I am using awk to grep ‘foo’ from a text file and cacluate sum of field # 7. But, result is rounded to an integer. I need exact result such as 385858.66 and not 385858 using the following command:

grep ‘foo’ 2012-2013.txt | awk ‘BEGIN{ sum=0.0}{ sub(“,”,””,$7); sum +=$7}END{ print “$” sum}’

$682444

I want $682444.57 as output. How can I force “awk” to do floating point math?

 

Floating-point numbers or “real” numbers are that have a fractional part. awk uses double-precision floating-point numbers to represent all numeric values. In other words, all numbers in awk are floating-point numbers i.e. all calculation done using floating-point numbers.

Example: Awk floating point calculation

The following example uses a file called trade.txt, which contains a list of week names as well as four profit values per week:

week1 12.5 12.5 13.5 18.5

week2 11.5 11.10 12.10 13.70

week3 8.5  8.10 8.5 12.5

week4 9.5 11.5 13.5 16.5

week5 8 7 13 17

The following awk program takes the file trade.txt and prints the sum of all four values:

awk  { sum = $2 + $3 + $4 + $5; print $1, sum }  trade.txt

Sample outputs:

week1 57

week2 48.4

week3 37.6

week4 51

week5 45

The following awk program takes the file trade.txt and prints the average of all four values:

awk  { sum = $2 + $3 + $4 + $5; avg = sum/4; print $1, sum , avg}  trade.txt

Sample outputs:

week1 57 14.25

week2 48.4 12.1

week3 37.6 9.4

week4 51 12.75

week5 45 11.25

To avoid surprises use printf to format text to make your output more beautiful and meaningful:

awk  { sum = $2 + $3 + $4 + $5; avg = sum/4; printf  %s: $%.2f ($%05.2f)\n ,$1, sum, avg}  trade.txt

OR

awk  { sum = $2 + $3 + $4 + $5; avg = sum/4; printf  %s: $%.2f ($%5.2f)\n ,$1, sum, avg}  trade.txt

Sample outputs:

week1: $57.00 ($14.25)

week2: $48.40 ($12.10)

week3: $37.60 ($ 9.40)

week4: $51.00 ($12.75)

week5: $45.00 ($11.25)

To fix your problem, replace the following awk code

grep  foo  2012-2013.txt  | awk  BEGIN{ sum=0.0}{ sub( , ,  ,$7); sum +=$7}END{ print  $  sum}

with:

grep  foo  2012-2013.txt  | awk  BEGIN{ sum=0.0}{ sub( , ,  ,$7); sum +=$7}END{  printf  $%.2f\n , sum}

You can skip the grep command and use awk as follows to match and perform sum of all $7:

awk  /foo/{ sub( , ,  ,$7); sum = old + $7; old=sum}END{ printf  $%.2f\n , sum}  2013-2014.txt

Recommended readings

See awk(1) for more info.