Reputation: 29
Good afternoon,
I would like to make a cumulative sum for each column and line in awk.
My in file is :
1 2 3 4
2 5 6 7
2 3 6 5
1 2 1 2
And I would like : per column
1 2 3 4
3 7 9 11
5 10 15 16
6 12 16 18
6 12 16 18
And I would like : per line
1 3 5 9 9
2 7 13 20 20
2 5 11 16 16
1 3 4 6 6
I did the sum per column as :
awk '{ for (i=1; i<=NF; ++i) sum[i] += $i}; END { for (i in sum) printf "%s ", sum[i]; printf "\n"; }' test.txt # sum
And per line .
awk '
BEGIN {FS=OFS=" "}
{
sum=0; n=0
for(i=1;i<=NF;i++)
{sum+=$i; ++n}
print $0,"sum:"sum,"count:"n,"avg:"sum/n
}' test.txt
But I would like to print all the lines and columns.
Do you have an idea?
Upvotes: 2
Views: 3269
Reputation: 67497
row sums with repeated last element
$ awk '{s=0; for(i=1;i<=NF;i++) $i=s+=$i; $i=s}1' file
1 3 6 10 10
2 7 13 20 20
2 5 11 16 16
1 3 4 6 6
$i=s
sets the index value (now incremented to NF+1) to the sum and 1
prints the line with that extra field.
columns sums with repeated last row
$ awk '{for(i=1;i<=NF;i++) c[i]=$i+=c[i]}1; END{print}' file
1 2 3 4
3 7 9 11
5 10 15 16
6 12 16 18
6 12 16 18
END{print}
repeats the last row
ps. your math seems to be wrong for the row sums
Upvotes: 2
Reputation: 26481
It looks like you have all the correct information available, all you are missing is the printout statements.
Is this what you are looking for?
accumulated sum of the columns:
% cat foo
1 2 3 4
2 5 6 7
2 3 6 5
1 2 1 2
% awk '{ for (i=1; i<=NF; ++i) {sum[i]+=$i; $i=sum[i] }; print $0}' foo
1 2 3 4
3 7 9 11
5 10 15 16
6 12 16 18
accumulated sum of the rows:
% cat foo
1 2 3 4
2 5 6 7
2 3 6 5
1 2 1 2
% awk '{ sum=0; for (i=1; i<=NF; ++i) {sum+=$i; $i=sum }; print $0}' foo
1 3 6 10
2 7 13 20
2 5 11 16
1 3 4 6
Both these make use of the following :
0
by default (if used numerically)$i
with what the sum
valueprint $0
Upvotes: 4