Correct data structure / sort method to use

[Niall Macpherson]

I have a problem sorting a hash by value rather than key. I have read
the FAQ concerning this but unfortunately it does not help . I am not
sure if this is due to with my misunderstanding of the way the sort
function works or due to me not using the correct data structure.

I have a number of database spaces each of which has a unique name
which I am using as the hash key . Each key has a size, used and free
attribute associated with it . So a typical entry would look like

$dbspaces{space67}{size} = 12345;
$dbspaces{space67}{used} = 123;
$dbspaces{space67}{free} = 12222;

I wish to sort the hash based on the 'used' value.

I do understand that I cannot sort the hash directly , I have to sort
the keys and then access the hash by the sorted keys - this is fine.

I do understand also that it may be better to use a hash of arrays ,
although I would prefer to be accessing $dbspaces{space67}{size} than
$dbspaces{space67}[0] for readability.

My problem is that I do not seem to be able to access the hash within
the sort function.

i.e if I use the format

my @sort_by_used_keys = sort sort_by_used (keys %dbspaces);

the sort function 'sort_by_used()' only has access to the $a and $b
variables, not the hash itself.

All the examples I have seen which access the hash itself seem to do so
using the 'inline' (hope that is the correct terminology) version of
sort - e.g

@sorted = sort { lc($hash{$a}) cmp lc($hash{$b}) } keys %hash;

Attached is a very cut down version of the code. It will compile and
run although it will not do anything since the 'sort_by_used' function
cannot access the hash and I therefore cannot perform the correct tests
within this function I would be grateful if someone could tell nme
what I am doing wrong.

Many thanks

use strict;
use warnings;
use Data:umper;

##--------------------------------------------------------------
sub sort_by_used
{
## How do I access the hash here ?

# This will not compile
#return ($dbspaces{$a}{used} <=> dbspaces{$b}{used})

return(0);
}
##--------------------------------------------------------------
##
## Note - cannot sort a hash directly.
##
## Need to sort the hash keys and then access the hash
## via the sorted array
##--------------------------------------------------------------

my %dbspaces = ();
my $key = "";

## Populate the hash
while(<DATA>)
{
chomp;
my @vals = split;
$key = $vals[0];
$dbspaces{$key}{size} = $vals[1];
$dbspaces{$key}{used} = $vals[2];
$dbspaces{$key}{free} = $vals[3];
}

print Dumper %dbspaces;

## Sort by the used attribute
my @sort_by_used_keys = sort sort_by_used (keys %dbspaces);

## Access the hash by the sorted keys, i.e used descending
foreach $key (@sort_by_used_keys)
{
print Dumper $dbspaces{$key};
}
exit(0);

__END__
space1 100 63 37
space2 200 12 188
space47 300 299 1
space99 400 375 25

 

[Anno Siegel]

I have a problem sorting a hash by value rather than key. I have read
the FAQ concerning this but unfortunately it does not help . I am not
sure if this is due to with my misunderstanding of the way the sort
function works or due to me not using the correct data structure.

I have a number of database spaces each of which has a unique name
which I am using as the hash key . Each key has a size, used and free
attribute associated with it . So a typical entry would look like

$dbspaces{space67}{size} = 12345;
$dbspaces{space67}{used} = 123;
$dbspaces{space67}{free} = 12222;

I wish to sort the hash based on the 'used' value.
[...]

My problem is that I do not seem to be able to access the hash within
the sort function.

i.e if I use the format

my @sort_by_used_keys = sort sort_by_used (keys %dbspaces);

the sort function 'sort_by_used()' only has access to the $a and $b
variables, not the hash itself.

Then make the hash(ref) a parameter of the function instead of (or in
addition to) the keys. Untested:

sub sort_by_used {
my $hr = shift;
sort { $hr->{ $a}->{ used} <=> $hr->{ $b}->{ used} } keys %$hr;
}

Anno

 

[Gunnar Hjalmarsson]

My problem is that I do not seem to be able to access the hash within
the sort function.

i.e if I use the format

my @sort_by_used_keys = sort sort_by_used (keys %dbspaces);

the sort function 'sort_by_used()' only has access to the $a and $b
variables, not the hash itself.

No. All you need to do is declaring the %dbspaces variable before
sort_by_used().

 

[Niall Macpherson]

Anno Siegel wrote:

Then make the hash(ref) a parameter of the function instead of (or in
addition to) the keys. Untested:

sub sort_by_used {
my $hr = shift;
sort { $hr->{ $a}->{ used} <=> $hr->{ $b}->{ used} } keys %$hr;
}

Anno
--

Thanks Anno . I would rather go with this than with Gunnars suggestion
since in my real program which is much larger, the %dbspaces hash is
not a global variable and I would prefer to pass it to the sort
function explicitly.

I had kind of figured this might be the besty way to go but I still
don't understand how $a and $b get passed through to the sort
subroutine though and therefore cannot figure out how to pass
aditional parameters.

I tried changing

my @sort_by_used_keys = sort sort_by_used ( keys %dbspaces);

to

my @sort_by_used_keys = sort sort_by_used ( \%dbspaces, keys
%dbspaces);

but then in the sort_by_used sub I don't appear to get the value ..

sub sort_by_used
{
my $hr = shift;

print Dumper $hr;
return(0);
}

gives me

$VAR1 = undef;

 

[Anno Siegel]

Anno Siegel wrote:



Thanks Anno . I would rather go with this than with Gunnars suggestion
since in my real program which is much larger, the %dbspaces hash is
not a global variable and I would prefer to pass it to the sort
function explicitly.

I had kind of figured this might be the besty way to go but I still
don't understand how $a and $b get passed through to the sort

The system takes care of $a and $b, they're none of your business.

subroutine though and therefore cannot figure out how to pass
aditional parameters.

What have $a and $b to do with additional parameters?

I tried changing

my @sort_by_used_keys = sort sort_by_used ( keys %dbspaces);

^^^^
What is the second sort() for? It will destroy the order sort_by_used()
has established.

to

my @sort_by_used_keys = sort sort_by_used ( \%dbspaces, keys
%dbspaces);

but then in the sort_by_used sub I don't appear to get the value ..

sub sort_by_used
{
my $hr = shift;

print Dumper $hr;
return(0);
}

gives me

$VAR1 = undef;

Not if you call sort_by_used as shown, it would show the content of
%dbspaces.

If you want to pass the keys as parameters, the sub should in some
way make use of them. Yours doesn't.

sub sort_by_used {
my $hr = shift;
sort { $hr->{ $a}->{ used} <=> $hr->{ $b}->{ used} @_;
}

Anno

 

[Niall Macpherson]

If you want to pass the keys as parameters, the sub should in some
way make use of them. Yours doesn't.

sub sort_by_used {
my $hr = shift;
sort { $hr->{ $a}->{ used} <=> $hr->{ $b}->{ used} @_;
}

Anno
--
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Thanks Anno - I realise what my mistake was and that is all now
working as expected. Many thank for the help !

 

[Dr.Ruud]

Niall Macpherson:

I have a problem sorting a hash by value rather than key.

You cannot sort a hash. You can access it in an ordered way, like via an
external index.

 

[Gunnar Hjalmarsson]

Niall Macpherson:

You cannot sort a hash. You can access it in an ordered way, like via an
external index.

Well, if you think that's worth mentioning, maybe you should suggest
that the FAQ is altered.

perldoc -q "sort a hash"

;-)

 

[Dr.Ruud]

Gunnar Hjalmarsson:

Dr.Ruud:

Well, if you think that's worth mentioning, maybe you should suggest
that the FAQ is altered.

perldoc -q "sort a hash"
;-)

The FAQ is OK with me: "Instead, you have ...".

 

[Gunnar Hjalmarsson]

Gunnar Hjalmarsson:

The FAQ is OK with me: "Instead, you have ...".

Now I see you are right. The FAQ entry does not contradict your statement.

 

[Dr.Ruud]

Gunnar Hjalmarsson:

Dr.Ruud:

Now I see you are right. The FAQ entry does not contradict your
statement.

Does a question have to be valid? No, but the answer has to be.
;-)