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How to count occurrences of text in a file?
The Next CEO of Stack OverflowCount duplicated words in a text fileuniq --count command is yields incorrect result?How to compare two (vague) file lists and print the duplicates?How to count occurrences of each character?How do I count text lines?How long it will take to sort uniq a 62GB file?How does uniq work?Count number of data points in fileWrong sorting a text fileHow to count number of partial occurrences of a string 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"
command-line bash sort uniq
edited 5 hours ago
dessert
25.2k673106
asked 6 hours ago
j0hj0h
6,5121657119
add a comment |
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"
command-line bash sort uniq
edited 5 hours ago
dessert
25.2k673106
asked 6 hours ago
j0hj0h
6,5121657119
With “bash”, do you mean the plain shell or the command line in general?
– dessert
6 hours ago
add a comment |
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"
command-line bash sort uniq
edited 5 hours ago
dessert
25.2k673106
asked 6 hours ago
j0hj0h
6,5121657119
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
command-line bash sort uniq
edited 5 hours ago
dessert
25.2k673106
asked 6 hours ago
j0hj0h
6,5121657119
edited 5 hours ago
dessert
25.2k673106
asked 6 hours ago
j0hj0h
6,5121657119
edited 5 hours ago
dessert
25.2k673106
edited 5 hours ago
dessert
25.2k673106
edited 5 hours ago
dessert
25.2k673106
25.2k673106
asked 6 hours ago
j0hj0h
6,5121657119
asked 6 hours ago
j0hj0h
6,5121657119
asked 6 hours ago
j0hj0h
6,5121657119
6,5121657119
With “bash”, do you mean the plain shell or the command line in general?
– dessert
6 hours ago
add a comment |
With “bash”, do you mean the plain shell or the command line in general?
– dessert
6 hours ago
With “bash”, do you mean the plain shell or the command line in general?
– dessert
6 hours ago
With “bash”, do you mean the plain shell or the command line in general?
– dessert
6 hours ago
add a comment |
4 Answers
4
active
oldest
votes
7
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
answered 5 hours ago
dessertdessert
25.2k673106
1
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 usinguniq -corawkonly need to read the file once,
– David
2 hours ago
@David this is true, but this would have been my first go at it as well, knowing that grep counts. Unless performance is measurably a problem... dont prematurely optimize?
– D. Ben Knoble
8 mins ago
add a comment |
14
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
answered 6 hours ago
Mikael FloraMikael Flora
1416
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.
2
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
5 hours ago
add a comment |
7
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]++ ENDfor(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]++ ENDprint 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
answered 5 hours ago
steeldriversteeldriver
70.3k11114186
add a comment |
5
Here is one possible solution:
IN_FILE="file.log"
for IP in $(awk 'print $1' "$IN_FILE" | sort -u)
do
echo -en "$IPtcount: "
grep -c "$IP" "$IN_FILE"
done
- replace
file.logwith 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 -cwill 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 "$IPtcount: "; grep -c "$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
answered 5 hours ago
pa4080pa4080
14.7k52872
Preferprintf...
– D. Ben Knoble
6 mins ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
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
answered 5 hours ago
dessertdessert
25.2k673106
1
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 usinguniq -corawkonly need to read the file once,
– David
2 hours ago
@David this is true, but this would have been my first go at it as well, knowing that grep counts. Unless performance is measurably a problem... dont prematurely optimize?
– D. Ben Knoble
8 mins ago
add a comment |
7
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
answered 5 hours ago
dessertdessert
25.2k673106
1
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 usinguniq -corawkonly need to read the file once,
– David
2 hours ago
@David this is true, but this would have been my first go at it as well, knowing that grep counts. Unless performance is measurably a problem... dont prematurely optimize?
– D. Ben Knoble
8 mins ago
add a comment |
7
7
7
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
answered 5 hours ago
dessertdessert
25.2k673106
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
answered 5 hours ago
dessertdessert
25.2k673106
edited 4 hours ago
answered 5 hours ago
dessertdessert
25.2k673106
answered 5 hours ago
dessertdessert
25.2k673106
answered 5 hours ago
dessertdessert
25.2k673106
25.2k673106
1
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 usinguniq -corawkonly need to read the file once,
– David
2 hours ago
@David this is true, but this would have been my first go at it as well, knowing that grep counts. Unless performance is measurably a problem... dont prematurely optimize?
– D. Ben Knoble
8 mins ago
add a comment |
1
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 usinguniq -corawkonly need to read the file once,
– David
2 hours ago
@David this is true, but this would have been my first go at it as well, knowing that grep counts. Unless performance is measurably a problem... dont prematurely optimize?
– D. Ben Knoble
8 mins ago
1
1
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
2 hours 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
2 hours ago
@David this is true, but this would have been my first go at it as well, knowing that grep counts. Unless performance is measurably a problem... dont prematurely optimize?
– D. Ben Knoble
8 mins ago
@David this is true, but this would have been my first go at it as well, knowing that grep counts. Unless performance is measurably a problem... dont prematurely optimize?
– D. Ben Knoble
8 mins ago
add a comment |
14
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
answered 6 hours ago
Mikael FloraMikael Flora
1416
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.
2
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
5 hours ago
add a comment |
14
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
answered 6 hours ago
Mikael FloraMikael Flora
1416
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.
2
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
5 hours ago
add a comment |
14
14
14
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
answered 6 hours ago
Mikael FloraMikael Flora
1416
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
answered 6 hours ago
Mikael FloraMikael Flora
1416
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.
edited 5 hours ago
answered 6 hours ago
Mikael FloraMikael Flora
1416
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 6 hours ago
Mikael FloraMikael Flora
1416
answered 6 hours ago
Mikael FloraMikael Flora
1416
1416
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.
2
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
5 hours ago
add a comment |
2
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
5 hours ago
2
2
One could use
sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.– dessert
5 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
5 hours ago
add a comment |
7
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]++ ENDfor(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]++ ENDprint 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
answered 5 hours ago
steeldriversteeldriver
70.3k11114186
add a comment |
7
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]++ ENDfor(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]++ ENDprint 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
answered 5 hours ago
steeldriversteeldriver
70.3k11114186
add a comment |
7
7
7
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]++ ENDfor(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]++ ENDprint 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
answered 5 hours ago
steeldriversteeldriver
70.3k11114186
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]++ ENDfor(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]++ ENDprint 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
answered 5 hours ago
steeldriversteeldriver
70.3k11114186
edited 5 hours ago
answered 5 hours ago
steeldriversteeldriver
70.3k11114186
answered 5 hours ago
steeldriversteeldriver
70.3k11114186
answered 5 hours ago
steeldriversteeldriver
70.3k11114186
70.3k11114186
add a comment |
add a comment |
5
Here is one possible solution:
IN_FILE="file.log"
for IP in $(awk 'print $1' "$IN_FILE" | sort -u)
do
echo -en "$IPtcount: "
grep -c "$IP" "$IN_FILE"
done
- replace
file.logwith 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 -cwill 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 "$IPtcount: "; grep -c "$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
answered 5 hours ago
pa4080pa4080
14.7k52872
Preferprintf...
– D. Ben Knoble
6 mins ago
add a comment |
5
Here is one possible solution:
IN_FILE="file.log"
for IP in $(awk 'print $1' "$IN_FILE" | sort -u)
do
echo -en "$IPtcount: "
grep -c "$IP" "$IN_FILE"
done
- replace
file.logwith 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 -cwill 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 "$IPtcount: "; grep -c "$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
answered 5 hours ago
pa4080pa4080
14.7k52872
Preferprintf...
– D. Ben Knoble
6 mins ago
add a comment |
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 "$IPtcount: "
grep -c "$IP" "$IN_FILE"
done
- replace
file.logwith 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 -cwill 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 "$IPtcount: "; grep -c "$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
answered 5 hours ago
pa4080pa4080
14.7k52872
Here is one possible solution:
IN_FILE="file.log"
for IP in $(awk 'print $1' "$IN_FILE" | sort -u)
do
echo -en "$IPtcount: "
grep -c "$IP" "$IN_FILE"
done
- replace
file.logwith 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 -cwill 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 "$IPtcount: "; grep -c "$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
answered 5 hours ago
pa4080pa4080
14.7k52872
edited 5 hours ago
answered 5 hours ago
pa4080pa4080
14.7k52872
answered 5 hours ago
pa4080pa4080
14.7k52872
answered 5 hours ago
pa4080pa4080
14.7k52872
14.7k52872
Preferprintf...
– D. Ben Knoble
6 mins ago
add a comment |
Preferprintf...
– D. Ben Knoble
6 mins ago
Prefer
printf...– D. Ben Knoble
6 mins ago
Prefer
printf...– D. Ben Knoble
6 mins ago
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With “bash”, do you mean the plain shell or the command line in general?
– dessert
6 hours ago