Re^2: is there a more simple way to have the intersection of two lists ?

my %seen; $seen{$_}=1 for @list1; my @intersection = grep ++$seen{}==2, @list2;
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You are missing a $_ there. And it doesn't work. (Consider @list1=(); @list2=qw(foo foo);.)

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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]