CareerCup

+ Hash to find the number of logs per IP address
+ Use a min-heap to find the top 10 from the hash

I solved this by scanning each line for the IP address, incrementing the respective count in a hash, then sorting the hash by value in the end.

#!/usr/bin/perl

open(FILE, "/path/to/log/file") or die "Sorry! $!";

my %hash = ();

while (my $line = <FILE>) {
	if ($line =~ m/(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})/) {
		$hash{$1}++;
	}
}

foreach (reverse sort {$hash{$a} cmp $hash{$b}} keys %hash) {
	print "$_: $hash{$_}\n";
}

code monkey is rite...we have to compare it with min value of the heap..in min heap it can be found in O(1) and replace it with new val whose count is more

Here is shell script code for doing it:

awk '{print $1}' access_log | sort -u > uniqueips.txt -- first get unique ips assuming ips is in one column.

for i in `cat uniqueips.txt `; do echo $i; grep $i access_log | wc -l; done

capture it in file and sort based on second field to get the top ten ips.

Here is shell script code for doing it:

awk '{print $1}' access_log | sort -u > uniqueips.txt -- first get unique ips assuming ips is in one column.

for i in 'cat uniqueips.txt'; do echo $i; grep $i access_log | wc -l; done

capture it in file and sort based on second field to get the top ten ips.

This is a question about perl.
Following code gives the list of IPaddress in descending order of its number of count.

use List::MoreUtils qw/uniq/;

open (FILE,"logfile.txt") || die "can't open: $!\n";
my $cnt=0;
my $i=0;
my @ipaddr;
my @all=<FILE>;
seek (FILE,0,0);
while (my $line = <FILE>){
my @new = split(" ",$line);
if ($cnt == 0){
foreach $str (@new){
if ($str =~ /\d+\.\d+\.\d+\.\d+/) {
$index=$cnt; last;
}
$cnt++;
}
}
$ipaddr[$i++]=$new[$index];
}
@ipaddr = uniq @ipaddr;

my @ipcount;
my $i=0;
foreach $str (@ipaddr) {
$num=grep ( /$str/,@all);
$ipcount[$i++]="$str $num";
$num=0;
}

@ipcount=reverse sort{(split " ",$a)[1] cmp (split " ",$b)[1]} @ipcount;
foreach $str (@ipcount){
print "$str\n";
}
close (FILE);


output is:
13.12.12.10 9
10.20.30.21 6
15.12.13.30 4

Let's start with a sample log file
192.12.4.5 data 10/12/1 blat
191.12.4.1 date 10/12/a bla
191.12.4.2 date 10/12/a bla
191.12.4.3 date 10/12/a bla
191.12.4.4 date 10/12/a bla
191.12.4.5 date 10/12/a bla
191.12.4.6 date 10/12/a bla
191.12.4.7 date 10/12/a bla
191.12.4.8 date 10/12/a bla
191.12.4.9 date 10/12/a bla
191.12.4.10 date 10/12/a bla
191.12.4.11 date 10/12/a bla
191.12.4.12 date 10/12/a bla
191.12.4.5 date 10/12/a bla
191.12.4.5 date 10/12/a bla
191.12.4.5 date 10/12/a bla
191.12.4.5 date 10/12/a bla
191.12.4.5 date 10/12/a bla
191.12.4.5 date 10/12/a bla
192.12.4.5 date 10/12/a blaa
190.12.4.5 data 10/12/1 blat
191.12.4.3 date 10/12/a bla

Now, since this is perl, we ignore for now complexity. For a long file (more than 1000 lines), one can do better without sort - just scan for the highest, remove it and repeat 10 times.

Note, that they ask for pseudo code , as sorting hash by values is somewhat tricky. I give here both the pseudo code (in a comment) and the real code.


#!/usr/bin/perl -w
use strict;

my %all_IPs;

# Read the log file
for (<>) {
	my @fields = split;
	$all_IPs{$fields[0]} ++;
}

# @best = %all_IPs sorted by keys in a reversed order
my @best = reverse sort  {$all_IPs{$a} cmp $all_IPs{$b}}(keys %all_IPs);

# print the best 10
for (1..10) {
	print $best[$_]."\n";
}

#!/usr/bin/perl -w

use strict;

my $file_path = "F:\\Thesis Work\\SLAMS_IP.csv";

open(FILE, $file_path) or die("Unable to open a file, $!");
open(HANDLE, ">file.txt") or die("Unable to open a file, $!");

my %ip_hash;
my $count;
while(defined(my $lines = <FILE>))
{
	if($lines =~ m/(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})/)
	{
		if(!(defined($ip_hash{$1})))
		{
			$count = 1;
		}
		else
		{
			$count++;
		}
		$ip_hash{$1} = $count;
	}
}

foreach my $key(sort {$ip_hash{$b} <=> $ip_hash{$a}}keys %ip_hash)
{
	print HANDLE "$key:$ip_hash{$key}\n";
}
close(HANDLE);
close(FILE);