UNIX (AIX) Command Help - Sed & Awk - unix

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)}'
`

Related

How to Remove Code Specific Code Lines using Unix

Could someone please help/advise how could I removed the first 4 line and the last 2 line of codes in my 3 JavaScript files using the Shell Script?
I tried using this guide: UNIX - delete specific lines but it will only work for the first 4 lines. All 3 Javascript files have different set of line of codes.
set -vx
lines2del="(1,2,3,4)"
sedCmds=${lines2del//,/d;}
sedCmds=${sedCmds/(/}
sedCmds=${sedCmds/)/}
sedCmds=${sedCmds}d
sed -i "$sedCmds" file
Any inputs are highly appreciated. Thanks
This might work for you (GNU sed):
sed -i '1,4d;N;$d;P;D' file
This deletes the lines 1 to 4 and then prints all other lines except the last two which it also deletes.
Add the following to your lines2del:
$(($(cat file | wc -l)-2)) // third last line
$(($(cat file | wc -l)-1)) // second last line
$(cat file | wc -l) // last line
$ seq 10 | tail -n +5 | head -n -2
5
6
7
8
$ seq 10 | awk '{p3=p2; p2=p1; p1=$0} NR>6{print p3}'
5
6
7
8
$ seq 10 | awk '{p[NR%6]=$0} NR>6{print p[(NR-2)%6]}'
5
6
7
8
$ seq 10 | awk -v b=4 -v a=2 'BEGIN{t=b+a} {p[NR%t]=$0} NR>t{print p[(NR-a)%t]}'
5
6
7
8
$ seq 10 | awk -v b=3 -v a=5 'BEGIN{t=b+a} {p[NR%t]=$0} NR>t{print p[(NR-a)%t]}'
4
5

Count lines that contains a repeated pattern

I have a file (File.txt) that contains 10 lines which may contains a repeated patterns at a certain position (14-17)
fsdf sfkljkl4565
fjjf lmlkfdm1235
fkljfgdfgdfg6583
eretjioijolj6933
ioj ijijsfoi4565
dgodiiopkpok6933
fsj opkjfiej4565
ihfzejijjijf4565
dfsdkfjlfeff1235
dijdijojijdz4565
The Desired Output is counting the lines that contains a pattern :
#occurences pattern
5 4565
2 1235
1 6583
2 6933
I have tried to filter the file
cat File.txt | cut -c14-17 | sort -n -K1,1-1,3 >> File_Filtered.txt
I need your help to add the first column (#occurences)
To get a count of repeats, use uniq -c. Thus, try:
$ cut -c13-17 File.txt | sort -n | uniq -c | sort -nr
5 4565
2 6933
2 1235
1 6583
The above was tested using Linux with GNU utilities. (Judging by your sample code, you may be using different tools.)
Including a header
The following includes the header and uses column -t to assure that everything lines up nicely:
$ { echo '#occurences pattern'; cut -c13-17 File.txt | sort -n | uniq -c | sort -nr; } | column -t
#occurences pattern
5 4565
2 6933
2 1235
1 6583
$ awk '{cnt[substr($0,13)]++} END{for (i in cnt) print cnt[i], i}' file
2 6933
1 6583
5 4565
2 1235

How can I get a range of line every nth interval using awk, sed, or other unix command?

I know how to get a range of lines by using awk and sed.
I also do know how to print out every nth line using awk and sed.
However, I don't know how to combined the two.
For example, I have a file with 1780000 lines.
For every 17800th line, I would like to print 17800th line plus the two after that.
So if I have a file with 1780000 lines and it starts from 1 and ends at 1780000, this will print:
1
2
3
17800
17801
17802
35600
35601
35602
# ... and so on.
Does anyone know how to get a range of line every nth interval using awk, sed, or other unix command?
Using GNU sed:
sed -n '0~17800{N;N;p}' input
Meaning,
For every 17800th line: 0~17800
Read two lines: {N;N;
And print these out: p}
We can also add the first three lines:
sed -n -e '1,3p' -e '0~17800{N;N;p}' input
Using Awk, this would be simpler:
awk 'NR%17800<3 || NR==3 {print}' input
$ cat file
1
2
3
4
5
6
7
8
9
10
$ awk '!(NR%3)' file
3
6
9
$ awk -v intvl=3 -v delta=2 '!(NR%intvl){print "-----"; c=delta} c&&c--' file
-----
3
4
-----
6
7
-----
9
10
$ awk -v intvl=4 -v delta=2 '!(NR%intvl){print "-----"; c=delta} c&&c--' file
-----
4
5
-----
8
9
$ awk -v intvl=4 -v delta=3 '!(NR%intvl){print "-----"; c=delta} c&&c--' file
-----
4
5
6
-----
8
9
10
seq -f %.0f 1780000 | awk 'NR < 4 || NR % 17800 < 3' | head
output:
1
2
3
17800
17801
17802
35600
35601
35602
53400
Explanation
The NR < 4 is for the first 3 lines because the requirement For every 17800th line, print 17800th line plus the two after that. doesn't fit the output you gave.
Here I use head for reducing the output size and you should remove it in your use case.
For GNU seq, you don't need -f %.0f.

Sort history on number of occurrences

Basically I want to print the 10
most used commands that are stored in the
bash history but they still have to be proceeded
by the number that indicates when it was used;
I got this far:
history | cut -f 2 | cut -d ' ' -f 3,5 | sort -k 2 -n
Which should sort the second column of the number of occurrences from the command in that row... But it doesn't do that. I know I can head -10 the pipe at the end to take the highest ten of them, but I'm kinda stuck with the sorting part.
The 10 most used commands stored in your history:
history | sed -e 's/ *[0-9][0-9]* *//' | sort | uniq -c | sort -rn | head -10
This gives you the most used command line entries by removing the history number (sed), counting (sort | uniq -c), sorting by frequency (sort -rn) and showing only the top ten entries.
If you just want the commands alone:
history | awk '{print $2;}' | sort | uniq -c | sort -rn | head -10
Both of these strip the history number. Currently, I have no idea, how to achieve that in one line.
If you want to find the top used commands in your history file, you will have to count the instances in your history. awk can be used to do this. In the following code, the awk segment will create a hashtable with commands as the key and the number of times they appear as the value. This is printed out with the last history number for that command and sorted:
history | cut -f 2 | cut -d ' ' -f 3,5 | awk '{a[$2]++;b[$2]=$1} END{for (i in a) {print b[i], i, a[i]}}' | sort -k3 -rn | head -n 10
Output looks like:
975 cd 142
972 vim 122
990 ls 118
686 hg 90
974 mvn 51
939 bash 39
978 tac 32
958 cat 28
765 echo 27
981 exit 17
If you don't want the last column you could pipe the output through cut -d' ' -f1,2.

Finding punctuation and counting the number of each from the Unix Command line

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

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