Re: sort of misunderstanding of sort

I assume you want the cross-product of a list with itself.

sort can't do this because it's designed to be fast with O(n*log(n))

There used to be a way to change the applied algorithm, but that's deprecated now https://perldoc.perl.org/sort

Cheers Rolf
_{(addicted to the 𐍀𐌴𐍂𐌻 Programming Language :)

Wikisyntax for the Monastery}

Comment on Re: sort of misunderstanding of sort Download Code

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Re^2: sort of misunderstanding of sort by LanX (Saint) on Mar 21, 2023 at 11:51 UTC
> I assume you want the cross-product `DB<34> sub cross (&$$) { my ($code,$list1,$list2) = @_; local ($a,$b +); my @res; for $a (@$list1) { for $b (@$list2) { push @res, &$code } +}; return @res; } DB<35> x cross { $a + $b } [0..2], [10,20,30] 0 10 1 20 2 30 3 11 4 21 5 31 6 12 7 22 8 32 DB<36>` [download] you can just let $list2 default to $list1 if undef to have a self-operation with only one list. `$list2 //= $list1` Cheers Rolf _{(addicted to the 𐍀𐌴𐍂𐌻 Programming Language :) Wikisyntax for the Monastery}	[reply] [d/l] [select]

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Re^2: sort of misunderstanding of sort
by LanX (Saint) on Mar 21, 2023 at 11:51 UTC

> I assume you want the cross-product

  DB<34> sub cross (&$$) { my ($code,$list1,$list2) = @_; local ($a,$b
+); my @res; for $a (@$list1) { for $b (@$list2) { push @res, &$code }
+}; return @res; }

  DB<35> x cross { $a + $b } [0..2], [10,20,30]
0  10
1  20
2  30
3  11
4  21
5  31
6  12
7  22
8  32
  DB<36>
[download]

you can just let $list2 default to $list1 if undef to have a self-operation with only one list.

$list2 //= $list1

Cheers Rolf
_{(addicted to the 𐍀𐌴𐍂𐌻 Programming Language :)

Wikisyntax for the Monastery}

[reply]
[d/l]
[select]