Hello,Chas,
Thanks advanced for your good suggestions.
I tidy up all the words said by you,and write the code as following.Is it
right?
while(<$sock>)
{
my ($key,$value) = split;
my $timestamp = time();
push @records, { time => $timestamp,
key => $key,
amount=> $value,
};
shift @records while $records[0]{time} < time() - 5*60;
my %sum;
for my $rec (@records) {
$sum{$rec->{key}} += $rec->{amount};
}
for (keys %sum)
{
do_something() if $sum{$_} > LIMIT;
}
}
-----Original Message-----
>From: Chas Owens <[EMAIL PROTECTED]>
>Sent: Jan 27, 2006 11:14 PM
>To: Jeff Pang <[EMAIL PROTECTED]>
>Cc: beginners@perl.org
>Subject: Re: count in continuous time piece
>
>On 1/27/06, Jeff Pang <[EMAIL PROTECTED]> wrote:
>> Now I'm still confused on this work.Maybe I have not described the problem
>> clearly.
>> Fox example,there are some items coming in continuous time piece:
>>
>> 00:00:01 itemA 200
>> 00:00:02 itemB 100
>> 00:00:03 itemC 150
>> 00:00:04 itemD 300
>> 00:00:05 itemE 250
>> ...
>> (the item appear as 'name => vaule' style. And most of the items's name are
>> different from others)
>>
>> In every 5 minutes, I'm doing it as following:
>>
>> {
>> sleep 5*60;
>> my %hash = calculate_from_the_items(); #for each item,I'll plus all
>> the historical records to a hash value,and the hash key is this item's name.
>> for (keys %hash){
>> do_something() if $hash{$_} > LIMIT;
>> }
>> clear_the_items_to_null(); #clear all the items's records to null
>> }
>>
>> But it's a very simple resolving method.I'm not satisfied with this way.
>> I want to get this result when each one item is coming:
>>
>> {
>> my $sock=shift;
>> while(<$sock>)
>> {
>> my $isTrue = do_judgement();#judge if it reach some a limit in last
>> 5 minutes relative the current time
>> if ($isTrue){
>> do_something();
>> }
>> clear(); #clear this item's historical record out of 5 minutes
>> relative the current time
>> }
>> }
>>
>> How can I do it?Any suggestion is welcome.Thanks.
>
>In your first example you are creating an aggregate based on a key.
>In your second example you want a similar aggregate, but you don't
>want to count items that are older than 5 minutes ago. This means you
>will need to keep the items separate until the check and you will need
>to store a timestamp (to know how old an item is). Assuming you
>cannot change what is coming in from the socket, you will need to add
>the timestamp that item was read (as opposed to when it was created).
>The ideal pattern for this is an array of either arrays or hashes
>depending on your preference. You will need to read in all available
>records from the socket attaching the time the item was read. At the
>end you should have a data structure that looks like this:
>
>my @records = (
> {
> time => 1138374300,
> key => 'joe',
> amount => 20
> },
> {
> time => 1138374362,
> key => 'mary',
> amount => 10
> },
> {
> time => 1138374462,
> key => 'joe',
> amount => 5
> },
> {
> time => 1138374500,
> key => 'joe',
> amount => 1
> },
> {
> time => 1138374530,
> key => 'mary',
> amount => 20
> },
>);
>
>Assuming it is now 1138374600 then the first record needs to be
>expunged with delete(). This could be achieved by doing something
>like this (warning untested code):
>
>for (@records) {
> delete $_ if $_->{time} < time() - 5*60;
>}
>
>You can then loop over the remaing records creating the aggregate like this
>
>my %sum;
>for my $rec (@records) {
> $sum{$rec->{key}} += $rec->{amount};
>}
--
http://home.earthlink.net/~pangj/
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