Reputation: 8318
I want to get the CPU and memory usage of a single process on Linux - I know the PID. Hopefully, I can get it every second and write it to a CSV using the 'watch' command. What command can I use to get this info from the Linux command-line?
Upvotes: 241
Views: 589581
Reputation: 384104
Launch a program and monitor it
This form is useful if you want to benchmark an executable easily:
topp() (
if [ -n "$O" ]; then
$* &
else
$* &>/dev/null &
fi
pid="$!"
trap "kill $pid" SIGINT
o='%cpu,%mem,vsz,rss'
printf '%s\n' "$o"
i=0
while s="$(ps --no-headers -o "$o" -p "$pid")"; do
printf "$i $s\n"
i=$(($i + 1))
sleep "${T:-0.1}"
done
)
Usage:
topp ./myprog arg1 arg2
Sample output:
%cpu,%mem,vsz
0 0.0 0.0 177584
1 0.0 0.1 588024
2 0.0 0.1 607084
3 0.0 0.2 637248
4 0.0 0.2 641692
5 68.0 0.2 637904
6 80.0 0.2 642832
where vsz is the total memory usage in KiB, e.g. the above had about 600MiB usage.
If your program finishes, the loop stops and we exit topp
.
Alternatively, if you git Ctrl + C, the program also stops due to the trap
: How do I kill background processes / jobs when my shell script exits?
The options are:
T=0.5 topp ./myprog
: change poll intervalO=1 topp ./myprog
: don't hide program stdout/stderr. This can be useful to help correlate at which point memory usage bursts with stdout.ps
vs top
on instantaneous CPU% usage
Note that the CPU usage given by ps
above is not "instantaneous" (i.e. over the last N seconds), but rather the average over the processes' entire lifetime as mentioned at: https://unix.stackexchange.com/questions/58539/top-and-ps-not-showing-the-same-cpu-result ps
memory measures should be fine however.
That thread as well as: How can I determine the current CPU utilization from the shell? suggest that the Linux kernel does not store any more intermediate usage statistics, so the only way to do that would be to poll and calculate for the previous period, which is what top
does.
We could therefore use top -n1
instead of ps
if we wanted that:
toppp() (
$* &>/dev/null &
pid="$!"
trap exit SIGINT
i=1
top -b n1 -d "${T:-0.1}" -n1 -p "$pid"
while true; do top -b n1 -d "${T:-0.1}" -n1 -p "$pid" | tail -1; printf "$i "; i=$(($i + 1)); done
)
as mentioned e.g. at: https://stackoverflow.com/a/62421136/895245 which produces output of type:
top - 17:36:59 up 9:25, 12 users, load average: 0.32, 1.75, 2.21
Tasks: 1 total, 1 running, 0 sleeping, 0 stopped, 0 zombie
%Cpu(s): 13.4 us, 2.5 sy, 0.0 ni, 84.0 id, 0.0 wa, 0.0 hi, 0.0 si, 0.0 st
MiB Mem : 31893.7 total, 13904.3 free, 15139.8 used, 2849.7 buff/cache
MiB Swap: 0.0 total, 0.0 free, 0.0 used. 16005.5 avail Mem
PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND
706287 ciro 20 0 590436 40352 20568 R 106.7 0.1 0:00.16 node
706287 ciro 20 0 607060 57172 21340 R 126.7 0.2 0:00.35 node
1 706287 ciro 20 0 642008 80276 21812 R 113.3 0.2 0:00.52 node
2 706287 ciro 20 0 641676 93108 21812 R 113.3 0.3 0:00.70 node
3 706287 ciro 20 0 647892 99956 21812 R 106.7 0.3 0:00.87 node
4 706287 ciro 20 0 655980 109564 21812 R 140.0 0.3 0:01.09 node
Some related threads:
top
just once (-b -n1
)
top
: no one has a better solution so we just tail
it:
My only problems with this is that top
is not as nice for interactive usage:
trap exit
is not working as it does with ps
. I have to kill the command Ctrl + \
, and then that does not kill the process itself which continues to run on the background, which means that if it is an infinite loop like a server, I have to ps aux
and then kill it.Maybe someone more shell savvy than me can find a solution for those.
ps
memory measurements should be the same as top
however if you're just after memory.
Related:
Tested on Ubuntu 21.10.
Upvotes: 24
Reputation: 1564
I use htop
sudo apt install htop
htop
Press F3 to search the process you are interested in and remember the PID. Quit with q and start htop again showing the process you want only
htop -p $PID
Upvotes: 1
Reputation: 2287
Based on this answer we can estimate the average CPU and memory utilization of a specific process for a specific amount of time by collecting N samples
with sampling period T
as follows:
N=3;
T=1;
PROCESS_NAME="my_proc";
top -b -c -n $(let tmp=N+1; echo $tmp) -d ${T} -p $(pgrep ${PROCESS_NAME}) |
grep ${PROCESS_NAME} |
tee /var/tmp/foo.log |
tail -n +2 |
awk -v N=$N 'BEGIN{
c=0;
m=0
}{
c=c+$9;
m=m+$10
}END{
printf("%s %s\n", c/N, m/N)
}';
In order to be able to evaluate the results we are collecting the output of the top into the /var/tmp/foo.log
file. The expected output is something like this:
2.33333 6.9
And the content of our log file:
196918 root 20 0 24.4g 1.3g 113872 S 0.0 6.9 39:58.15 my_proc
196918 root 20 0 24.4g 1.3g 113872 S 2.0 6.9 39:58.17 my_proc
196918 root 20 0 24.4g 1.3g 113872 S 3.0 6.9 39:58.20 my_proc
196918 root 20 0 24.4g 1.3g 113872 S 2.0 6.9 39:58.22 my_proc
Note that we ignore (tail -n +2
) the first execution of the top command.
Upvotes: 1
Reputation: 9285
Based on @Neon answer, my two cents here:
pidstat -h -r -u -v -p $(ps aux | grep <process name> | awk '{print $2}' | tr '\n' ',')
Upvotes: 0
Reputation: 22797
Based on @caf's answer, this working nicely for me.
Calculate average for given PID:
measure.sh
times=100
total=0
for i in $(seq 1 $times)
do
OUTPUT=$(top -b -n 1 -d 0.1 -p $1 | tail -1 | awk '{print $9}')
echo -n "$i time: ${OUTPUT}"\\r
total=`echo "$total + $OUTPUT" | bc -l`
done
#echo "Average: $total / $times" | bc
average=`echo "scale=2; $total / $times" | bc`
echo "Average: $average"
Usage:
# send PID as argument
sh measure.sh 3282
Upvotes: 1
Reputation: 2399
The following command gets the average of CPU and memory usage every 40 seconds for a specific process(pid)
pidstat 40 -ru -p <pid>
Output for my case(first two lines for CPU usage, second two lines for memory):
02:15:07 PM PID %usr %system %guest %CPU CPU Command
02:15:47 PM 24563 0.65 0.07 0.00 0.73 3 java
02:15:07 PM PID minflt/s majflt/s VSZ RSS %MEM Command
02:15:47 PM 24563 6.95 0.00 13047972 2123268 6.52 java
Upvotes: 4
Reputation: 11
This is a nice trick to follow one or more programs in real time while also watching some other tool's output:
watch "top -bn1 -p$(pidof foo),$(pidof bar); tool"
Upvotes: 1
Reputation: 4915
For those who struggled for a while wonderring why the selected answer does not work:
ps -p <pid> -o %cpu,%mem
No SPACE ibetween %cpu,
and %mem
.
Upvotes: 5
Reputation: 2033
ps
command (should not use):
top
command (should use):
Use top
to get CPU usage in real time(current short interval):
top -b -n 2 -d 0.2 -p 6962 | tail -1 | awk '{print $9}'
will echo like: 78.6
-b
: Batch-mode-n 2
: Number-of-iterations, use 2
because: When you first run it, it has no previous
sample to compare to, so these initial values are the percentages since boot.-d 0.2
: Delay-time(in second, here is 200ms)-p 6962
: Monitor-PIDstail -1
: the last rowawk '{print $9}'
: the 9-th column(the cpu usage number)Upvotes: 76
Reputation: 159
All of the answers here show only the memory percentage for the PID.
Here's an example of how to get the rss memory usage in KB for all apache processes, replace "grep apache" with "grep PID" if you just want to watch a specific PID:
watch -n5 "ps aux -y | grep apache | awk '{print \$2,\$6}'"
This prints:
Every 5.0s: ps aux -y | grep apache | awk '{print $2,$6}'
Thu Jan 25 15:44:13 2018
12588 9328
12589 8700
12590 9392
12591 9340
12592 8700
12811 15200
15453 9340
15693 3800
15694 2352
15695 1352
15697 948
22896 9360
With CPU %:
watch -n5 "ps aux -y | grep apache | awk '{print \$2,\$3,\$6}'"
Output:
Every 5.0s: ps aux -y | grep apache | awk '{print $2,$3,$6}'
Thu Jan 25 15:46:00 2018
12588 0.0 9328
12589 0.0 8700
12590 0.0 9392
12591 0.0 9340
12592 0.0 8700
12811 0.0 15200
15453 0.0 9340
15778 0.0 3800
15779 0.0 2352
15780 0.0 1348
15782 0.0 948
22896 0.0 9360
Upvotes: 2
Reputation: 211
Above list out the top cpu and memory consuming process
ps axo %cpu,%mem,command | sort -nr | head
Upvotes: 2
Reputation: 239301
ps -p <pid> -o %cpu,%mem,cmd
(You can leave off "cmd" but that might be helpful in debugging).
Note that this gives average CPU usage of the process over the time it has been running.
Upvotes: 310
Reputation: 103
As commented in caf's answer above, ps and in some cases pidstat will give you the lifetime average of the pCPU. To get more accurate results use top. If you need to run top once you can run:
top -b -n 1 -p <PID>
or for process only data and header:
top -b -n 1 -p <PID> | tail -3 | head -2
without headers:
top -b -n 1 -p <PID> | tail -2 | head -1
Upvotes: 6
Reputation: 21
ps axo pid,etime,%cpu,%mem,cmd | grep 'processname' | grep -v grep
PID - Process ID
etime - Process Running/Live Duration
%cpu - CPU usage
%mem - Memory usage
cmd - Command
Replace processname with whatever process you want to track, mysql nginx php-fpm etc ...
Upvotes: 2
Reputation: 1090
Use pidstat (from sysstat - Refer Link).
e.g. to monitor these two process IDs (12345 and 11223) every 5 seconds use
$ pidstat -h -r -u -v -p 12345,11223 5
Upvotes: 45
Reputation: 506
(If you are in MacOS 10.10, try the accumulative -c option of top:
top -c a -pid PID
(This option is not available in other linux, tried with Scientific Linux el6 and RHEL6)
Upvotes: 1
Reputation: 1515
ps aux|awk '{print $2,$3,$4}'|grep PID
where the first column is the PID,second column CPU usage ,third column memory usage.
Upvotes: 1
Reputation: 5559
You can get the results by the name of the process using
ps -C chrome -o %cpu,%mem,cmd
the -C
option allows you to use process name without knowing it's pid.
Upvotes: 47
Reputation: 139
ps aux | awk '{print $4"\t"$11}' | sort | uniq -c | awk '{print $2" "$1" "$3}' | sort -nr
or per process
ps aux | awk '{print $4"\t"$11}' | sort | uniq -c | awk '{print $2" "$1" "$3}' | sort -nr |grep mysql
Upvotes: 3
Reputation: 31580
You could use top -b
and grep out the pid you want (with the -b
flag top runs in batch mode), or also use the -p
flag and specify the pid without using grep.
Upvotes: 5
Reputation: 3909
A variant of caf's answer:
top -p <pid>
This auto-refreshes the CPU usage so it's good for monitoring.
Upvotes: 87
Reputation: 28769
To get the memory usage of just your application (as opposed to the shared libraries it uses, you need to use the Linux smaps interface). This answer explains it well.
Upvotes: 1