At 05:59 PM 05/05/2002, you wrote:
>Hi Steve,
>
>This may not be the most eficient way, but it seems to work.
>
>Someone will undoubtedly offer another (probably better) way to do it.
This is good, but more than necessary...
>Toby
>
><snip>
>my @numbers = qw(01 02 03 03 05 08 09 12 14 13 11 18 17 12 15 16 15 16 12 13
>14 16 17 22 23 24 25 25 23 22 21 20);
>my %counts;
for my $num ( @numbers ) {
$counts{$num}++;
}
for my $num ( sort keys %counts ) {
print "There were $counts{$num} occurences of $num\n";
}
Notably, no need to check for the existence of the hash element before incrementing
it, even with strict.
>&occur_in_array(\@numbers);
>
>sub occur_in_array
>{
> my $aRef = shift;
>
> foreach (@$aRef)
> {
> if (!exists($counts{$_}))
> {
> $counts{$_} = 1;
> }
> else
> {
> $counts{$_}++;
> }
> }
>
>}
>
>while (($key,$value) = each %counts)
>{
> print "$key = $value\n";
>}
></snip>
>
>-----Original Message-----
>From: steve silvers [mailto:[EMAIL PROTECTED]]
>Sent: Monday, May 06, 2002 10:51 AM
>To: [EMAIL PROTECTED]
>Subject: Regular Expression
>
>
>I have seen a couple of good regular expression questions asked and
>answered. Now I have a question. This is probably easy but i'm stuck on it.
>All I want to do is take.
>
>
>my @numbers =
>(01,02,03,03,05,08,09,12,14,13,11,18,17,12,15,16,15,16,12,13,14,16,17,22,23,
>24,25,25,23,22,21,20);
> so on, could be up to 99. Read them in and get a count of all the numbers.
>
>How many 01 or 17 or 22 there are in the array. I saw an answer for doing
>this with single numbers such as 2897, but how with double numbers. The out
>put i'm looking for is similar.
>
>one - however many times.
>two - however many times.
>...
>...
>twenty - however many times.
>...
>...
>
>You get the picture :-)
>
>Any help would be great.
>Thanks in advance. Steve.
>
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