bash Script to Sum Numerical Output

Question

I'd like to write a Bash script that reports a sum. I'm starting with 20 text files that have the text "LOC: 54" somewhere in them (the "54" could be any integer). So here are the results of searching for this line with grep:

# grep LOC *
ceil.c: * LOC: 38
count_newlines.c: * LOC: 28
even.c: * LOC: 35
every_product.c: * LOC: 48
expand_factors.c: * LOC: 54
factor.c: * LOC: 41
fibonacci.c: * LOC: 49
get_element.c: * LOC: 37
is_composite.c: * LOC: 43
isprime.c: * LOC: 36
largest.c: * LOC: 37
max_product.c: * LOC: 68
mult_list.c: * LOC: 38
nlist.c: * LOC: 37
palindrome.c: * LOC: 72
prime_factors.c: * LOC: 57
remove_dups.c: * LOC: 50
select_products.c: * LOC: 36
square_list.c: * LOC: 31
sum_list.c: * LOC: 38

What could I do to pull just the numerical information together to produce a single number, the sum of the above numbers? I believe in the above example it would be 873.

amit_g · Accepted Answer · 2011-04-21 17:28:22Z

5

awk -F: '/LOC/ {sum += $3;} END {print sum;}' InputFileName

edited Apr 21, 2011 at 17:28

answered Apr 21, 2011 at 17:04

amit_g

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3 Comments

dvanaria Over a year ago

How does this work? It's not looking for the specific text "LOC"

dvanaria Over a year ago

Ah, I see: grep LOC * | awk -F: '{sum += $3;} END {print sum;}'

amit_g Over a year ago

No need for a separate grep. Just add a /LOC/ in the sum statement. Updated the post. You can make it even more specific if needed i.e. / * LOC/

bash-o-logist · Accepted Answer · 2011-04-21 23:14:28Z

1

This is a pure bash script. Does not work with floating point numbers though

#!/bin/bash


shopt -s nullglob
declare -i sum=0
for file in file*
do
    if [ -f "$file" ];then
    while read -r line
    do
        case "$line" in
            *"LOC: "* )
            [[ $line =~ "LOC: ([0-9]+)" ]]
            ((sum+=${BASH_REMATCH[1]}))
            ;;
        esac
    done < "$file"
    fi
done
echo "Sum: $sum"

answered Apr 21, 2011 at 23:14

bash-o-logist

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Comments

odrm · Accepted Answer · 2011-04-21 17:17:37Z

0

This awk script will add those lines that contain LOC:, and end with a number:

 grep LOC * | awk 'BEGIN { FS= "LOC: "; sum = 0 } NF==2 && /[0-9]+$/ { sum+=$2 } END { print sum }'

answered Apr 21, 2011 at 17:17

odrm

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Comments

Fritz G. Mehner · Accepted Answer · 2011-04-21 18:02:18Z

0

Pure Bash. The following script sum.sh does it:

declare sum=0
while read -a line ; do
  (( sum += ${line[@]: -1} ))               # sum up last elements
done < <(grep --exclude=*.sh LOC *)         # exclude script files
echo -e "sum = $sum"

answered Apr 21, 2011 at 18:02

Fritz G. Mehner

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1 Comment

bash-o-logist Over a year ago

you have grep :). not really pure bash

glenn jackman · Accepted Answer · 2011-04-21 21:17:51Z

0

Just an exercise, not the best solution

set -- $(awk '/LOC:/ {print $NF}' filename)
IFS='+'
answer=$( echo "$*" | bc )

answered Apr 21, 2011 at 21:17

glenn jackman

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