Array question

udinakar has asked for the wisdom of the Perl Monks concerning the following question:

Replies are listed 'Best First'.
Re: Array question by moritz (Cardinal) on Sep 05, 2007 at 10:47 UTC
Store the data in a hash, with Date+IP as the key and the repetition count as the value. When you enter an array item into the hash, you just add the count in the last column to the current value.	[reply]
Re^2: Array question by injunjoel (Priest) on Sep 05, 2007 at 17:40 UTC
Great suggestion, `<shameless-plug>`Have a look at The Uniqueness of hashes. to see some examples of just this sort of thing.`</shameless-plug>` -InjunJoel "I do not feel obliged to believe that the same God who endowed us with sense, reason and intellect has intended us to forego their use." -Galileo	[reply] [d/l] [select]
Re: Array question by andreas1234567 (Vicar) on Sep 05, 2007 at 10:48 UTC
Something like: use strict; use warnings; my %hash = (); for (<DATA>) { chomp; my ($day, $ip, $i, $j) = split(q{,}, $_); $hash{$day}{$ip}{i} += $i; $hash{$day}{$ip}{j} += $j; } foreach my $day (sort keys %hash) { foreach my $ip (sort keys %{$hash{$day}}) { print qq{$day,$ip,} . $hash{$day}{$ip}{i} . q{,}. $hash{$day}{$ip}{j}; } } __DATA__ 01 Jul 2007,221.6.19.196,9,19 01 Jul 2007,221.6.19.197,1,3 01 Jul 2007,221.6.19.196,12,12 01 Jul 2007,221.6.19.197,2 ,2 02 Jul 2007,202.119.104.22,1,1 02 Jul 2007,221.6.19.196,20,45 03 Jul 2007,202.119.104.12,1,11 03 Jul 2007,202.119.110.236,1,2 03 Jul 2007,210.29.132.9,1,2 03 Jul 2007,210.29.141.188,1,2 03 Jul 2007,221.6.19.195,3,7 03 Jul 2007,221.6.19.196,4,7 03 Jul 2007,222.192.2.213,1,0 03 Jul 2007,202.119.108.42,3,3 [download] `$ perl -l 637112.pl 01 Jul 2007,221.6.19.196,21,31 01 Jul 2007,221.6.19.197,3,5 02 Jul 2007,202.119.104.22,1,1 02 Jul 2007,221.6.19.196,20,45 03 Jul 2007,202.119.104.12,1,11 03 Jul 2007,202.119.108.42,3,3 03 Jul 2007,202.119.110.236,1,2 03 Jul 2007,210.29.132.9,1,2 03 Jul 2007,210.29.141.188,1,2 03 Jul 2007,221.6.19.195,3,7 03 Jul 2007,221.6.19.196,4,7 03 Jul 2007,222.192.2.213,1,0` [download] Update: Renamed variables from `$a, $b` to `$i, $j` as adviced by GrandFather. -- Andreas	[reply] [d/l] [select]
Re^2: Array question by GrandFather (Saint) on Sep 05, 2007 at 10:53 UTC
Please avoid using $a and $b as general purpose variables. They are "magic" variables used by sort. DWIM is Perl's answer to Gödel	[reply]
Re^3: Array question by narainhere (Monk) on Sep 05, 2007 at 10:59 UTC
Any good pointers to read about this magic variables.Really wanna know grainy details about them The world is so big for any individual to conquer	[reply]
Re^4: Array question by GrandFather (Saint) on Sep 05, 2007 at 11:04 UTC
Re: Array question by GrandFather (Saint) on Sep 05, 2007 at 10:49 UTC
We like to see that you have made a little effort, partly so that we know you have actually thought about the problem and are not just hoping someone will save you having to do any work, and partly to gage your level of understanding of Perl and programming. To get you started you should take a look at perlretut and perlre to gain a little understanding of pattern matching using regular expressions. You will also find it helpful to go back to basics by looking at hashes and their application for managing data accessed by unique keys (see perldata). DWIM is Perl's answer to Gödel	[reply]
Re^2: Array question by udinakar (Novice) on Sep 05, 2007 at 11:03 UTC
Actually this is a table reterived from the database and i tried this my @new_duplicateip_results; foreach my $first (@$select_result) { foreach my $second (@$select_result) { if ((@$first->[0] eq @$second->[0]) && (@$first->[1] eq @$seco +nd->[1]) && (@$first->[$#{$first}] eq $self->product_id) && (@$second +->[$#{$second}] eq "WS-".$self->product_id)) { my $ip_string = "@$first->[0],@$first->[1],"; foreach (2..$#{$first}) { $ip_string .= @$first->[$_]+@$second->[$_].","; } my @split_array = split(/,/,$ip_string) ; push @new_duplicateip_results,\@split_array; } } } foreach (@new_duplicateip_results) { print STDERR join "#",@$_; print STDERR "\n"; } [download] However i need the values that are not matching the creiria. If i include else then multiple rows are fetched. Thanks, Dinakar	[reply] [d/l]
Re^3: Array question by GrandFather (Saint) on Sep 05, 2007 at 11:37 UTC
I apologize for leaping to the worst interpretation of your post! I hope the code below helps. It will muck up the date ordering, but I've run out of time tonight to clean that up. use strict; use warnings; use Data::Dump::Streamer; my @data = ( ['01 Jul 2007','221.6.19.196',9,19], ['01 Jul 2007','221.6.19.197',1,3], ['01 Jul 2007','221.6.19.196',12,12], ['01 Jul 2007','221.6.19.197',2,2], ['02 Jul 2007','202.119.104.22',1,1], ['02 Jul 2007','221.6.19.196',20,45], ['03 Jul 2007','202.119.104.12',1,11], ['03 Jul 2007','202.119.110.236',1,2], ['03 Jul 2007','210.29.132.9',1,2], ['03 Jul 2007','210.29.141.188',1,2], ['03 Jul 2007','221.6.19.195',3,7], ['03 Jul 2007','221.6.19.196',4,7], ['03 Jul 2007','222.192.2.213',1,0], ['03 Jul 2007','202.119.108.42',3,3], ); my %sums; $sums{$_->[0]}{a} += $_->[1], $sums{$_->[0]}{b} += $_->[2] for map {["$_->[0],$_->[1]", @{$_}[2,3]]} @data; my @result = map {[@{$_}[1, 2, 3, 4]]} sort {$a->[0] cmp $b->[0]} map {[$_, split (',', $_), @{$sums{$_}}{'a', 'b'}]} keys %sums; Dump (\@result); [download] Prints: $ARRAY1 = [ [ '01 Jul 2007', '221.6.19.196', 21, 31 ], [ '01 Jul 2007', '221.6.19.197', 3, 5 ], [ '02 Jul 2007', '202.119.104.22', ( 1 ) x 2 ], [ '02 Jul 2007', '221.6.19.196', 20, 45 ], [ '03 Jul 2007', '202.119.104.12', 1, 11 ], [ '03 Jul 2007', '202.119.108.42', ( 3 ) x 2 ], [ '03 Jul 2007', '202.119.110.236', 1, 2 ], [ '03 Jul 2007', '210.29.132.9', 1, 2 ], [ '03 Jul 2007', '210.29.141.188', 1, 2 ], [ '03 Jul 2007', '221.6.19.195', 3, 7 ], [ '03 Jul 2007', '221.6.19.196', 4, 7 ], [ '03 Jul 2007', '222.192.2.213', 1, 0 ] ]; [download] DWIM is Perl's answer to Gödel	[reply] [d/l] [select]
Re^4: Array question by Anonymous Monk on Sep 05, 2007 at 13:02 UTC
Re^5: Array question by GrandFather (Saint) on Sep 05, 2007 at 20:37 UTC
Re^4: Array question by udinakar (Novice) on Sep 06, 2007 at 08:58 UTC
Re^5: Array question by GrandFather (Saint) on Sep 06, 2007 at 11:28 UTC
Some notes below your chosen depth have not been shown here
Re^3: Array question by eric256 (Parson) on Sep 05, 2007 at 21:34 UTC
If this is a table why don't you let the database do the work? `SELECT date, ip, sum(a), sum(b) FROM table GROUP BY date, ip` ___________ Eric Hodges	[reply] [d/l]
Re: Array question by narainhere (Monk) on Sep 05, 2007 at 10:57 UTC
Build a hash with date and ip adress as key,and keep the values in an array After reading the first two entries the hash would look like this, `%myhash{ "01 Jul 2007,221.6.19.196"=>[9,19], "01 Jul 2007,221.6.19.197"=>[1,3] }` [download] Now the third ipaddress and date is same as first one,so add up the new values with the already existing values.Once you have parsed the entire list,can get you o/p by printing the key's and value's of the array.Please throw in some sentences on you next post's, your post looks like a command to a computer ;) The world is so big for any individual to conquer	[reply] [d/l]
Re^2: Array question by narainhere (Monk) on Sep 05, 2007 at 11:03 UTC
Sorry for being redundant.Didn't thought so many replies would have come before I post.	[reply]
Re: Array question by Cristoforo (Curate) on Sep 05, 2007 at 23:43 UTC
I thought I'd post this solution because it has the two different sorts needed. Chris #!/usr/bin/perl use strict; use warnings; use Socket; # program uses the inet_aton and inet_ntoa functions my %month; @month{ qw/ Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec/} = '01'.. +'12'; my %data; while (<DATA>) { my ($date, $ip, $count1, $count2) = split /,/; $data{$date}{ inet_aton($ip) }{count1} += $count1; $data{$date}{ inet_aton($ip) }{count2} += $count2; } for my $date (sort by_date keys %data) { for my $ip (sort keys %{ $data{$date} }) { my $c1 = $data{$date}{$ip}{count1}; my $c2 = $data{$date}{$ip}{count2}; print join(",", $date, inet_ntoa($ip), $c1, $c2), "\n"; } } sub by_date { my ($d_a, $m_a, $y_a) = split ' ', $a; my ($d_b, $m_b, $y_b) = split ' ', $b; my $A = "$y_a$month{$m_a}$d_a"; my $B = "$y_b$month{$m_b}$d_b"; $A cmp $B; } __DATA__ 01 Jul 2007,221.6.19.196,9,19 01 Jul 2007,221.6.19.197,1,3 01 Jul 2007,221.6.19.196,12,12 01 Jul 2007,221.6.19.197,2 ,2 02 Jul 2007,202.119.104.22,1,1 02 Jul 2007,221.6.19.196,20,45 03 Jul 2007,202.119.104.12,1,11 03 Jul 2007,202.119.110.236,1,2 03 Jul 2007,210.29.132.9,1,2 03 Jul 2007,210.29.141.188,1,2 03 Jul 2007,221.6.19.195,3,7 03 Jul 2007,221.6.19.196,4,7 03 Jul 2007,222.192.2.213,1,0 03 Jul 2007,202.119.108.42,3,3 03 Aug 2007,221.6.110.9,1,2 03 Aug 2007,221.6.110.9,10,10 ***prints C:\perlp>perl t7.pl 01 Jul 2007,221.6.19.196,21,31 01 Jul 2007,221.6.19.197,3,5 02 Jul 2007,202.119.104.22,1,1 02 Jul 2007,221.6.19.196,20,45 03 Jul 2007,202.119.104.12,1,11 03 Jul 2007,202.119.108.42,3,3 03 Jul 2007,202.119.110.236,1,2 03 Jul 2007,210.29.132.9,1,2 03 Jul 2007,210.29.141.188,1,2 03 Jul 2007,221.6.19.195,3,7 03 Jul 2007,221.6.19.196,4,7 03 Jul 2007,222.192.2.213,1,0 03 Aug 2007,221.6.110.9,11,12 [download]	[reply] [d/l]
Re^2: Array question by Anonymous Monk on Sep 06, 2007 at 04:56 UTC
Thanks to one and all. My apology for delayed response. I have taken the logic of GrandFather and tweaked it to my requirment it WORKED. I was caught up with that work. Special thanks to GrandFather. I cannot do it from the database as there is a constraint in the logic built. Thanks, Dinakar	[reply]


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Array question

The world is so big for any individual to conquer

The world is so big for any individual to conquer