How to count occurrences of text in a file?












8

















I have a log file sorted by IP addresses,
I want to find the number of occurrences of each unique IP address.
How can I do this with bash? Possibly listing the number of occurrences next to an ip, such as:



5.135.134.16 count: 5
13.57.220.172: count 30
18.206.226 count:2


and so on.



Here’s a sample of the log:



5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:56 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:06 -0400] "POST /wp-login.php HTTP/1.1" 200 3985 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:08 -0400] "POST /wp-login.php HTTP/1.1" 200 3833 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:09 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:11 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:12 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:15 -0400] "POST /wp-login.php HTTP/1.1" 200 3837 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:17 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] "GET / HTTP/1.1" 200 25160 "-" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/61.0.3163.100 Safari/537.36"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"









share|improve this question

























  • With “bash”, do you mean the plain shell or the command line in general?

    – dessert
    4 hours ago
















8

















I have a log file sorted by IP addresses,
I want to find the number of occurrences of each unique IP address.
How can I do this with bash? Possibly listing the number of occurrences next to an ip, such as:



5.135.134.16 count: 5
13.57.220.172: count 30
18.206.226 count:2


and so on.



Here’s a sample of the log:



5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:56 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:06 -0400] "POST /wp-login.php HTTP/1.1" 200 3985 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:08 -0400] "POST /wp-login.php HTTP/1.1" 200 3833 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:09 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:11 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:12 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:15 -0400] "POST /wp-login.php HTTP/1.1" 200 3837 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:17 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] "GET / HTTP/1.1" 200 25160 "-" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/61.0.3163.100 Safari/537.36"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"









share|improve this question

























  • With “bash”, do you mean the plain shell or the command line in general?

    – dessert
    4 hours ago














8












8








8


4








I have a log file sorted by IP addresses,
I want to find the number of occurrences of each unique IP address.
How can I do this with bash? Possibly listing the number of occurrences next to an ip, such as:



5.135.134.16 count: 5
13.57.220.172: count 30
18.206.226 count:2


and so on.



Here’s a sample of the log:



5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:56 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:06 -0400] "POST /wp-login.php HTTP/1.1" 200 3985 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:08 -0400] "POST /wp-login.php HTTP/1.1" 200 3833 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:09 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:11 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:12 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:15 -0400] "POST /wp-login.php HTTP/1.1" 200 3837 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:17 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] "GET / HTTP/1.1" 200 25160 "-" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/61.0.3163.100 Safari/537.36"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"









share|improve this question


















I have a log file sorted by IP addresses,
I want to find the number of occurrences of each unique IP address.
How can I do this with bash? Possibly listing the number of occurrences next to an ip, such as:



5.135.134.16 count: 5
13.57.220.172: count 30
18.206.226 count:2


and so on.



Here’s a sample of the log:



5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:56 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:06 -0400] "POST /wp-login.php HTTP/1.1" 200 3985 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:08 -0400] "POST /wp-login.php HTTP/1.1" 200 3833 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:09 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:11 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:12 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:15 -0400] "POST /wp-login.php HTTP/1.1" 200 3837 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:17 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] "GET / HTTP/1.1" 200 25160 "-" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/61.0.3163.100 Safari/537.36"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"






command-line bash sort uniq






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share|improve this question








edited 3 hours ago









dessert

25.2k673106




25.2k673106










asked 4 hours ago









j0hj0h

6,5121657119




6,5121657119













  • With “bash”, do you mean the plain shell or the command line in general?

    – dessert
    4 hours ago



















  • With “bash”, do you mean the plain shell or the command line in general?

    – dessert
    4 hours ago

















With “bash”, do you mean the plain shell or the command line in general?

– dessert
4 hours ago





With “bash”, do you mean the plain shell or the command line in general?

– dessert
4 hours ago










4 Answers
4






active

oldest

votes


















6














You can use grep and uniq for the list of addresses, loop over them and grep again for the count:



for i in $(<log grep -o '^[^ ]*' | uniq); do
printf '%s count %dn' "$i" $(<log grep -c "$i")
done


grep -o '^[^ ]*' outputs every character from the beginning (^) until the first space of each line, uniq removes repeated lines, thus leaving you with a list of IP addresses. Thanks to command substitution, the for loop loops over this list printing the currently processed IP followed by “ count ” and the count. The latter is computed by grep -c, which counts the number of lines with at least one match.



Example run



$ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
5.135.134.16 count 5
13.57.220.172 count 9
13.57.233.99 count 1
18.206.226.75 count 2
18.213.10.181 count 3





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  • This solution iterates over the input file repeatedly, once for each IP address, which will be very slow if the file is large. The other solutions using uniq -c or awk only need to read the file once,

    – David
    25 mins ago



















11














You can use cut and uniq tools:



cut -d ' ' -f1 test.txt  | uniq -c
5 5.135.134.16
9 13.57.220.172
1 13.57.233.99
2 18.206.226.75
3 18.213.10.181


Explanation :





  • cut -d ' ' -f1 : extract first field (ip address)


  • uniq -c : report repeated lines and display the number of occurences






share|improve this answer










New contributor




Mikael Flora is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1





    One could use sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.

    – dessert
    3 hours ago



















6














If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer



If you really need the given output format, a single-pass way to do it in Awk would be



awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log


This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) would be:



awk '
NR==1 {last=$1}
$1 != last {print last, "count: " c[last]; last = $1}
{c[$1]++}
END {print last, "count: " c[last]}
'


Ex.



$ awk 'NR==1 {last=$1} $1 != last {print last, "count: " c[last]; last = $1} {c[$1]++} END{print last, "count: " c[last]}' log
5.135.134.16 count: 5
13.57.220.172 count: 9
13.57.233.99 count: 1
18.206.226.75 count: 2
18.213.10.181 count: 3





share|improve this answer

































    5














    Here is one possible solution:





    IN_FILE="file.log"
    for IP in $(awk '{print $1}' "$IN_FILE" | sort -u)
    do
    echo -en "${IP}tcount: "
    grep -c "$IP" "$IN_FILE"
    done



    • replace file.log with the actual file name.

    • the command substitution expression $(awk '{print $1}' "$IN_FILE" | sort -u) will provide a list of the unique values of the first column.

    • then grep -c will count each of these values within the file.




    $ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$IP" "$IN_FILE"; done
    13.57.220.172 count: 9
    13.57.233.99 count: 1
    18.206.226.75 count: 2
    18.213.10.181 count: 3
    5.135.134.16 count: 5





    share|improve this answer


























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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6














      You can use grep and uniq for the list of addresses, loop over them and grep again for the count:



      for i in $(<log grep -o '^[^ ]*' | uniq); do
      printf '%s count %dn' "$i" $(<log grep -c "$i")
      done


      grep -o '^[^ ]*' outputs every character from the beginning (^) until the first space of each line, uniq removes repeated lines, thus leaving you with a list of IP addresses. Thanks to command substitution, the for loop loops over this list printing the currently processed IP followed by “ count ” and the count. The latter is computed by grep -c, which counts the number of lines with at least one match.



      Example run



      $ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
      5.135.134.16 count 5
      13.57.220.172 count 9
      13.57.233.99 count 1
      18.206.226.75 count 2
      18.213.10.181 count 3





      share|improve this answer


























      • This solution iterates over the input file repeatedly, once for each IP address, which will be very slow if the file is large. The other solutions using uniq -c or awk only need to read the file once,

        – David
        25 mins ago
















      6














      You can use grep and uniq for the list of addresses, loop over them and grep again for the count:



      for i in $(<log grep -o '^[^ ]*' | uniq); do
      printf '%s count %dn' "$i" $(<log grep -c "$i")
      done


      grep -o '^[^ ]*' outputs every character from the beginning (^) until the first space of each line, uniq removes repeated lines, thus leaving you with a list of IP addresses. Thanks to command substitution, the for loop loops over this list printing the currently processed IP followed by “ count ” and the count. The latter is computed by grep -c, which counts the number of lines with at least one match.



      Example run



      $ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
      5.135.134.16 count 5
      13.57.220.172 count 9
      13.57.233.99 count 1
      18.206.226.75 count 2
      18.213.10.181 count 3





      share|improve this answer


























      • This solution iterates over the input file repeatedly, once for each IP address, which will be very slow if the file is large. The other solutions using uniq -c or awk only need to read the file once,

        – David
        25 mins ago














      6












      6








      6







      You can use grep and uniq for the list of addresses, loop over them and grep again for the count:



      for i in $(<log grep -o '^[^ ]*' | uniq); do
      printf '%s count %dn' "$i" $(<log grep -c "$i")
      done


      grep -o '^[^ ]*' outputs every character from the beginning (^) until the first space of each line, uniq removes repeated lines, thus leaving you with a list of IP addresses. Thanks to command substitution, the for loop loops over this list printing the currently processed IP followed by “ count ” and the count. The latter is computed by grep -c, which counts the number of lines with at least one match.



      Example run



      $ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
      5.135.134.16 count 5
      13.57.220.172 count 9
      13.57.233.99 count 1
      18.206.226.75 count 2
      18.213.10.181 count 3





      share|improve this answer















      You can use grep and uniq for the list of addresses, loop over them and grep again for the count:



      for i in $(<log grep -o '^[^ ]*' | uniq); do
      printf '%s count %dn' "$i" $(<log grep -c "$i")
      done


      grep -o '^[^ ]*' outputs every character from the beginning (^) until the first space of each line, uniq removes repeated lines, thus leaving you with a list of IP addresses. Thanks to command substitution, the for loop loops over this list printing the currently processed IP followed by “ count ” and the count. The latter is computed by grep -c, which counts the number of lines with at least one match.



      Example run



      $ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
      5.135.134.16 count 5
      13.57.220.172 count 9
      13.57.233.99 count 1
      18.206.226.75 count 2
      18.213.10.181 count 3






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 3 hours ago

























      answered 4 hours ago









      dessertdessert

      25.2k673106




      25.2k673106













      • This solution iterates over the input file repeatedly, once for each IP address, which will be very slow if the file is large. The other solutions using uniq -c or awk only need to read the file once,

        – David
        25 mins ago



















      • This solution iterates over the input file repeatedly, once for each IP address, which will be very slow if the file is large. The other solutions using uniq -c or awk only need to read the file once,

        – David
        25 mins ago

















      This solution iterates over the input file repeatedly, once for each IP address, which will be very slow if the file is large. The other solutions using uniq -c or awk only need to read the file once,

      – David
      25 mins ago





      This solution iterates over the input file repeatedly, once for each IP address, which will be very slow if the file is large. The other solutions using uniq -c or awk only need to read the file once,

      – David
      25 mins ago













      11














      You can use cut and uniq tools:



      cut -d ' ' -f1 test.txt  | uniq -c
      5 5.135.134.16
      9 13.57.220.172
      1 13.57.233.99
      2 18.206.226.75
      3 18.213.10.181


      Explanation :





      • cut -d ' ' -f1 : extract first field (ip address)


      • uniq -c : report repeated lines and display the number of occurences






      share|improve this answer










      New contributor




      Mikael Flora is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.
















      • 1





        One could use sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.

        – dessert
        3 hours ago
















      11














      You can use cut and uniq tools:



      cut -d ' ' -f1 test.txt  | uniq -c
      5 5.135.134.16
      9 13.57.220.172
      1 13.57.233.99
      2 18.206.226.75
      3 18.213.10.181


      Explanation :





      • cut -d ' ' -f1 : extract first field (ip address)


      • uniq -c : report repeated lines and display the number of occurences






      share|improve this answer










      New contributor




      Mikael Flora is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.
















      • 1





        One could use sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.

        – dessert
        3 hours ago














      11












      11








      11







      You can use cut and uniq tools:



      cut -d ' ' -f1 test.txt  | uniq -c
      5 5.135.134.16
      9 13.57.220.172
      1 13.57.233.99
      2 18.206.226.75
      3 18.213.10.181


      Explanation :





      • cut -d ' ' -f1 : extract first field (ip address)


      • uniq -c : report repeated lines and display the number of occurences






      share|improve this answer










      New contributor




      Mikael Flora is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.










      You can use cut and uniq tools:



      cut -d ' ' -f1 test.txt  | uniq -c
      5 5.135.134.16
      9 13.57.220.172
      1 13.57.233.99
      2 18.206.226.75
      3 18.213.10.181


      Explanation :





      • cut -d ' ' -f1 : extract first field (ip address)


      • uniq -c : report repeated lines and display the number of occurences







      share|improve this answer










      New contributor




      Mikael Flora is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this answer



      share|improve this answer








      edited 3 hours ago





















      New contributor




      Mikael Flora is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      answered 4 hours ago









      Mikael FloraMikael Flora

      1116




      1116




      New contributor




      Mikael Flora is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Mikael Flora is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Mikael Flora is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.








      • 1





        One could use sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.

        – dessert
        3 hours ago














      • 1





        One could use sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.

        – dessert
        3 hours ago








      1




      1





      One could use sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.

      – dessert
      3 hours ago





      One could use sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.

      – dessert
      3 hours ago











      6














      If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer



      If you really need the given output format, a single-pass way to do it in Awk would be



      awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log


      This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) would be:



      awk '
      NR==1 {last=$1}
      $1 != last {print last, "count: " c[last]; last = $1}
      {c[$1]++}
      END {print last, "count: " c[last]}
      '


      Ex.



      $ awk 'NR==1 {last=$1} $1 != last {print last, "count: " c[last]; last = $1} {c[$1]++} END{print last, "count: " c[last]}' log
      5.135.134.16 count: 5
      13.57.220.172 count: 9
      13.57.233.99 count: 1
      18.206.226.75 count: 2
      18.213.10.181 count: 3





      share|improve this answer






























        6














        If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer



        If you really need the given output format, a single-pass way to do it in Awk would be



        awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log


        This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) would be:



        awk '
        NR==1 {last=$1}
        $1 != last {print last, "count: " c[last]; last = $1}
        {c[$1]++}
        END {print last, "count: " c[last]}
        '


        Ex.



        $ awk 'NR==1 {last=$1} $1 != last {print last, "count: " c[last]; last = $1} {c[$1]++} END{print last, "count: " c[last]}' log
        5.135.134.16 count: 5
        13.57.220.172 count: 9
        13.57.233.99 count: 1
        18.206.226.75 count: 2
        18.213.10.181 count: 3





        share|improve this answer




























          6












          6








          6







          If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer



          If you really need the given output format, a single-pass way to do it in Awk would be



          awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log


          This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) would be:



          awk '
          NR==1 {last=$1}
          $1 != last {print last, "count: " c[last]; last = $1}
          {c[$1]++}
          END {print last, "count: " c[last]}
          '


          Ex.



          $ awk 'NR==1 {last=$1} $1 != last {print last, "count: " c[last]; last = $1} {c[$1]++} END{print last, "count: " c[last]}' log
          5.135.134.16 count: 5
          13.57.220.172 count: 9
          13.57.233.99 count: 1
          18.206.226.75 count: 2
          18.213.10.181 count: 3





          share|improve this answer















          If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer



          If you really need the given output format, a single-pass way to do it in Awk would be



          awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log


          This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) would be:



          awk '
          NR==1 {last=$1}
          $1 != last {print last, "count: " c[last]; last = $1}
          {c[$1]++}
          END {print last, "count: " c[last]}
          '


          Ex.



          $ awk 'NR==1 {last=$1} $1 != last {print last, "count: " c[last]; last = $1} {c[$1]++} END{print last, "count: " c[last]}' log
          5.135.134.16 count: 5
          13.57.220.172 count: 9
          13.57.233.99 count: 1
          18.206.226.75 count: 2
          18.213.10.181 count: 3






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 3 hours ago

























          answered 4 hours ago









          steeldriversteeldriver

          70.3k11114186




          70.3k11114186























              5














              Here is one possible solution:





              IN_FILE="file.log"
              for IP in $(awk '{print $1}' "$IN_FILE" | sort -u)
              do
              echo -en "${IP}tcount: "
              grep -c "$IP" "$IN_FILE"
              done



              • replace file.log with the actual file name.

              • the command substitution expression $(awk '{print $1}' "$IN_FILE" | sort -u) will provide a list of the unique values of the first column.

              • then grep -c will count each of these values within the file.




              $ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$IP" "$IN_FILE"; done
              13.57.220.172 count: 9
              13.57.233.99 count: 1
              18.206.226.75 count: 2
              18.213.10.181 count: 3
              5.135.134.16 count: 5





              share|improve this answer






























                5














                Here is one possible solution:





                IN_FILE="file.log"
                for IP in $(awk '{print $1}' "$IN_FILE" | sort -u)
                do
                echo -en "${IP}tcount: "
                grep -c "$IP" "$IN_FILE"
                done



                • replace file.log with the actual file name.

                • the command substitution expression $(awk '{print $1}' "$IN_FILE" | sort -u) will provide a list of the unique values of the first column.

                • then grep -c will count each of these values within the file.




                $ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$IP" "$IN_FILE"; done
                13.57.220.172 count: 9
                13.57.233.99 count: 1
                18.206.226.75 count: 2
                18.213.10.181 count: 3
                5.135.134.16 count: 5





                share|improve this answer




























                  5












                  5








                  5







                  Here is one possible solution:





                  IN_FILE="file.log"
                  for IP in $(awk '{print $1}' "$IN_FILE" | sort -u)
                  do
                  echo -en "${IP}tcount: "
                  grep -c "$IP" "$IN_FILE"
                  done



                  • replace file.log with the actual file name.

                  • the command substitution expression $(awk '{print $1}' "$IN_FILE" | sort -u) will provide a list of the unique values of the first column.

                  • then grep -c will count each of these values within the file.




                  $ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$IP" "$IN_FILE"; done
                  13.57.220.172 count: 9
                  13.57.233.99 count: 1
                  18.206.226.75 count: 2
                  18.213.10.181 count: 3
                  5.135.134.16 count: 5





                  share|improve this answer















                  Here is one possible solution:





                  IN_FILE="file.log"
                  for IP in $(awk '{print $1}' "$IN_FILE" | sort -u)
                  do
                  echo -en "${IP}tcount: "
                  grep -c "$IP" "$IN_FILE"
                  done



                  • replace file.log with the actual file name.

                  • the command substitution expression $(awk '{print $1}' "$IN_FILE" | sort -u) will provide a list of the unique values of the first column.

                  • then grep -c will count each of these values within the file.




                  $ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$IP" "$IN_FILE"; done
                  13.57.220.172 count: 9
                  13.57.233.99 count: 1
                  18.206.226.75 count: 2
                  18.213.10.181 count: 3
                  5.135.134.16 count: 5






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 4 hours ago

























                  answered 4 hours ago









                  pa4080pa4080

                  14.7k52872




                  14.7k52872






























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