What's the best way to get the last N elements of a Perl array?
If the array has less than N, I don't want a bunch of undefs in the return value.
views:
458answers:
4
A:
my $size = (scalar @list) - 1; my @newList = @list[$size-$n..$size];
Tim Howland
2009-03-04 18:05:57
Doesn't work. You need the .. sigil, not comma, and $size is too large by one.
chaos
2009-03-04 18:15:26
you're right, too much time in groovy- I'll edit to match
Tim Howland
2009-03-04 18:17:08
May as well just say $#list like Nathan instead of putting scalar(@list) - 1 in a variable.
chaos
2009-03-04 18:31:24
I thought about that- since the OP was new to perl, I thought it would be better to avoid the $# construction, in favor of the more straightforward scalar call- I remember hating all the wacky special variables when I was getting started.
Tim Howland
2009-03-04 19:04:15
Who said I'm new to Perl?
mike
2009-03-06 22:18:13
+15
A:
@last_n = @source[-$n..-1];
If you require no undefs, then:
@last_n = $n >= @source ? @source : @source[-$n..-1];
chaos
2009-03-04 18:06:40
It work okay. undefs go into @last_n in the positions for which @source has no values, which is correct for some not-entirely-unreasonable semantics of what it means to "take the last N elements".
chaos
2009-03-04 18:13:53
oh, I've never used negative subscripts like that, I learned something today!
Nathan
2009-03-04 23:49:28
@Chaos: I'm putting this here since you wouldn't see a comment on the main post. What's with all the scrubbing and polishing of old, answered questions lately? You've got too much rep to be looking for more (and I know it's not your style). Bored at work? :)
Telemachus
2009-08-27 20:23:48
A:
@a = (a .. z);
@last_five = @a[ $#a - 4 .. $#a ];
say join " ", @last_five;
outputs:
v w x y z
Nathan
2009-03-04 18:06:52