Shows cpu load in percent

sed -e 's/ .*//' -e 's/\.//' -e 's/^0*//' /proc/loadavg
Show the current load of the CPU as a percentage. Read the load from /proc/loadavg and convert it using sed: Strip everything after the first whitespace: sed -e 's/ .*//' Delete the decimal point: sed -e 's/\.//' Remove leading zeroes: sed -e 's/^0*//'
Sample Output
$ cat /proc/loadavg
7.73 7.51 3.42 1/416 8168
$ sed -e 's/ .*//' -e 's/\.//' -e 's/^0*//' /proc/loadavg
773

5
By: flatcap
2014-04-18 19:12:05

1 Alternatives + Submit Alt

  • This version is precise and requires one second to collect statistics. Check sample output for a more generic version and also a remote computer invocation variant. It doesn't work with the busybox version of the 'top' command but can be adjusted Show Sample Output


    1
    top -bn2|awk -F, '/Cpu/{if (NR>4){print 100-gensub(/.([^ ]+).*/,"\\1","g",$4)}}'
    ichbins · 2014-04-18 17:48:05 0

What Others Think

echo $(sed -e 's/ .*//' -e 's/\.//' -e 's/^0*//' /proc/loadavg)%
mpb · 234 weeks and 6 days ago
@mqb: Or you could just alter the first sed clause: sed -e 's/ .*/%/' -e 's/\.//' -e 's/^0*//' /proc/loadavg
flatcap · 234 weeks and 6 days ago
Load average is not a percentage. From `man 1 uptime` : "...a load average of 1 means a single CPU system is loaded all the time while on a 4 CPU system it means it was idle 75% of the time."
Mozai · 233 weeks and 3 days ago

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