Processing apache logs quickly

Question

I'm currently running an awk script to process a large (8.1GB) access-log file, and it's taking forever to finish. In 20 minutes, it wrote 14MB of the (1000 +- 500)MB I expect it to write, and I wonder if I can process it much faster somehow.

Here is the awk script:

#!/bin/bash

awk '{t=$4" "$5; gsub("[\[\]\/]"," ",t); sub(":"," ",t);printf("%s,",$1);system("date -d \""t"\" +%s");}' $1

---

EDIT:

For non-awkers, the script reads each line, gets the date information, modifies it to a format the utility date recognizes and calls it to represent the date as the number of seconds since 1970, finally returning it as a line of a .csv file, along with the IP.

Example input: 189.5.56.113 - - [22/Jan/2010:05:54:55 +0100] "GET (...)"

Returned output: 189.5.56.113,124237889

Answer 1

@OP, your script is slow mainly due to the excessive call of system date command for every line in the file, and its a big file as well (in the GB). If you have gawk, use its internal mktime() command to do the date to epoch seconds conversion

awk 'BEGIN{
   m=split("Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec",d,"|")
   for(o=1;o<=m;o++){
      date[d[o]]=sprintf("%02d",o)
    }
}
{
    gsub(/\[/,"",$4); gsub(":","/",$4); gsub(/\]/,"",$5)
    n=split($4, DATE,"/")
    day=DATE[1]
    mth=DATE[2]
    year=DATE[3]
    hr=DATE[4]
    min=DATE[5]
    sec=DATE[6]
    MKTIME= mktime(year" "date[mth]" "day" "hr" "min" "sec)
    print $1,MKTIME

}' file

output

$ more file
189.5.56.113 - - [22/Jan/2010:05:54:55 +0100] "GET (...)"
$ ./shell.sh    
189.5.56.113 1264110895