running ps aux returns :
USER 131 2.1 0.1 23423 423 FFF/5 R 10:12 0:00 -bash
USER 131 2.1 0.1 23423 423 FFF/5 R 10:12 0:00 -test
USER 131 2.1 0.1 23423 423 FFF/5 R 10:12 0:00 -test1
Attempting to filter on bash with wildcards so just
USER 131 2.1 0.1 23423 423 FFF/5 R 10:12 0:00 -bash
is returned :
ps aux|grep "*bash*"
which returns :
invalid option :
grep: invalid option -- 'p'
Usage: grep [OPTION]... PATTERN [FILE]...
Try `grep --help' for more information.
How to filter the output for bash ?
You should just use ps aux|grep 'bash' and it will work the way you want.The * when used in the grep command actually refers to the regex repetition operator of "zero or more" , not the * wildcard character.
ps aux | grep bash | grep -v bash
to return all bash process
Some versions of ps support this directly. For example, to list all processes whose name is bash, run ps like this:
ps -C bash
Related
I am using R in my Ubuntu machine with latest configuration
In R, I get below result:
> read.fwf(pipe('ps -ef | grep /var/lib/docker/'), width = 60)
V1
1 root 29155 29151 0 11:18 pts/0 00:00:00 sh -c ps -ef
2 root 29157 29155 0 11:18 pts/0 00:00:00 grep /var/li
However in Ubuntu console I get different result
ps -ef | grep /var/lib/docker/
root 29150 2509 0 11:17 pts/0 00:00:00 grep --color=auto /var/lib/docker/
I wanted R to fetch PID of /var/lib/docker/, which is according to Ubuntu 2509
Can anyone help me understand why I am getting different result and how to fetch the PID number correctly?
Thanks,
Use ps() in the ps package. This function outputs a data.frame with the process id information.
library(ps)
pid_df <- ps()
pid_df$pid[grep("docker", pid_df$name)]
or in one line:
subset(ps(), grep("docker", name))$pid
I'm running this on an AIX 6.1.
The intended purpose of this command is to display the following information in the following format:
GetUsedRAM:GetUsedSwap:CPU_0_System:CPU_0_User:…CPU_N_System:CPU_N_User
The command is composed of several sub commands:
echo `vmstat 1 2 | tr -s ' ' ':' | cut -d':' -f4,5,14-15 | tail -1 | sed 's/\([0-9]*:[0-9]*:\)\([0-9]*:[0-9]*\)/\1/'``mpstat -a 1 1 | tr -s ' ' '|' | head -8 | tail -4 | cut -d'|' -f 25,27 | awk -F "|" '{printf "%.0f:%.0f:",$2,$1}' | sed '$s/.$//'| sed -e "s/ \{1,\}$//"| awk '{int a[10];split($1, a,":");printf("%d:%d:%d:%d:%d:%d:%d:%d",a[0],a[1],a[2],a[3],a[4],a[5],a[6],a[7])}'`
Which I'll re format for clarity:
echo \
`vmstat 1 2 |
tr -s ' ' ':' |
cut -d':' -f4,5,14-15 |
tail -1 |
sed 's/\([0-9]*:[0-9]*:\)\([0-9]*:[0-9]*\)/\1/' \
` \
`mpstat -a 1 1 |
tr -s ' ' '|' |
head -8 |
tail -4 |
cut -d'|' -f 25,27 |
awk -F "|" '{printf "%.0f:%.0f:",$2,$1}' |
sed '$s/.$//' |
sed -e "s/ \{1,\}$//" |
awk '{int a[10];split($1, a,":");printf("%d:%d:%d:%d:%d:%d:%d:%d",a[0],a[1],a[2],a[3],a[4],a[5],a[6],a[7])}' \
`
I understand all of the tr, cut, head tail, and (roughly) vmstat/mpstat commands. The first sed is where I get lost, I've tried running the command in smaller segments and not quite sure why it seems to work as a whole but not when I truncate the command before the next tr.
I'm also not so sure on the awk command although I understand the premise vaguely, as a function allowing formatted output.
Similarly, I have a vague understanding of sed being a command allowing certain strings/characters being replaced in some file.
I'm not able to make out what this specific implementation in the above case is.
Could anyone provide some clarity or direction as to exactly what is happening at each sed and awk step within the context of the entire command?
Thanks for your help.
Simplification
This two simpler commands will get the exact same output:
# GetUsedRAM:GetUsedSwap:CPU_0_System:CPU_0_User:…CPU_N_System:CPU_N_User
# Select fields 4,5 of last line, and format with :
comm1=`vmstat 1 2 |
awk '$4~/[0-9]/{avm=$4;fre=$5} END{printf "%s:%s",avm,fre}'
`
# Select fields 27 (sy) and 25 (us) for four cpu, print as decimal.
comm2=`mpstat -A 1 1 |
awk -v firstline=6 -v cpus=4 '
BEGIN{start=firstline-1; end=firstline+cpus;}
NR>start && NR<end {printf( ":%d:%d", $27,$25)}'
`
echo "${comm1}${comm2}"
Description.
Description of original commands
The whole command is the concatenation of two commands.
The first command:
The output of the vmstat is shown in this link.
The columns 4 and 5 are 'avm' and 'fre'. The output in columns 14 and 15,
seem to be 'us' (user) and 'sy' (system). And I say seem as no output
from the user is available to confirm.
The first command
`vmstat 1 2 | # Execute the command vmstat.
tr -s ' ' ':' | # convert all spaces to colon (:).
cut -d':' -f4,5,14-15 | # select fields 4,5,14,and 15
tail -1 | # select last line.
sed 's/\([0-9]*:[0-9]*:\)\([0-9]*:[0-9]*\)/\1/' \ # See below.
`
The sed command selects inside braces all digits [0-9]* before a colon
repeated twice. And then again (without the last colon). That's the whole
string in two parts: « (dd:dd:)(dd:dd) » (d means digit).
And finally, it replaces such whole string by what was selected inside
the first braces /\1/.
All this complexity just removes fields 14 and 15 as selected by cut.
A simpler command with exactly the same output is:
Select fields 4,5 of last line, and format with (:).
`vmstat 1 2 | awk '
$4~/[0-9]/{avm=$4;fre=$5} END{printf "%s:%s:",avm,fre}'
`
The second command:
The output of mpstat -A is similar to this one from Linux.
And also similar to this AIX mpstat -d output.
However, the exact output of AIX 6.1 for mpstat -a (ALL) on the computer
used could have several variations. Anyway, guided by the intended final
output desired: CPU_0_System:CPU_0_User:…CPU_N_System:CPU_N_User.
It seems that the columns to be selected should be us (user) and sy
(sys) percent of time that used the cpu for all cpu in use,
which seem to be four on the computer measured.
The manual for AIX 6.1 mpstat is here.
It has a list of all the 40 columns that are presented when the option
-a ALL is used:
CPU min maj mpcs mpcr dev soft dec ph cs ics bound rq push
S3pull S3grd S0rd S1rd S2rd S3rd S4rd S5rd S3hrd S4hrd S5hrd
sysc us sy wa id pc %ec ilcs vlcs lcs %idon %bdon %istol %bstol %nsp
us and sy are listed as the fields 27 and 28, however the command presented
by the user selects fields number 25 and 27. Close but not the same. The
only way to confirm would be to receive the output of the command from the user.
For testing I will be using the output of mpstat 5 1 from here.
# mpstat 5 1
System configuration: lcpu=4 ent=1.0 mode=Uncapped
cpu min maj mpc int cs ics rq mig lpa sysc us sy wt id pc %ec lcs
0 4940 0 1 632 685 268 0 320 100 263924 42 55 0 4 0.57 35.1 277
1 990 0 3 1387 2234 805 0 684 100 130290 28 47 0 25 0.27 16.6 649
2 3943 0 2 531 663 223 0 389 100 276520 44 54 0 3 0.57 34.9 270
3 1298 0 2 1856 2742 846 0 752 100 82141 31 40 0 29 0.22 13.4 650
ALL 11171 0 8 4406 6324 2142 0 2145 100 752875 39 51 0 10 1.63 163.1 1846
The second command
`mpstat -A 1 1 | # execute command
tr -s ' ' '|' | # replace all spaces with (|).
head -8 | # select 8 first lines.
tail -4 | # select last four lines.
cut -d'|' -f 25,27 | # select fields 25 and 27
awk -F "|" '{printf "%.0f:%.0f:",$2,$1}' | # print the fields as integers.
sed '$s/.$//' | # on the last line ($), substitute the last character (.$) by nothing.
sed -e "s/ \{1,\}$//" | # remove trailing space(s).
awk '{
int a[10];
split($1, a,":");
printf("%d:%d:%d:%d:%d:%d:%d:%d",a[0],a[1],a[2],a[3],a[4],a[5],a[6],a[7])
}' \
`
About the int: For older versions of awk, calling a function without the parentheses is equivalent to call the function on $0. int is equivalent to int($0), which is not printed, nor used. The same happens to the value of a[10].
The split sets each value of the command in a[i]. Then, all values of a[i] are printed as decimals.
The equivalent, and way simpler is:
Command #2
`mpstat -A 1 1 |
awk -v firstline=6 -v cpus=4 '
BEGIN{start=firstline-1; end=firstline+cpus;}
NR>start && NR<end {printf( ":%d:%d", $27,$25)}'
`
How do I tell grep to only print out lines if the "filename" matches when I'm piping through ls? I want it to ignore everything on each line until after the timestamp. There must be some easy way to do this on a single command.
As you can see, without it, if I searched for the file "rwx", it would return not only the line with rwx.c, but also the first three lines because of permissions. I was going to use AWK but I want it to display the whole last line if I search for "rwx".
Any ideas?
EDIT: Thanks for the hacks below. However, it would be great to have a more bug-free method. For example, if I had a file named "rob rob", I wouldn't be able to use the stated solutions.
drwxrwxr-x 2 rob rob 4096 2012-03-04 18:03 .
drwxrwxr-x 4 rob rob 4096 2012-03-04 12:38 ..
-rwxrwxr-x 1 rob rob 13783 2012-03-04 18:03 a.out
-rw-rw-r-- 1 rob rob 4294 2012-03-04 18:02 function1.c
-rw-rw-r-- 1 rob rob 273 2012-03-04 12:54 function1.c~
-rw-rw-r-- 1 rob rob 16 2012-03-04 18:02 rwx.c
-rw-rw-r-- 1 rob rob 16 2012-03-04 18:02 rob rob
The following will list only file name, and one file in each row.
$ ls -1
To include . files
$ ls -1a
Please note that the argument is number "1", not letter "l".
Why don't you use grep and match the file name following the timestamp?
grep -P "[0-9]{2}:[0-9]{2} $FILENAME(\.[a-zA-Z0-9]+)?$"
The [0-9]{2}:[0-9]{2} is for the time, the $FILENAME is where you'd put rob rob or rwx, and the trailing (\.[a-zA-Z0-9]+)? is to allow for an optional extension.
Edit: #JonathanLeffler below points out that when files are older than bout 6 months the time column gets replaced by a year - this is what happens on my computer anyhow. You could do ([0-9]{2}:[0-9]{2}|(19|20)[0-9]{2}) to allow time OR year, but you may be best of using awk (?).
[foo#bar ~/tmp]$ls -al
total 8
drwxrwxr-x 2 foo foo 4096 Mar 5 09:30 .
drwxr-xr-- 83 foo foo 4096 Mar 5 09:30 ..
-rw-rw-r-- 1 foo foo 0 Mar 5 09:30 foo foo
-rw-rw-r-- 1 foo foo 0 Mar 5 09:29 rwx.c
-rw-rw-r-- 1 foo foo 0 Mar 5 09:29 tmp
[foo#bar ~/tmp]$export filename='foo foo'
[foo#bar ~/tmp]$echo $filename
foo foo
[foo#bar ~/tmp]$ls -al | grep -P "[0-9]{2}:[0-9]{2} $filename(\.[a-zA-Z0-9]+)?$"
-rw-rw-r-- 1 cha66i cha66i 0 Mar 5 09:30 foo foo
(You could additionally extend to matching the whole line if you wanted:
^ # start of line
[d-]([r-][w-][x-]){3} + # permissions & space (note: is there a 't' or 's'
# sometimes where the 'd' can be??)
[0-9]+ # whatever that number is
[\w-]+ [\w-]+ + # user/group (are spaces allowed in these?)
[0-9]+ + # file size (modify for -h switch??)
(19|20)[0-9]{2}- # yyyy (modify if you want to allow <1900)
(1[012]|0[1-9])- # mm
(0[1-9]|[12][0-9]|3[012]) + # dd
([01][0-9]|2[0-3]):[0-6][0-9] +# HH:MM (24hr)
$filename(\.[a-zA-Z0-9]+)? # filename & optional extension
$ # end of line
. You get the point, tailor to your needs.)
Assuming that you aren't prepared to do:
ls -ld $(ls -a | grep rwx)
then you need to exploit the fact that there are 8 columns with space separation before the file name starts. Using egrep (or grep -E), you could do:
ls -al | egrep "^([^ ]+ +){8}.*rwx"
This looks for 'rwx' after the 8th column. If you want the name to start with rwx, omit the .*. If you want the name to end with rwx, add a $ at the end. Note that I used double quotes so you could interpolate a variable in place of the literal rwx.
This was tested on Mac OS X 10.7.3; the ls -l command consistently gives three columns for the date field:
-r--r--r-- 1 jleffler staff 6510 Mar 17 2003 README,v
-r--r--r-- 1 jleffler staff 26676 Mar 3 21:44 ccs.nmd
Your ls -l seems to be giving just two columns, so you'd need to change the {8} to {7} for your machine - and beware migrating between systems.
Well, if you're working with filenames that don't have spaces in them, you could do something like this:
grep 'rwx\S*$'
Aside frrm the fact that you can use pattern matching with ls, exaple ksh and bash,
which is probably what you should do, you can use the fact that filename occur in a
fixed position. awk (gawk, nawk or whaever you have) is a better choice for this.
If you have to use grep it smells like homework to me. Please tag it that way.
Assume the filename starting position is based on this output from ls -l in linux: 56
-rwxr-xr-x 1 Administrators None 2052 Feb 28 20:29 vote2012.txt
ls -l | awk ' substr($0,56) ~/your pattern even with spaces goes here/'
e.g.,
ls -l | awk ' substr($0,56) ~/^val/'
will find files starting with "val"
As a simple hack, just add a space before your filename so you don't match the beginning of the output:
ls -al | grep '\srwx'
Edit: OK, this is not as robust as it should be. Here's awk:
ls -l | awk ' $9 ~ /rwx/ { print $0 }'
This works for me, unlike ls -l & others as some folks pointed out. I like this because its really generic & gives me the base file name, which removes the path names before the file.
ls -1 /path_name |awk -F/ '{print $NF}'
Only one command you needed for this --
ls -al | gawk '{print $9}'
You can use this:
ls -p | grep -v /
this is super old, but i needed the answer and had a hard time finding it. i didn't really care about the one-liner part; i just needed it done. this is down and dirty and requires that you count the columns. i'm not looking for an upvote here, just leaving some options for future searcher-ers.
the helpful awk trick is here -- Using awk to print all columns from the nth to the last
if
YOUR_FILENAME="rob rob"
and
WHERE_FILENAMES_START=8
ls -al | while read x; do
y=$(echo "$x" | awk '{for(i=$WHERE_FILENAMES_START; i<=NF; ++i) printf $i""FS; print ""}')
[[ "$YOUR_FILENAME " = "$y" ]] && echo "$x"
done
if you save it as a bash script and swap out the vars with $2 and $1, throw the script in your usr bin... then you'll have your clean simple one-liner ;)
output will be:
> -rw-rw-r-- 1 rob rob 16 2012-03-04 18:02 rob rob
the question was for a one-liner so...
ls -al | while read x; do [[ "$YOUR_FILENAME " = "$(echo "$x" | awk '{for(i=WHERE_FILENAMES_START; i<=NF; ++i) printf $i""FS; print ""}')" ]] && echo "$x" ; done
(lol ;P)
on another note: mathematical.coffee your answer was rad. it didn't solve my version of this problem, so i didn't upvote, but i liked your regex breakdown :D
I want find all of the punctuation marks used my .txt file and give a count of the number of occurrences of each one. How would I go about doing this?? I am new at this but I am trying to learn! This is not homework! I have been doing research on grep and sed right now.
$ perl -CSD -nE '$seen{$1}++ while /(\pP)/g; END { say "$_ $seen{$_}" for keys %seen }' sometextfile.utf8
As in
$ perl -CSD -nE '$seen{$1}++ while /(\pP)/g; END { say "$_ $seen{$_}" for keys %seen }' programming_perl_4th_edition.pod | sort -k2rn
, 21761
. 19578
; 10986
( 8856
) 8853
- 7606
: 7420
" 7300
_ 5305
’ 4906
/ 4528
{ 2966
} 2947
\ 2258
# 2121
# 2070
* 1991
' 1715
“ 1406
” 1404
[ 1007
] 1003
% 881
! 838
? 824
& 555
— 330
‑ 72
– 41
‹ 16
› 16
‐ 10
⁂ 10
… 8
· 3
「 2
」 2
« 1
» 1
‒ 1
― 1
‘ 1
• 1
‥ 1
⁃ 1
･ 1
If you want not just punctuation but punctuation and symbols, use [\pP\pS] in your pattern. Don’t use old-style POSIX classes whatever you do, though.
Use sed, tr, sort and uniq (and no perl):
sed -E 's/[^[:punct:]]//g;s/(.)/\1x/g' myfile.txt | tr 'x' '\n' | sort | uniq -c
I did it this way (sed + tr) so it will work on both unix and mac. Mac needs an imbedded linefeed in the sed command, but unix can use \n. This way it works everywhere.
This will work on non-mac unix:
sed -E 's/[^[:punct:]]//g;s/(.)/\1\n/g' myfile.txt | sort | uniq -c
I wish to copy the top 1000 lines in a text file containing more than 50 million entries, to another new file, and also delete these lines from the original file.
Is there some way to do the same with a single shell command in Unix?
head -1000 input > output && sed -i '1,+999d' input
For example:
$ cat input
1
2
3
4
5
6
$ head -3 input > output && sed -i '1,+2d' input
$ cat input
4
5
6
$ cat output
1
2
3
head -1000 file.txt > first100lines.txt
tail --lines=+1001 file.txt > restoffile.txt
Out of curiosity, I found a box with a GNU version of sed (v4.1.5) and tested the (uncached) performance of two approaches suggested so far, using an 11M line text file:
$ wc -l input
11771722 input
$ time head -1000 input > output; time tail -n +1000 input > input.tmp; time cp input.tmp input; time rm input.tmp
real 0m1.165s
user 0m0.030s
sys 0m1.130s
real 0m1.256s
user 0m0.062s
sys 0m1.162s
real 0m4.433s
user 0m0.033s
sys 0m1.282s
real 0m6.897s
user 0m0.000s
sys 0m0.159s
$ time head -1000 input > output && time sed -i '1,+999d' input
real 0m0.121s
user 0m0.000s
sys 0m0.121s
real 0m26.944s
user 0m0.227s
sys 0m26.624s
This is the Linux I was working with:
$ uname -a
Linux hostname 2.6.18-128.1.1.el5 #1 SMP Mon Jan 26 13:58:24 EST 2009 x86_64 x86_64 x86_64 GNU/Linux
For this test, at least, it looks like sed is slower than the tail approach (27 sec vs ~14 sec).
This is a one-liner but uses four atomic commands:
head -1000 file.txt > newfile.txt; tail +1000 file.txt > file.txt.tmp; cp file.txt.tmp file.txt; rm file.txt.tmp
Perl approach:
perl -ne 'if($i<1000) { print; } else { print STDERR;}; $i++;' in 1> in.new 2> out && mv in.new in
Using pipe:
cat en-tl.100.en | head -10