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How do I make a Perl "array of arrays of hashes"?

Answer 1

+5 A:

You can address elements of such a data structure thus:

$x->[1][2]{foo} = "hello"

You don't even have to pre-define the structure. Just start working with the elements as if they're already there.

Marcelo Cantos 2010-09-09 11:39:50

This feature of Perl is known as **autovivification**.

Alan Haggai Alavi 2010-09-09 11:50:21

Does it work nicely with `use warnings` and `use strict`?

Dummy00001 2010-09-09 13:27:46

@Dummy000001: yes.

Marcelo Cantos 2010-09-09 13:42:45

Please don't use the variable `$a` (or `$b`) in examples that don't involve `sort`. `$a` and `$b` are special variables in Perl and can cause odd errors when used outside of a function passed to `sort`.

Ven'Tatsu 2010-09-09 13:59:23

Answer 2

+4 A:

my $aah =
        [ # outer array
                [ # first inner array
                        { # first inner hash
                                foo => 'bar',
                        },
                        { # second inner hash
                                bar => 'baaz',
                        },
                ],
                [ # secnd inner array
                        #...
                ],
                # ...
        ];

You can access the elements like this:

$aah->[0]->[1]->{bar} # => 'baaz'

jkramer 2010-09-09 11:40:11

Or `$aah->[0][1]{bar}`... everything in an array or hash is a scalar, thus any substructures can be assumed to be references, thus only the first arrow is ever needed.

Axeman 2010-09-09 12:45:37

Or `$$arr[0][1]{bar}`. I personally prefer that over the `->` notation.

Dummy00001 2010-09-09 13:27:07

This doesn't need to be an array reference, replace the first [ with ( the last ] with ) and $aah with @aah. Now you can access the data like this: $aah[0][1]{bar}; no de-referencing syntax needed, for the reason stated by Axeman.

MkV 2010-09-09 13:55:17

Answer 3

A:

my $arr = 
  [
    [
      {key1 => $value1, key2 => $value2},
      {key1 => $value3}
    ],
    [
      {rubbish => 'nonsense'},
    ]
   ];

etc.

Colin Fine 2010-09-09 11:40:51

Just because its a nested data structure doesn't mean it needs to be an arrayref, it can be an array of arrayrefs of hashrefs.

MkV 2010-09-09 13:57:20

@MkV: Most programmers would prefer the arrayref, since it's cheaper and simpler to pass around as a data structure than a list would be. BTW, did you downvote this answer? If so, that's a pretty lame reason to do so.

Marcelo Cantos 2010-09-09 14:16:47

Of course it can be an array rather than an arrayref at the top. So what? It doesn't change the answer in any significant way. I don't know why you felt it necessary to add that comment to my and jkramer's answers.

Colin Fine 2010-09-10 15:02:35

Answer 4

+6 A:

perldoc perldsc is a good document to read to get an idea of data structures in Perl.

Alan Haggai Alavi 2010-09-09 11:45:29

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How do I make a Perl "array of arrays of hashes"?

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