Re: Re: Re: Getting last valid index from a particular list in a @LoL

#!/usr/bin/perl -wT
use strict;

my @aoa = ([1,2,3],
           [1,2,3,4,5],
           [1,2,3,4,5,6,7,8],
           [1,2,3,4],
           [1,2,3,4,5,6],
          );

my $i;
my $index = (sort{$$b[1]-$$a[1]}map{[$i++,$#$_]}@aoa)[0]->[0];
print "\$aoa[$index] has the most elements\n";

=OUTPUT
$aoa[2] has the most elements
[download]

-Blake

Comment on Re: Re: Re: Getting last valid index from a particular list in a @LoL Download Code

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Re4: Getting last valid index from a particular list in a @LoL by Hofmator (Curate) on Oct 17, 2001 at 15:59 UTC
`$index = (sort{$$b[1]-$$a[1]}map{[$i++,$#$_]}@aoa)[0]->[0];` That's nice, very nice indeed, blakem ... but some people seem to have problems understanding it, so allow me to take this apart a bit: `@tmp = map {[$i++,$#$_]} @aoa;`This creates a list of two element arrays (yes, to be precise of references to two element arrays). The first being the position (index) in @aoa, the second the last index of the current array in @aoa. So for the example given, this produces the following array: `( [0,2], [1,4], [2,7], [3,3], [4,5] )` `@sorted = sort {$$b[1]-$$a[1]} @tmp;`This created list is now sorted on the 2nd (last index) element, with the biggest element sorted to the head of the list. `$$b[1]-$$a[1]` is just a fancy way of writing `$$b[1] <=> $$a[1]` which might look more familiar. So the list now looks like this: `( [2,7], [4,5], [1,4], [3,3], [0,2] )` `$index = $sorted[0]->[0];`As a last step, the first element of the sorted list is taken (i.e. `[2,7]` in our example, a reference to a 2 element array) and the first (`->[0]`) item is extracted. This is the index we were looking for. -- Hofmator	[reply] [d/l] [select]
Re: Re4: Getting last valid index from a particular list in a @LoL by blakem (Monsignor) on Oct 17, 2001 at 21:35 UTC
Thanks for expounding on my uncharacteristically terse answer. I was a bit grumpy when I wrote it, having just gone a few rounds over at Varying Variable Names..... -Blake	[reply]