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Yes, that function is unchanged, but it writes the result in column 4, not 1 as column 1 is always the bin (interval) number -- I assume that 1 was a typo.

Small tweeks to the original awk to show number of new IPs in the current bin compared to the previous.


#!/usr/bin/env ksh

awk -v bin_size=${1:-5} '
function dump( )
{
if(...

??? Try:
paste -d' ' file1 file2

Have a go with this:


#!/usr/bin/env ksh

awk -v bin_size=${1:-5} '
function dump( )
{
if( NR == 1 )
return;
printf( "%3d %3d %3d\n", bin+1, total,...

It works perfectly and it's a very good script (I tried to write one but I couldn't). Get your INPUTFILE from stdin like this:
./this_script 1 <INPUTFILE

OK, try this. It copies the current IPs array to PIPs (previous IPs) once a line is output. This array is used to check each for IPs not in the previous range.

awk -F"[:, ]" ' { now=mktime("2000...

sed 's/\..*//' inputfile > newfile

awk -F '.' '{print $1}' inputfile > newfile

This is a one-liner version of agama's approach which may speed thing up a little bit.
awk -F: '{t=$1*3600+$2*60+$3} NR>1{printf( "%.6f\n", t-p)}{p=t}' infile

There's no check for a clock roll-over, but this might get you started:


awk '
{
split( $1, a, ":" )
t = a[1] * 3600;
t += a[2] * 60;...

Using awk
awk 'NF{a[$NF]++}END{for(i in a)print i,a[i]}' file | sort

You can run agama's code directly, no need add it in awk command again.

awk -v window=${1:-7200} '
{
if( $1 > end_window ) # reached the end of the time window
...

This should get you started. You'll have to add conversion if you want to allow the user to supply 2 hours rather than 7200 seconds. Not sure how fast this will be, don't have the patience tonight...