Re: inverting keys in a 2-D Hash

[Athanasius]

Hello melmoth, and welcome to the Monastery!

I don’t know about the “cleanest” way, but delete works nicely:

#! perl
use strict;
use warnings;
use Data::Dump;

my %HASH;
$HASH{keyA}{keyB} = [1 .. 4];
dd \%HASH;

my $array_ref = $HASH{keyA}{keyB};

delete $HASH{keyB};
delete $HASH{keyA};

$HASH{keyB}{keyA} = $array_ref;
dd \%HASH;

Output:

23:53 >perl 1360_SoPW.pl
{ keyA => { keyB => [1 .. 4] } }
{ keyB => { keyA => [1 .. 4] } }

23:53 >

Hope that helps,

[LanX]

Maybe you should write to a new hash if you can't be sure about the risk of losing information overwriting 2D symmetries. (Imagine keys A and B being from the same set of values, like indices of a matrix)

Afaik should delete return the deleted value, such that you can do it in a one-liner

Otoh no need to delete when writing a new hash.

[Laurent_R]

delete $HASH{keyB};

What is the point of this code line, since $HASH{keyB} probably doesn't exist?

Personally, I would prefer, if possible, to make a copy of the original hash.

[sundialsvc4]

Can’t say that I agree with you on this one:   there’s really no compelling requirement to attempt to modify the structure “in place,” and plenty of good reasons not to . . .

Especially when we consider that “what the hashref-of-hashrefs actually contains is ...” arrayrefs.   We aren’t copying any arrays:   what we’re actually doing is creating a second (inverted ...) hashref-of-hashrefs structure, which contains references to ... the same arrays that the “old” one refers-to.   (And we are, thereafter, entirely free to keep both of these hashrefs-of-hashrefs, without penalty, since they are both effectively just “indexes” to a single collection of anonymous-arrays.)