), so it can split it accordingly.
-Original Message-
From: David T-G
To: perl beginners
Cc: Timothy Johnson
Sent: 6/9/02 7:51 PM
Sub
Timothy, et al --
...and then Timothy Johnson said...
%
% Ok, I finally got a chance to test it, and the problem with my code is that
% split expects a scalar as the second argument. This does work:
%
% ($temp) = $fullpath =~ m:/mp3/(.+):;
% @working = split /\//,$temp;
%
% because it is
Try this:
@working = split(/\//,($fullpath =~ m:/mp3/(.+):)[0]);
-Original Message-
From: David T-G [mailto:[EMAIL PROTECTED]]
Sent: Sunday, June 09, 2002 1:43 PM
To: perl beginners
Cc: Timothy Johnson
Subject: Re: shifting solved (was "Re: chomp-ing DOS lines, shifting,
and a variabl
Timothy, et al --
...and then David T-G said...
%
% ...and then Timothy Johnson said...
% %
% % I can't test this where I am right now, but would something like this work?
% %
% % @working = split /\//,($fullpath =~ m|/mp3/(.+)|); #changed match delimiter
%
% I'm surprised to find that it doe
Timothy, et al --
...and then Timothy Johnson said...
%
% I can't test this where I am right now, but would something like this work?
%
% @working = split /\//,($fullpath =~ m|/mp3/(.+)|); #changed match delimiter
I'm surprised to find that it does, but I'm glad I tested it. I thought
that ma
Elias Assmann
Subject: Re: shifting solved (was "Re: chomp-ing DOS lines, shifting,
and a variable variable")
Elias, et al --
and then Elias Assmann said...
%
% Oh my, what a bad day for my poor little brain... Sorry for all that
% confusion.
*grin* No problem; it made me c
Elias, et al --
...and then Elias Assmann said...
%
% Oh my, what a bad day for my poor little brain... Sorry for all that
% confusion.
*grin* No problem; it made me check my answers :-)
%
% On Sun, 9 Jun 2002, David T-G wrote:
%
% > ...and then Elias Assmann said...
% >
% > % be two lines
Oh my, what a bad day for my poor little brain... Sorry for all that
confusion.
On Sun, 9 Jun 2002, David T-G wrote:
> Elias, et al --
>
> ...and then Elias Assmann said...
>
> % be two lines, so how about this: @working = m'/mp3(/[^/]+)+';?
It seems I have suffered a misconception about what
Elias, et al --
...and then Elias Assmann said...
%
% On Sat, 8 Jun 2002, David T-G wrote:
%
% > % I can't modify $fullpath because I'll use it later, but for parsing I
% > % don't need the leading "/mp3/" part, and the only way I've found to get
...
% > While it may not be a perfect solution,
Elias --
...and then Elias Assmann said...
%
% On Sun, 9 Jun 2002, Elias Assmann wrote:
%
% > be two lines, so how about this: @working = m'/mp3(/[^/]+)+';?
%
% So much for posting code without trying it... This isn't working, but
*grin*
% it isn't obvious to me why, and I don't have time r
On Sun, 9 Jun 2002, Elias Assmann wrote:
> be two lines, so how about this: @working = m'/mp3(/[^/]+)+';?
So much for posting code without trying it... This isn't working, but
it isn't obvious to me why, and I don't have time right now :-(
Elias
--
Gefängnis für Hans Mustermann wegen
On Sat, 8 Jun 2002, David T-G wrote:
> % I can't modify $fullpath because I'll use it later, but for parsing I
> % don't need the leading "/mp3/" part, and the only way I've found to get
> % rid of it elegantly is
> %
> % ...
> % @working = split(/\//,$fullpath) ; # cut path in
Hello Craig,
Saturday, December 08, 2001, Craig Inman <[EMAIL PROTECTED]> wrote:
CI> As I'm new to perl, I'm more or less trying to write a 'nested while
CI> read' loop (atleast that is what my attempts come out to look like so
CI> far).
CI> trying something like
CI> open (A, $list1) or di
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