Print unique words, total number of occurrences and sum using `awk`

Question

How can I print unique words, number of their occurrences and the sum of their values in the relevant column using a single array in awk?

I'm using awk like:

awk -F, '{sum[$1]+=$2} END{for (x in sum) print x, sum[x]}' inFile

Can I modify the command above to print the total number of occurrences of unique words as well? Something like the below result for the following sample input:

Result (the order of the printed results doesn't matter):

A 2 25 
B 1 12 
C 3 18

Input:

A,15
C,13
C,4
A,10
B,12
C,1

I can add another array to count them separately but I think there should be another way to print it just using the same array.

Is there any index of the array sum which stores the total words seen?

Answer 1

This should do:

awk -F, '{x[$1]["count"]++;x[$1]["sum"]+=$2}END{for(y in x){print y,x[y]["count"],x[y]["sum"]}}' in

Basically you replace the array with a multidimensional array in order to store both the count of the occurences of each unique first field and the sum of their relative second fields.

% cat in
A,15
C,13
C,4
A,10
B,12
C,1
% awk -F, '{x[$1]["count"]++;x[$1]["sum"]+=$2}END{for(y in x){print y,x[y]["count"],x[y]["sum"]}}' in
A 2 25
B 1 12
C 3 18

Answer 2

No, there is no such index. Array values don't keep count of how many times they have been incremented. The most natural thing to do here is to use a second array:

$ awk -F, '{sum[$1]+=$2;seen[$1]++} END{for(x in sum) print x,seen[x],sum[x]}' file
A 2 25
B 1 12
C 3 18

You can also use a two dimensional array as shown in Kos's answer but, as you can see, that really doesn't simplify things in any way. Alternatively, you could use some perl magic:

$ perl -F, -lane 'push @{$k{$F[0]}},${$k{$F[0]}}[-1]+$F[1]; 
            END{print "$_ ",$#{$k{$_}}+1," ${$k{$_}}[-1]" for keys(%k)}' file
C 3 18
B 1 12
A 2 25

No, that is not line noise and yes, it uses a single array to print everything.