Multiple counts in grep? -

so have big log file each line contains date. count number of lines containing each date.

i came awful solution, consisting of manually typing each of following commands:

grep -c "2014-01-01" big.log grep -c "2014-01-02" big.log grep -c "2014-01-03" big.log

i have written small python script, seems overkill. there quicker / more elegant solution?

you can maybe use of regex , uniq -c count results.

see example:

$ cat 2014-01-03 aaa 2014-01-03 aaa 2014-01-02 aaa 2014-01-01 aaa 2014-01-04 aaa hello 2014-01-01 aaa

and let's 2014-01-0x, being x digit, , count them:

$ grep -o "2014-01-0[0-9]" | sort | uniq -c       2 2014-01-01       1 2014-01-02       2 2014-01-03       1 2014-01-04

note piping sort needed make uniq -c work properly. can see more info in answer what meaning of delimiter in cut , why in command sorting twice?.

Silver