awk to count using each unique value

Shell Programming and Scripting
1

Im looking for an awk script that will take the unique values in column 5, then print and count the unique values in column 6.

CA001011500  11111 11111 -9999 201301 AAA
CA001012040  11111 11111 -9999 201301 AAA
CA001012573  11111 11111 -9999 201301 BBB
CA001012710  11111 11111 -9999 201301 BBB
CA001015105  11111 11111 -9999 201301 BBB
CA001015638  11111 11111 -9999 201301 CCC
CA001011500  11111 11111  -9999 201302 AAA
CA001012040  11111 11111  -9999 201302 BBB
CA001012573  11111 11111 -9999 201302 BBB
CA001012710  11111  11111 -9999 201302 CCC
CA001015105  11111 11111  -9999 201302 CCC
CA001015638  11111 11111  -9999 201302 CCC

Output will need to be:

201301 AAA 2
201301 BBB 3
201301 CCC 1
201302 AAA 1
201302 BBB 2
201302 CCC 3

I know this will print the unique values of column 5, but how do i tie in the count of unique values in column 6?

 awk '{ a[$5]++ } END { for (b in a) { print b } }'

Thanks in advance!

2

Use field 5 and 6 as index:

awk '{A[$5 OFS $6]++}END{for(k in A) print k, A[k]}' file

to get sorted output:

awk '{A[$5 OFS $6]++}END{for(k in A) print k, A[k]}' file | sort
1 Like
3

This works as expected. Thanks!

4

try also:

awk '$0=$5 FS $6' infile | uniq -c
5

As long as the unique values you want to count are on adjacent lines in your input file, you can easily do it all in awk without needing sort or uniq :

awk '
last != $5 OFS $6 {
        if(c) print last, c
        c = 1
        last = $5 OFS $6
        next
}
{       c++}
END {   print last, c}' file

If you want to try this on a Solaris/SunOS system, change awk to /usr/xpg4/bin/awk , /usr/xpg6/bin/awk , or nawk .