How do I iterate through each line of a command's output in bash?

Question

I have a script that reads from /proc/stat and calculates CPU usage. There are three relevant lines in /proc/stat:

cpu  1312092 24 395204 12582958 77712 456 3890 0 0 0
cpu0 617029 12 204802 8341965 62291 443 2718 0 0 0
cpu1 695063 12 190402 4240992 15420 12 1172 0 0 0

Currently, my script only reads the first line and calculates usage from that:

cpu=($( cat /proc/stat | grep '^cpu[^0-9] ' ))
unset cpu[0]
idle=${cpu[4]}
total=0
for value in "${cpu[@]}"; do
  let total=$(( total+value ))
done

let usage=$(( (1000*(total-idle)/total+5)/10 ))
echo "$usage%"

This works as expected, because the script only parses this line:

cpu  1312092 24 395204 12582958 77712 456 3890 0 0 0

It's easy enough to get only the lines starting with cpu0 and cpu1

cpu=$( cat /proc/stat | grep '^cpu[0-9] ' )

but I don't know how to iterate over each line and apply this same process. Ive tried resetting the internal field separator inside a subshell, like this:

cpus=$( cat /proc/stat | grep '^cpu[0-9] ' )
(
IFS=$'\n'
for cpu in $cpus; do
    cpu=($cpu)
    unset cpu[0]
    idle=${cpu[4]}
    total=0
    for value in "${cpu[@]}"; do
        let total=$(( total+value ))
    done
    let usage=$(( (1000*(total-idle)/total+5)/10 ))
    echo -n "$usage%"
done
)

but this gets me a syntax error

line 18: (1000*(total-idle)/total+5)/10 : division by 0 (error token is "+5)/10 ")

If I echo the cpu variable in the loop it looks like it's separating the lines properly. I looked at this thread and I think Im assigning the cpu variable to an array properly but is there another error Im not seeing?

I put my script into whats wrong with my script and it doesnt show me any errors apart from a warning about using cat within $(), s o I'm stumped.

@Staven I'll look into that. Is something like this pretty much a one liner in awk? — Michael A, Dec 09 '13 at 04:53
@janos Thanks for the reminder. Your bash answer is helpful, although the `awk` script in the other answer is superior to the one-line `awk` script in your answer. — Michael A, Dec 17 '13 at 22:23
@user30588 you're welcome. I don't know what you mean by "superior" though, his solution is equivalent, only the syntax is different. — janos, Dec 17 '13 at 23:01
@janos The non-one-liner code is more readable, and takes advantage of the fact that `awk` allows for floating-point division. The `(1000 * (total - idle) / total + 5) / 10` hack that I used in my bash code (and that's in your `awk` code) isn't necessary in `awk`. It just adds clutter. — Michael A, Dec 17 '13 at 23:03

janos · Accepted Answer · 2013-12-09T05:02:15.540

6

Change this line in the middle of your loop:

IFS=' ' cpu=($cpu)

You need this because outside of your loop you're setting IFS=$'\n', but with that settingcpu($cpu)` won't do what you expect.

Btw, I would write your script like this:

#!/bin/bash -e

grep ^cpu /proc/stat | while IFS=$'\n' read cpu; do
    cpu=($cpu)
    name=${cpu[0]}
    unset cpu[0]
    idle=${cpu[4]}
    total=0
    for value in "${cpu[@]}"; do
        ((total+=value))
    done
    ((usage=(1000 * (total - idle) / total + 5) / 10))
    echo "$name $usage%"
done

The equivalent using awk:

awk '/^cpu/ { total=0; idle=$5; for (i=2; i<=NF; ++i) { total += $i }; print $1, int((1000 * (total - idle) / total + 5) / 10) }' < /proc/stat

edited Dec 09 '13 at 05:02

answered Dec 09 '13 at 04:48

janos

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3

Everything is a one-liner if you delete the newlines. – Staven Dec 09 '13 at 05:08
Don't know why you said that. It was natural for me to write this on one line, I'm not arguing it's better to write on one line. I see you wrote pretty much the same thing but expanded to multiple lines, which is more readable. – janos Dec 17 '13 at 22:59

Staven · Answer 2 · 2013-12-09T05:35:44.733

3

Because the OP asked, an awk program.

awk '
    /cpu[0-9] .*/ {
        total = 0
        idle = $5
        for(i = 0; i <= NF; i++) { total += $i; }
        printf("%s: %f%%\n", $1, 100*(total-idle)/total);
    }
' /proc/stat

The /cpu[0-9] .*/ means "execute for every line matching this expression". The variables like $1 do what you'd expect, but the 1st field has index 1, not 0: $0 means the whole line in awk.

edited Dec 09 '13 at 05:35

answered Dec 09 '13 at 05:07

Staven

3,113
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Does awk not have the rounding issues that bash has? That's why I used `(1000 * (total - idle) / total + 5) / 10` instead of just `100*(total-idle)/total` – Michael A Dec 09 '13 at 05:26
@user30588 I don't know. What rounding issues does Bash have? Regardless though, I've put the wrong format letter in the printf, fixed now. – Staven Dec 09 '13 at 05:36
`bash` doesn't have rounding issues so much as it only performs integer division: `a/b` is always an integer, with the remainder thrown out. – chepner Dec 09 '13 at 16:15
1

Then no, `awk` doesn't have such a limitation. Try `awk 'BEGIN { print 1/2; }' /dev/null`, for example. – Staven Dec 10 '13 at 14:35

How do I iterate through each line of a command's output in bash?

2 Answers2