my $last = ( grep { ... } LIST )[-1];
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Thanks for your swift reply and I got the solution.

This is largely redundant now, but:

>perl -wMstrict -le
"use List::AllUtils qw(lastval);
 my @call_names = (
   'test AB 25',
   'test BC 29',
   'test AC 25',
   'test AA 30',
   'test AA 31',
   'test AA 25',
   'test AA 20',
   );
 my $last_25 = lastval { m{(\d+)}xms and $1 == 25 } @call_names;
 print qq{'$last_25'};
"
'test AA 25'
[download]

Or even (Larry help us)
    my $last_25 = lastval { (m{\d+}xmsg)[0] == 25 } @call_names;

if you provide enough detail that we can understand what you're doing!

Please answer these questions by posting the code that generates the unsatisfactory output:

• How and where is @Last_Element declared
• what are $i's initial value and limit?
• How are you incrementing $i?

I -- for one -- am having a hard time crystal-balling code that would behave as you say. One standard bit of advice here is that those with questions post code that actually compiles under strict and warnings.

Note (just for starters) that @STATION_TYPE[$i] offends the warnings pragma (which we hope you are using along with use strict;) instigating the message: Scalar value @STATION_TYPE[$i] better written as $STATION_TYPE[$i]

Oh, yes: the standard way to find the last element in an array is to use the subscript -1 ("print $CALL_NAME[-1];") ... which, of course is not the element you actually want. But if you were to iterate backward from $CALL_NAME[-1] looking for the FIRST "25" you could save a lot of processing time.

And, uh, your original code is not very good. For example, it'll match "2250" too.