how to compute CPU usage with bash?

Asked Jun 23 '16 at 10:44

Active Jun 23 '16 at 10:49

Viewed 217 times

I tried this:

top -bn1 | grep "Cpu(s)" | \
>            sed "s/.*, *\([0-9.]*\)%* id.*/\1/" | \
>            awk '{print 100 - $1"%"}'
2.1%

what's the meaning ?

edited Jun 23 '16 at 10:49

Andreas Louv

asked Jun 23 '16 at 10:44

LeoChu

I think it gets the CPU usage of the system idle process, then inverts it. – Jun 23 '16 at 10:50
On my Debian, it seems to extract the idle %, then subtract that from 100, which should get all the "non-idle" fields in one number. – tripleee Jun 23 '16 at 10:55
Refactoring the [useless use of `grep | sed | awk`](http://www.iki.fi/era/unix/award.html#grep) should reduce the load, albeit only slightly. – tripleee Jun 23 '16 at 10:56
@tripleee Dead link – 123 Jun 23 '16 at 10:58
@123 Sorry, should be back up later today. – tripleee Jun 23 '16 at 11:02

0 Answers0