Re: is there a more simple way to have the intersection of two lists ?

If neither list contains duplicate elements:

my %seen;
my @intersection = grep ++$seen{$_}==2, @list1, @list2;
[download]

Or if they could

my %seen;
$seen{$_}=1 for @list1;
my @intersection = grep ++$seen{$_}==2, @list2;
[download]

Update: $seen{} ⇒ $seen{$_}

Comment on Re: is there a more simple way to have the intersection of two lists ? Select or Download Code

Replies are listed 'Best First'.
Re^2: is there a more simple way to have the intersection of two lists ? by ysth (Canon) on Mar 19, 2008 at 06:40 UTC
`my %seen; $seen{$_}=1 for @list1; my @intersection = grep ++$seen{}==2, @list2;` [download] You are missing a $_ there. And it doesn't work. (Consider `@list1=(); @list2=qw(foo foo);`.) -- Online Fortune Cookie Search	[reply] [d/l] [select]
Re^3: is there a more simple way to have the intersection of two lists ? by ikegami (Patriarch) on Mar 19, 2008 at 06:48 UTC
Hum, right. I over-optimized. I should have stuck to the general solution (i.e. accepts more than two sets) I first wrote. `my %seen; $seen{$_}\|=1<<0 for @list1; $seen{$_}\|=1<<1 for @list2; my @intersection = grep $seen{$_}==(1<<2)-1, keys %seen;` [download] It loses any order, though.	[reply] [d/l]