preserve sort order in another list

[Brett]

I have two arrays and i wish to sort the first one numerically, but after
sorting, I would like the second array to be in the same matching order as
the first array.

ie.

@l1={3,1,2};
@l2={'a','b','c'};

@l1 = sort {$a <=> $b} (@l1); # sort list @l1 to be {1,2,3}

# do something to @l2 to make order {'b','c','a'} (preserving the original
mapping with the first list)

There's probably an easy way to do this that i'm not aware of.

Thanks in advance,
Brett.

 

[Jürgen Exner]

I have two arrays and i wish to sort the first one numerically, but
after sorting, I would like the second array to be in the same
matching order as the first array.

ie.

@l1={3,1,2};
@l2={'a','b','c'};

@l1 = sort {$a <=> $b} (@l1); # sort list @l1 to be {1,2,3}

# do something to @l2 to make order {'b','c','a'} (preserving the
original mapping with the first list)

There's probably an easy way to do this that i'm not aware of.

Indeed, there is: use a data structure that matches your problem better.
Instead of having a pair of unrelated arrays use a single array of pairs.
And then sort that single array by the value of the first component of each
pair.

jue

 

[Brett]

Indeed, there is: use a data structure that matches your problem better.
Instead of having a pair of unrelated arrays use a single array of pairs.
And then sort that single array by the value of the first component of each
pair.

jue

I'm new to this, and tried to give it a go.

struct ID =>
{
number => '$',
name => '$',
};

my @IDs = ID->new();
#fill data...

@IDs = sort {$a->number <=> $b->number} (@IDs); # numeric sort on number

but that failed to sort the list as i expected. How can i use sort to do
what I want?

 

[Jim Gibson]

I'm new to this, and tried to give it a go.

struct ID =>
{
number => '$',
name => '$',
};

my @IDs = ID->new();
#fill data...

@IDs = sort {$a->number <=> $b->number} (@IDs); # numeric sort on number

but that failed to sort the list as i expected. How can i use sort to do
what I want?

Please post a complete program so we can see where you are going wrong,
but post it to comp.lang.perl.misc because this newsgroup is defunct.

Here is a version that uses an array of array references to an array of
two elements, as Jürgen suggested:

#!/usr/local/bin/perl

use strict;
use warnings;

my @l1 = qw/ 3 1 2 /;
my @l2 = qw/ a b c /;

my @ids = map { [ $l1[$_], $l2[$_] ] } (0..$#l1);

print "Unsorted Array:\n";
for my $r ( @ids ) {
print " @$r\n";
}
print "\nSorted array:\n";
foreach my $r ( sort { $a->[0] <=> $b->[0] } @ids ) {
print " @$r\n";
}

__OUTPUT__

Unsorted Array:
3 a
1 b
2 c

Sorted array:
1 b
2 c
3 a

 

[Brett]

I'm new to this, and tried to give it a go.

struct ID =>
{
number => '$',
name => '$',
};

my @IDs = ID->new();
#fill data...

@IDs = sort {$a->number <=> $b->number} (@IDs); # numeric sort on number

but that failed to sort the list as i expected. How can i use sort to do
what I want?

Please post a complete program so we can see where you are going wrong,
but post it to comp.lang.perl.misc because this newsgroup is defunct.

Here is a version that uses an array of array references to an array of
two elements, as Jürgen suggested:

#!/usr/local/bin/perl

use strict;
use warnings;

my @l1 = qw/ 3 1 2 /;
my @l2 = qw/ a b c /;

my @ids = map { [ $l1[$_], $l2[$_] ] } (0..$#l1);

print "Unsorted Array:\n";
for my $r ( @ids ) {
print " @$r\n";
}
print "\nSorted array:\n";
foreach my $r ( sort { $a->[0] <=> $b->[0] } @ids ) {
print " @$r\n";
}

__OUTPUT__

Unsorted Array:
3 a
1 b
2 c

Sorted array:
1 b
2 c
3 a

Thanks, i'll report to the other group. I'd like to stick with the
structures, since they look a bit easier.

Brett.