I would like to search from the output of ps command by using grep. The criteria is: it contains the specific username and command.
For instance, with the following ps result:
suh1 48980 0.0 0.0 144396 20372 ? Ss Mar16 0:20 SCREEN -DR l2
tprossi 52257 0.0 0.0 127344 3264 ? Ss Mar13 0:26 SCREEN
mannerh1 54331 0.0 0.0 125368 1288 ? Ss 14:32 0:00 SCREEN
thakorv1 54791 0.0 0.0 126096 2116 ? Ss Mar16 0:00 SCREEN
valiman2 58699 0.0 0.0 125364 1244 ? Ss Mar17 0:00 SCREEN
rxue 60121 4.0 0.0 108344 1120 pts/176 R+ 14:39 0:00 ps aux
rxue 60122 0.0 0.0 103252 920 pts/176 S+ 14:39 0:00 grep -e rxue -e SCREEN
I would like to get the process list with user "tprossi" and with command SCREEN.
One solution is:
ps aux | grep tprossi | grep -i screen
But here the grep is called twice, which is not so elegant. Is there any shortcut?
I tried also:
ps aux | grep -i "tprossi*screen"
but it doesn't work :<
Anyone can give some suggestion? Thanks in advance!
You miss a .
ps aux | grep -i "tprossi.*screen"
Without it, you just say repeat i zero or more times.
The . represent any character.
You can also use awk
ps aux | awk '/tprossi/ && /screen/' # in any order
or
ps aux | awk '/tprossi.*screen/' # in this order
Related
When I pipe to grep after a ps aux command it isn't showing the categories at the top of the list (USER, PID, %CPU, %MEM, etc)
Is there something I can do about this?
ps aux --sort -rss | grep $USER | head -n 4
user01 1610 0.0 0.3 17968 10156 ? Ss Jan19 0:01 /lib/systemd/systemd --user
user01 1611 0.0 0.0 104400 2108 ? S Jan19 0:00 (sd-pam)
user01 1617 0.0 0.1 48216 4812 ? S<sl Jan19 0:00 /usr/bin/pipewire
user01 1618 0.0 0.1 32108 4256 ? Ssl Jan19 0:00 /usr/bin/pipewire-media-session
Expecting to see the following at the top of the list:
USER PID %CPU %MEM VSZ RSS TTY STAT START TIME COMMAND
I'm assuming it has something to do with grep filtering them out?
A lot of times when I use ps and pipe it into grep, I put one of the grep characters in [ ] square brackets. I do this do the actual call to grep is not included in the ps readout. It helps when I am scripting. Like if I redirect the output to a file, and that file is empty, the process I am looking at is not on. I do not know why bracketing one character of the grep works, and was amused to see that using the brackets is actually faster and takes less processing time.
-bash-3.2$ time ps -ef | grep *.[j]il
real 0m0.027s
user 0m0.012s
sys 0m0.017s
-bash-3.2$ time ps -ef | grep *.jil
zksuy7k 11528 18285 0 20:54 pts/7 00:00:00 grep *.jil
real 0m0.040s
user 0m0.015s
sys 0m0.016s
-bash-3.2$
I have a Docker image based on Ubuntu that runs a supervisor script as the CMD at the end of the Dockerfile. This successfully runs uwsgi and nginx in the container on start up. However, the following appended at the end of the supervisor-app.conf does not work:
[program:Xvfb]
command=/usr/bin/Xvfb :1 -screen 0 1024x768x16 &> xvfb.log &
When I open a shell into a running docker instance there is no X instance running:
root#9221694363ea:/# ps aux | grep X
root 39 0.0 0.0 8868 784 ? S+ 15:32 0:00 grep --color=auto X
However, running exactly the same command as in the supervisor-app.conf works
root#9221694363ea:/# /usr/bin/Xvfb :1 -screen 0 1024x768x16 &> xvfb.log &
[1] 40
root#9221694363ea:/# ps aux | grep X
root 40 1.2 0.1 170128 21604 ? Sl 15:33 0:00 /usr/bin/Xvfb :1 -screen 0 1024x768x16
root 48 0.0 0.0 8868 792 ? S+ 15:33 0:00 grep --color=auto X
so what's wrong with the line in the supervisor-app.conf?
Supervisor does not handle bash specific operators such as the-run-in-the -background '&' or redirections like '>' as per my original failing config line.
I solved it by using bash -c thus:
[program:Xvfb]
command=bash -c "/usr/bin/Xvfb :1 -screen 0 1024x768x16 &> xvfb.log"
Now when I get into the docker bash shell the Xvfb window is created waiting for me to use it elsewhere in the code.
I am trying to write a Fortran program which will eat up a lot of memory (for the reasoning behind this, please see the note at the end of this question). I am doing this by allocating a 3 dimensional array of size (n,n,n) and then deallocating it - continually increasing n until I run out of memory (this should happen when ~16 GB of memory is used). Unfortunately, it seems as if my program is running out of memory long before I see the system resources get up to 16 GB.
Here is my sample code:
1 program fill_mem
2 implicit none
3 integer, parameter :: ikind = selected_int_kind(8)
4 integer, parameter :: rkind = 8
5
6 integer(kind = ikind) :: nfiles = 100
7 integer(kind = ikind) :: n = 1200
8 integer(kind = ikind) :: i, nn
9
10 real(kind = rkind), allocatable :: real_arr(:,:,:)
11
12 character(500) :: sysline
13
14
15 call system('echo ''***no_allocation***'' > outfile')
16 call system('ps aux | grep fill_mem.exe >> outfile')
17 !call system('smem | grep fill_mem.exe >> sm.out')
18 allocate(real_arr(n, n, n))
19
20 nn = 100000
21 do i = 1,nn
22 deallocate(real_arr)
23 n = n + 10
24 print*, 'n = ', n
25 allocate(real_arr(n, n, n))
26 call system('echo ''*************'' >> outfile')
27 write(sysline, *) 'allocation', i, '... n = ', n
28
29 write(*, '(f10.5, a)') 100.0*real(i)/real(nn), '%'
30
31 call system(trim(adjustl('echo '//sysline//'>> outfile')))
32 call system('ps aux | grep fill_mem.exe >> outfile')
33 enddo
34
35 end program fill_mem
and here is the sample output:
1 ***no_allocation***
2 1000 12350 0.0 0.0 12780 760 pts/1 S+ 13:32 0:00 ./fill_mem.exe
3 1000 12352 0.0 0.0 4400 616 pts/1 S+ 13:32 0:00 sh -c ps aux | grep fill_mem.exe >> outfile
4 1000 12354 0.0 0.0 9384 920 pts/1 S+ 13:32 0:00 grep fill_mem.exe
5 *************
6 allocation 1 ... n = 1210
7 1000 12350 0.0 0.0 13853104 796 pts/1 S+ 13:32 0:00 ./fill_mem.exe
8 1000 12357 0.0 0.0 4400 616 pts/1 S+ 13:32 0:00 sh -c ps aux | grep fill_mem.exe >> outfile
9 1000 12359 0.0 0.0 9384 920 pts/1 S+ 13:32 0:00 grep fill_mem.exe
10 *************
11 allocation 2 ... n = 1220
12 1000 12350 0.0 0.0 14199096 952 pts/1 S+ 13:32 0:00 ./fill_mem.exe
13 1000 12362 0.0 0.0 4400 612 pts/1 S+ 13:32 0:00 sh -c ps aux | grep fill_mem.exe >> outfile
14 1000 12364 0.0 0.0 9384 920 pts/1 S+ 13:32 0:00 grep fill_mem.exe
15 *************
16 allocation 3 ... n = 1230
17 1000 12350 0.0 0.0 14550804 956 pts/1 S+ 13:32 0:00 ./fill_mem.exe
18 1000 12367 0.0 0.0 4400 612 pts/1 S+ 13:32 0:00 sh -c ps aux | grep fill_mem.exe >> outfile
19 1000 12369 0.0 0.0 9384 920 pts/1 S+ 13:32 0:00 grep fill_mem.exe
20 *************
21 allocation 4 ... n = 1240
22 1000 12350 0.0 0.0 14908284 956 pts/1 S+ 13:32 0:00 ./fill_mem.exe
23 1000 12372 0.0 0.0 4400 612 pts/1 S+ 13:32 0:00 sh -c ps aux | grep fill_mem.exe >> outfile
24 1000 12374 0.0 0.0 9384 920 pts/1 S+ 13:32 0:00 grep fill_mem.exe
25 *************
26 allocation 5 ... n = 1250
27 1000 12350 0.0 0.0 15271572 956 pts/1 S+ 13:32 0:00 ./fill_mem.exe
28 1000 12377 0.0 0.0 4400 612 pts/1 S+ 13:32 0:00 sh -c ps aux | grep fill_mem.exe >> outfile
29 1000 12379 0.0 0.0 9384 916 pts/1 S+ 13:32 0:00 grep fill_mem.exe
30 *************
31 allocation 6 ... n = 1260
32 1000 12350 0.0 0.0 15640720 956 pts/1 S+ 13:32 0:00 ./fill_mem.exe
33 1000 12382 0.0 0.0 4400 616 pts/1 S+ 13:32 0:00 sh -c ps aux | grep fill_mem.exe >> outfile
34 1000 12384 0.0 0.0 9384 920 pts/1 S+ 13:32 0:00 grep fill_mem.exe
35 *************
36 allocation 7 ... n = 1270
37 1000 12350 0.0 0.0 16015776 956 pts/1 S+ 13:32 0:00 ./fill_mem.exe
38 1000 12387 0.0 0.0 4400 616 pts/1 S+ 13:32 0:00 sh -c ps aux | grep fill_mem.exe >> outfile
39 1000 12389 0.0 0.0 9384 920 pts/1 S+ 13:32 0:00 grep fill_mem.exe
Now, I see that the VSZ portion gets up to ~15 GB so I am assuming when I try to address more, it fails with
Operating system error: Cannot allocate memory
Allocation would exceed memory limit
because there is not that much RAM. Why is it that RSS is so far below that, though? When I actually look on my system resources I see about 140 MB being used up (I am running this in a Linux VM and monitoring the system resources through Windows - I have given the GM 16 GB of RAM to use though, so I should see the VM memory increasing until it reaches the 16 GB mark - for what it's worth, the VM has VT-x/Nested Paging/PAE/NX so it should use the physical architecture just like the native OS).
Can anyone explain why I do not see my program actually using up the full 16 GB of RAM and how I can write my code to keep these arrays I am creating in RAM - fully utilizing my available hardware?
NOTE: The reason I am trying to write a sample program which reads a lot of memory is that I am working with data which takes up around 14 GB of space in ascii text. I will need to be working with data A LOT throughout the course of this program, so I want to read it all in at once and then reference it from RAM throughout the duration of the program. To make sure I am doing this correctly, I am trying to write a simple program which will store a very large array (~15 GB) in memory all at once.
(Caveat: The Fortran standard doesn't say how such thing ought to be implemented etc., the description below refers to how Fortran compilers are typically implemented on current operating systems.)
When you execute an ALLOCATE statement (or equivalently, calling malloc() in C, FWIW), you're not actually reserving physical memory, but only mapping address space for your process. That's why the VSZ goes up, but not the RSS. Actually reserving physical memory for your process happens only when you first access the memory (typically at page size granularity, that is, 4 KB on most current hw). So only once you start putting some data into your array does the RSS begin to climb. E.g. a statement like
real_arr = 42.
ought to bump up your RSS to the vicinity of the VSZ.
You probably need to increase the memory allocated to the stack. For example, see http://software.intel.com/en-us/articles/intel-fortran-compiler-increased-stack-usage-of-80-or-higher-compilers-causes-segmentation-fault
I have a list of number in a file with format: {integer}\n . So a possible list is:
3
12
53
23
18
32
1
4
i want to use grep to get the count of a specific number, but grep -c "1" file results 3 because it takes into account except the 1, the 12, 18 also. How can i correct this?
Although all the answers until now are logical, and i thought of them and tested before, actually nothing works:
username#domain2:~/code/***/project/random/r2$ cat out.txt
2
16
11
1
13
2
1
16
16
9
username#domain2:~/code/***/project/random/r2$ grep -Pc "^1$" out.txt
0
username#domain2:~/code/***/project/random/r2$ grep -Pc ^1$ out.txt
0
username#domain2:~/code/***/project/random/r2$ grep -c ^1$ out.txt
0
username#domain2:~/code/***/project/random/r2$ grep -c "^1$" out.txt
0
username#domain2:~/code/***/project/random/r2$ grep -xc "^1$" out.txt
0
username#domain2:~/code/***/project/random/r2$ grep -xc "1" out.txt
0
Use the -x flag:
grep -xc 1 file
This is what it means:
-x, --line-regexp
Select only those matches that exactly match the whole line.
There a some other ways you can do this besides grep
$ cat file
3 1 2 100
12 x x x
53
23
18
32
1
4
$ awk '{for(i=1;i<=NF;i++) if ($i=="1") c++}END{print c}' file
2
$ ruby -0777 -ne 'puts $_.scan(/\b1\b/).size' file
2
$ grep -o '\b1\b' file | wc -l
2
$ tr " " "\n" < file | grep -c "\b1\b"
2
Use this regex...
\D1\D
...or ^1$ with multiline mode on.
Tested with RegExr and they both work.
Use e.g. ^123$ to match "Beginning of line, 123, End of line"
grep -wc '23' filename.txt
It will count the number of exact matches of digit 23.