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Hi,
I'm trying to get an array into order i've got this array,
$number[1] = "5";
$number[2] = "7";
$number[3] = "4";
$number[4] = "8";
$number[5] = "7";

i'm trying to get it into another array it looks like this,
$number2[1] = "8";
$number2[2] = "7";
$number2[3] = "7";
$number2[4] = "5";
$number2[5] = "4";

So far i've got some code like this,
for($i=1;$i<=$count;$i++) {
for($ii=1;$ii<=$count;$ii++) {
if ($number[$ii]>=$highest[$i]) {
$number2 = $number[$i];
print "\n$number2[$i]";
}
}
}

I've tried the pseudo but could figure out the logic. This is the first
time i've had to do ordering of this sort so i'm a bit stuck at which
way to go next. I've checked out cpan but couldn't find anything that
looked like it would do what i wanted.

Has anyone got any ideas please.

Thanks

Gary

Gary Mayor <> writes:

> Hi,
> I'm trying to get an array into order i've got this array,
> $number[1] = "5";
> $number[2] = "7";
> $number[3] = "4";
> $number[4] = "8";
> $number[5] = "7";
>
> i'm trying to get it into another array it looks like this,
> $number2[1] = "8";
> $number2[2] = "7";
> $number2[3] = "7";
> $number2[4] = "5";
> $number2[5] = "4";

I think I would do something like this...

@number2 = reverse sort @number1;

Ryan
--
perl -e '$;=q,BllpZllla_nNanfc]^h_rpF,;@;=split//,
$;;$^R.=--$=*ord for split//,$~;sub _{for(1..4){$=
=shift;$=--if$=!=4;while($=){print chr(ord($;[$%])
+shift);$%++;$=--;}print " ";}}_(split//,$^R);q;;'

Gary Mayor wrote:
> I'm trying to get an array into order i've got this array,

That's called sorting, and there is a function to help you do it:

http://www.perldoc.com/perl5.8.0/pod/func/sort.html

This is what I think you are trying to do:

my @number = (5, 7, 4, 8, 7);
my @number2 = sort { $b <=> $a } @number;
print "$_\n" for @number2;

But you obviously need to study some basics about Perl:

http://www.perldoc.com/perl5.8.0/pod/perlintro.html

Good luck!

--
Gunnar Hjalmarsson
Email: http://www.gunnar.cc/cgi-bin/contact.pl

Now that is wicked five minutes later i had an answer cheers mate that
works a treat.

Thanks

Gary

Ryan Shondell wrote:
> Gary Mayor <> writes:
>
>
>>Hi,
>>I'm trying to get an array into order i've got this array,
>>$number[1] = "5";
>>$number[2] = "7";
>>$number[3] = "4";
>>$number[4] = "8";
>>$number[5] = "7";
>>
>>i'm trying to get it into another array it looks like this,
>>$number2[1] = "8";
>>$number2[2] = "7";
>>$number2[3] = "7";
>>$number2[4] = "5";
>>$number2[5] = "4";
>
>
> I think I would do something like this...
>
> @number2 = reverse sort @number1;
>
> Ryan

Ryan Shondell wrote:
> Gary Mayor writes:
>>I'm trying to get an array into order i've got this array,
>>$number[1] = "5";
>>$number[2] = "7";
>>$number[3] = "4";
>>$number[4] = "8";
>>$number[5] = "7";
>>
>>i'm trying to get it into another array it looks like this,
>>$number2[1] = "8";
>>$number2[2] = "7";
>>$number2[3] = "7";
>>$number2[4] = "5";
>>$number2[5] = "4";
>
> I think I would do something like this...
>
> @number2 = reverse sort @number1;

A lexical sort of numbers? Not a good advice, is it? (It happens to
give the same result as a numerical sort applied on the above data,
since all the values are integers with the same number of digits.)

perldoc -f sort

--
Gunnar Hjalmarsson
Email: http://www.gunnar.cc/cgi-bin/contact.pl

Gunnar Hjalmarsson <> writes:

> A lexical sort of numbers? Not a good advice, is it? (It happens to
> give the same result as a numerical sort applied on the above data,
> since all the values are integers with the same number of digits.)

You're right, of course. I imagine my brain saw the quotes around the
numbers, and went on "lexical autopilot".

@number2 = sort {$b <=> $a} @number1;

is a much better solution.

Ryan
--
perl -e '$;=q,BllpZllla_nNanfc]^h_rpF,;@;=split//,
$;;$^R.=--$=*ord for split//,$~;sub _{for(1..4){$=
=shift;$=--if$=!=4;while($=){print chr(ord($;[$%])
+shift);$%++;$=--;}print " ";}}_(split//,$^R);q;;'

Gary Mayor wrote:
>
> I'm trying to get an array into order i've got this array,
> $number[1] = "5";
> $number[2] = "7";
> $number[3] = "4";
> $number[4] = "8";
> $number[5] = "7";
>
> i'm trying to get it into another array it looks like this,
> $number2[1] = "8";
> $number2[2] = "7";
> $number2[3] = "7";
> $number2[4] = "5";
> $number2[5] = "4";

Others have suggested:

@number2 = sort { $b <=> $a } @number;

However that populates $number[0] which I see you don't want. Or did
you think that Perl's arrays started at [1]? Anyway, this will do what
you want:

@number2 = ( undef, sort { $b <=> $a } @number );

Or:

@number2[ 1 .. $#number ] = sort { $b <=> $a } @number;



John
--
use Perl;
program
fulfillment

[This followup was posted to comp.lang.perl.misc]

In article <brqhgo$gc8$>,
says...
> Hi,
> I'm trying to get an array into order i've got this array,
> $number[1] = "5";
> $number[2] = "7";
> $number[3] = "4";
> $number[4] = "8";
> $number[5] = "7";
>
> i'm trying to get it into another array it looks like this,
> $number2[1] = "8";
> $number2[2] = "7";
> $number2[3] = "7";
> $number2[4] = "5";
> $number2[5] = "4";
>
> So far i've got some code like this,
> for($i=1;$i<=$count;$i++) {
> for($ii=1;$ii<=$count;$ii++) {
> if ($number[$ii]>=$highest[$i]) {
> $number2 = $number[$i];
> print "\n$number2[$i]";
> }
> }
> }
>
> I've tried the pseudo but could figure out the logic. This is the first
> time i've had to do ordering of this sort so i'm a bit stuck at which
> way to go next. I've checked out cpan but couldn't find anything that
> looked like it would do what i wanted.
>
> Has anyone got any ideas please.
>
> Thanks
>
> Gary


You should use perl's builtin "sort" function. See "perldoc -f sort"
for further details.