Grep count multiple occurrences

bashgrep

Is it possible to do a grep count of multiple occurrences in a file in one single command? For example:

$ cat > file
blah alfa
beta blah
blah blahgamma
gamma

I can do:

grep -c 'alfa' file 
1 
grep -c 'beta' file  
1
grep -c 'gamma' file  
2

But is it possible to so domething like:

grep -c -e 'alfa' -e 'beta' -e 'gamma' -somemoreblackmagic file

and get the counts for each of them? 

alfa 1
beta 1
gamma 2

Best Answer

I don't think grep is capable of what you want to do.

Just use awk instead:-)

This solution may not work well for large files (is not optimized). And works for plain words only - not regexps. But it's easy to add some features if so desired.

Low end version with restrictions outlined in comments below:

awk '
{
    split($0, b); for (i in b) ++A[b[i]]
}
END {
    split("'"$*"'", a)
    for (i in a) print sprintf("%s %d", a[i], A[a[i]])
}
'

just give the search strings directly to the script

[EDIT]
fixed version with regex support (see comment below). Please tell me if there still are any open issues.

# ---- my favorite ----
awk -F' ?-c ' '
BEGIN { split("'"$*"'", a) }
{ for (i = 2; a[i]; ++i) if (match($0, a[i])) ++A[i] }
END { for (i = 2; a[i]; ++i) if (A[i]) print a[i] " " A[i] }
'
# ---- my favorite ----

sample usage:

script_name -c alfa -c beta -c gamma << !
alfa
beta
gamma
gamma
!

gives:

alfa 1
beta 1
gamma 2

regex usage:

script_name -c   "^al"    -c "beta" -c gamma -c "m.$" << !
alfa
beta
gamma
gamma
!

gives:

^al 1
beta 1
gamma 2
m.$ 2

[/EDIT]

Related Question