Two arrays into hash

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Re: Two arrays into hash by moritz (Cardinal) on Jul 16, 2009 at 08:29 UTC
perlreftut describes how to construct such data structures. If you have problems understanding something in there, or you hit some errors while trying, feel free to ask and explain what's the problem.	[reply]
Re: Two arrays into hash by Bloodnok (Vicar) on Jul 16, 2009 at 10:31 UTC
Methinx you could use rudimentary algebraic methods to arrive at your goal, consider: `@array1=qw(A B C); @array2=qw(1 2 3);` [download] is equivalent to/can be re-written as: `@array1=('A', 'B', 'C'); @array2=('1', '2', '3');` [download] Also, the requirement can be re-written as: `%hash = ( array1=>[('A','B','C')], array2=>[('1','2','3')] };` [download] So, substituting for `('A','B','C')` & `('1','2','3')`, into the re-written required result, gives: `%hash = ( array1=>[@array1], array2=>[@array2] };` [download] But, `[@array]` can be/usually is written as `\@array`, hence, `%hash = ( array1=>\@array1, array2=>\@array2 };` [download] As required - also discoverable by reference to perlreftut - as moritz has previously suggested. A user level that continues to overstate my experience :-))	[reply] [d/l] [select]
Re^2: Two arrays into hash by davorg (Chancellor) on Jul 16, 2009 at 10:40 UTC
`[@array]` can be/usually is written as `\@array` Careful. The two are not the same. `\@array` returns a reference to the existing array. `[@array]` creates a new (anonymous) copy of the original array and returns a reference to the copy. It's a subtle difference, but an important one. `#!/usr/bin/perl use strict; use warnings; my @arr = (1 .. 5); my %hash = ( aref => \@arr, copy => [ @arr ], ); print "@{$hash{copy}}\n"; print "@{$hash{aref}}\n"; @arr = ('a' .. 'e'); print "@{$hash{copy}}\n"; # original values print "@{$hash{aref}}\n"; # changed values` [download] -- See the Copyright notice on my home node. Perl training courses	[reply] [d/l]
Re^3: Two arrays into hash by Bloodnok (Vicar) on Jul 16, 2009 at 10:43 UTC
Good point, well made - I rather overlooked that didn't I ? Thanx davorg. A user level that continues to overstate my experience :-))	[reply]


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