awk, sed, grep specific strings from a file in Linux - linux

Here is part of the complete file that I am trying to filter:
Hashmode: 13761 - VeraCrypt PBKDF2-HMAC-SHA256 + XTS 512 bit + boot-mode (Iterations: 200000)
Speed.#2.........: 2038 H/s (56.41ms) # Accel:128 Loops:32 Thr:256 Vec:1
Speed.#3.........: 2149 H/s (53.51ms) # Accel:128 Loops:32 Thr:256 Vec:1
Speed.#*.........: 4187 H/s
The aim is to print the following:
13761 VeraCrypt PBKDF2-HMAC-SHA256 4187 H/s
Here is what I tried.
The complete file is called complete.txt
cat complete.txt | grep Hashmode | awk '{print $2,$4,$5}' > mode.txt
Output:
13761 VeraCrypt PBKDF2-HMAC-SHA256
Then:
cat complete.txt | grep Speed.# | awk '{print $2,$3}' > speed.txt
Output:
4187 H/s
Then:
paste mode.txt speed.txt
The issue is that the lines do not match. There are approx 200 types of modes to filter within the file 'complete.txt'
I also have a feeling that this can be done using a much simpler command with sed or awk.

I am guessing you are looking for something like the following.
awk '/Hashmode:/ { if(label) print label, speed; label = $2 " " $4 " " $5 }
/Speed\.#/ { speed = $2 " " $ 3 }
END { if (label) print label, speed }' complete.txt
We match up the Hashmode line with the last Speed.# line which follows, then print when we see a new Hashmode, or reach end of file. (Failing to print the last one is a common beginner bug.)

This might work for you (GNU sed):
sed -E '/Hashmode:/{:a;x;s/^[^:]*: (\S+) -( \S+ \S+ ).*\nSpeed.*:\s*(\S+ \S+).*/\1\2\3/p;x;h;d};H;$!d;ba' file
If a line contains Hashmode, swap to the hold space and using pattern matching, manipulate its contents to the desired format and print, swap back to the pattern space, copy the current line to the hold space and delete the current line.
Otherwise, append the current line to the hold space and delete the current line, unless the current line is the last line in the file, in which case process the line as if it contained Hashmode.
N.B. The first time Hashmode is encountered, nothing is output. Subsequent matches and the end-of-file condition will be the only times printing occurs.

Related

extracting text from one file and creating new file with that text using linux/bash

I have a sequence file that has a repeated pattern that looks like this:
$>g34 | effector probability: 0.6
GPCKPRTSASNTLTTTLTTAEPTPTTIATETTIATSDSSKTTTIDNITTTTSEAESNTKTESSTIAQTRTTTDTSEHESTTASSVSSQPTTTEGITTTSIAQTRTTTDTSEHESTTASSVSSQPTTTEGITTTS"
$>g104 | effector probability: 0.65
GIFSSLICATTAVTTGIICHGTVTLATGGTCALATLPAPTTSIAQTRTTTDTSEH
$>g115 | effector probability: 0.99
IAQTRTTTDTSEHESTTASSVSSQPTTTEGITTTSIAQTRTTTDTSEHESTTASSVSSQPTTTEGITTTS
and so on.
I want to extract the text between and including each >g## and create a new file titled protein_g##.faa
In the above example it would create a file called "protein_g34.faa" and it would be:
$>g34 | effector probability: 0.6
GPCKPRTSASNTLTTTLTTAEPTPTTIATETTIATSDSSKTTTIDNITTTTSEAESNTKTESSTIAQTRTTTDTSEHESTTASSVSSQPTTTEGITTTSIAQTRTTTDTSEHESTTASSVSSQPTTTEGITTTS
I was trying to use sed but I am not very experienced using it. My guess was something like this:
$ sed -n '/^>g*/s///p; y/ /\n/' file > "g##"
but I can clearly tell that that is wrong... maybe the right thing is using awk?
Thanks!
Yeah, I would use awk for that. I don't think sed can write to more than one different output stream.
Here's how I would write that:
< input.txt awk '/^\$>/{fname = "protein_" substr($1, 3) ".faa"; print "sending to " fname} {print $0 > fname}'
Breaking it down into details:
< input.txt This part reads in the input file.
awk Runs awk.
/^\$>/ On lines which start with the literal string $>, run the piece of code in brackets.
(If previous step matched) {fname = "protein_" substr($1, 3) ".faa"; print "sending to " fname} Take the first field in the previous line. Remove the first two characters of that field. Surround that with protein_ .faa. Save it as the variable fname. Print a message about switching files.
This next block has no condition before it. Implicitly, that means that it matches every line.
{print $0 > fname} Take the entire line, and send it to the filename held by fname. If no file is selected, this will cause an error.
Hope that helps!
If awk is an option:
awk '/\|/ {split($1,a,">"); fname="protein_"a[2]".faa"} {print $0 >> fname}' src.dat
awk is better than sed for this problem. You can implement it in sed with
sed -rz 's/(\$>)(g[^ ]*)([^\n]*\n[^\n]*)\n/echo '\''\1\2\3'\'' > protein_\2.faa/ge' file
This solution is nice for showing some sed tricks:
-z for parsing fragments that span several lines
(..) for remembering strings
\$ matching a literal $
[^\n]* matching until end of line
'\'' for a single quote
End single quoted string, escape single quote and start new single quoted string
\2 for recalling the second remembered string
Write a bash command in the replacement string
e execute result of replacement
awk procedure
awk allows records to be extracted between empty (or white space only) lines by setting the record separator to an empty string RS=""
Thus the records intended for each file can be got automatically.
The id to be used in the filename can be extracted from field 1 $1 by splitting the (default white-space-separated) field at the ">" mark, and using element 2 of the split array (named id in this example).
The file is written from awk before closing the file to prevent errors is you have many lines to process.
The awk procedure
The example data was saved in a file named all.seq and the following procedure used to process it:
awk 'BEGIN{RS="";} {split($1,id,">"); fn="protein_"id[2]".faa"; print $0 > fn; close(fn)}' all.seq
tests results
(terminal listings/outputs)
$ ls
all.seq protein_g104.faa protein_g115.faa protein_g34.faa
$ cat protein_g104.faa
$>g104 | effector probability: 0.65
GIFSSLICATTAVTTGIICHGTVTLATGGTCALATLPAPTTSIAQTRTTTDTSEH
$ cat protein_g115.faa
$>g115 | effector probability: 0.99
IAQTRTTTDTSEHESTTASSVSSQPTTTEGITTTSIAQTRTTTDTSEHESTTASSVSSQPTTTEGITTTS
$ cat protein_g34.faa
$>g34 | effector probability: 0.6
GPCKPRTSASNTLTTTLTTAEPTPTTIATETTIATSDSSKTTTIDNITTTTSEAESNTKTESSTIAQTRTTTDTSEHESTTASSVSSQPTTTEGITTTSIAQTRTTTDTSEHESTTASSVSSQPTTTEGITTTS"
Tested using GNU Awk 5.1.0

Output only the first pattern-line and its following line

I need to filter the output of a command.
I tried this.
bpeek | grep nPDE
My problem is that I need all matches of nPDE and the line after the found file. So the output would be like:
iteration nPDE
1 1
iteration nPDE
2 4
The best case would be if it would show me the found line only once and then only the line after it.
I found solutions with awk, But as far as I know awk can only read files.
There is an option for that.
grep --help
...
-A, --after-context=NUM print NUM lines of trailing context
Therefore:
bpeek | grep -A 1 'nPDE'
With awk (for completeness since you have grep and sed solutions):
awk '/nPDE/{c=2} c&&c--'
grep -A works if your grep supports it (it's not in POSIX grep). If it doesn't, you can use sed:
bpeek | sed '/nPDE/!d;N'
which does the following:
/nPDE/!d # If the line doesn't match "nPDE", delete it (starts new cycle)
N # Else, append next line and print them both
Notice that this would fail to print the right output for this file
nPDE
nPDE
context line
If you have GNU sed, you can use an address range as follows:
sed '/nPDE/,+1!d'
Addresses of the format addr1,+N define the range between addr1 (in our case /nPDE/) and the following N lines. This solution is easier to adapt to a different number of context lines, but still fails with the example above.
A solution that manages cases like
blah
nPDE
context
blah
blah
nPDE
nPDE
context
nPDE
would like like
sed -n '/nPDE/{$p;:a;N;/\n[^\n]*nPDE[^\n]*$/!{p;b};ba}'
doing the following:
/nPDE/ { # If the line matches "nPDE"
$p # If we're on the last line, just print it
:a # Label to jump to
N # Append next line to pattern space
/\n[^\n]*nPDE[^\n]*$/! { # If appended line does not contain "nPDE"
p # Print pattern space
b # Branch to end (start new loop)
}
ba # Branch to label (appended line contained "nPDE")
}
All other lines are not printed because of the -n option.
As pointed out in Ed's comment, this is neither readable nor easily extended to a larger amount of context lines, but works correctly for one context line.

Use sed or awk to replace line after match

I'm trying to create a little script that basically uses dig +short to find the IP of a website, and then pipe that to sed/awk/grep to replace a line. This is what the current file looks like:
#Server
123.455.1.456
246.523.56.235
So, basically, I want to search for the '#Server' line in a text file, and then replace the two lines underneath it with an IP address acquired from dig.
I understand some of the syntax of sed, but I'm really having trouble figuring out how to replace two lines underneath a match. Any help is much appreciated.
Based on the OP, it's not 100% clear exactly what needs to replaced where, but here's a a one-liner for the general case, using GNU sed and bash. Replace the two lines after "3" with standard input:
echo Hoot Gibson | sed -e '/3/{r /dev/stdin' -e ';p;N;N;d;}' <(seq 7)
Outputs:
1
2
3
Hoot Gibson
6
7
Note: sed's r command is opaquely documented (in Linux anyway). For more about r, see:
"5.9. The 'r' command isn't inserting the file into the text" in this sed FAQ.
here's how in awk:
newip=12.34.56.78
awk -v newip=$newip '{
if($1 == "#Server"){
l = NR;
print $0
}
else if(l>0 && NR == l+1){
print newip
}
else if(l==0 || NR != l+2){
print $0
}
}' file > file.tmp
mv -f file.tmp file
explanation:
pass $newip to awk
if the first field of the current line is #Server, let l = current line number.
else if the current line is one past #Server, print the new ip.
else if the current row is not two past #Server, print the line.
overwrite original file with modified version.

Remove all text from last dot in bash

I have a file named test.txt which has:
abc.cde.ccd.eed.12345.5678.txt
abcd.cdde.ccdd.eaed.12346.5688.txt
aabc.cade.cacd.eaed.13345.5078.txt
abzc.cdae.ccda.eaed.29345.1678.txt
abac.cdae.cacd.eead.18145.2678.txt
aabc.cdve.cncd.ened.19945.2345.txt
If I want to remove everything beyond the first . like:
cde.ccd.eed.12345.5678.txt
cdde.ccdd.eaed.12346.5688.txt
cade.cacd.eaed.13345.5078.txt
cdae.ccda.eaed.29345.1678.txt
cdae.cacd.eead.18145.2678.txt
cdve.cncd.ened.19945.2345.txt
Then I will do
for i in `cat test.txt`; do echo ${i#*.}; done
but If I want to remove everything after the last . like:
abc.cde.ccd.eed.12345.5678
abcd.cdde.ccdd.eaed.12346.5688
aabc.cade.cacd.eaed.13345.5078
abzc.cdae.ccda.eaed.29345.1678
abac.cdae.cacd.eead.18145.2678
aabc.cdve.cncd.ened.19945.2345
what should I do?
With awk:
awk 'BEGIN{FS=OFS="."} NF--' file
In case there are no empty lines, this works. It sets input and output field separators to the dot .. Then, decreases the number of fields in one, so that the last one is kept out. Then it performs the default awk action: {print $0}, that is, print the line.
With sed:
sed 's/\.[^.]*$//' file
This catches the last block of . + text + end of line and replaces it with nothing. That is, it removes it.
With rev and cut:
rev file | cut -d'.' -f2- | rev
rev reverses the line, so that cut can print from the 2nd word to the end. Then, rev back to get the correct output.
With bash:
while ISF= read -r line
do
echo "${line%.*}"
done < file
This perform a string operation consisting in replacing the shortest match of .* from the end of the variable $line content.
With grep:
grep -Po '.*(?=\.)' file
Look-ahead to print just what is before the last dot.
All of them return:
abc.cde.ccd.eed.12345.5678
abcd.cdde.ccdd.eaed.12346.5688
aabc.cade.cacd.eaed.13345.5078
abzc.cdae.ccda.eaed.29345.1678
abac.cdae.cacd.eead.18145.2678
aabc.cdve.cncd.ened.19945.2345

Extract K-th Line from Chunks Using Sed/AWK/Perl

I have some data that looks like this. It comes in chunk of four lines. Each chunk starts with a # character.
#SRR037212.1 FC30L5TAA_102708:7:1:741:1355 length=27
AAAAAAAAAAAAAAAAAAAAAAAAAAA
+SRR037212.1 FC30L5TAA_102708:7:1:741:1355 length=27
::::::::::::::::::::::::;;8
#SRR037212.2 FC30L5TAA_102708:7:1:1045:1765 length=27
TATAACCAGAAAGTTACAAGTAAACAC
+SRR037212.2 FC30L5TAA_102708:7:1:1045:1765 length=27
88888888888888888888888888
What I want to do is to extract last line of each chunk. Yielding:
::::::::::::::::::::::::;;8
888888888888888888888888888
Note that the last line of the chunk may contain any standard ASCII character
including #.
Is there an effective one-liner to do it?
The following sed command will print the 3rd line after the pattern:
sed -n '/^#/{n;n;n;p}' file.txt
If there are no blank lines:
perl -ne 'print if $. % 4 == 0' file
$ awk 'BEGIN{RS="#";FS="\n"}{print $4 } ' file
::::::::::::::::::::::::;;8
88888888888888888888888888
If you always have those 4 lines in a chunk, some other ways
$ ruby -ne 'print if $.%4==0' file
::::::::::::::::::::::::;;8
88888888888888888888888888
$ awk 'NR%4==0' file
::::::::::::::::::::::::;;8
88888888888888888888888888
It also seems like your line is always after the line that start with "+", so
$ awk '/^\+/{getline;print}' file
::::::::::::::::::::::::;;8
88888888888888888888888888
$ ruby -ne 'gets && print if /^\+/' file
::::::::::::::::::::::::;;8
88888888888888888888888888
This prints the lines before lines that starts with #, and also the last line. It can work with non uniform sized chunks, but assumes that only a chunk leading line starts with #.
sed -ne '1d;$p;/^#/!{x;d};/^#/{x;p}' file
Some explanation is in order:
First you don't need the first line so delete it 1d
Next you always need the last line, so print it $p
If you don't have a match swap it into the hold buffer and delete it x;d
If you do have match swap it out of the hold buffer, and print it x;p
This works similarly to dogbane's answer
awk '/^#/ {mark = NR} NR == mark + 3 {print}' inputfile
And, like that answer, will work regardless of the number of lines in each chunk (as long as there are at least 4).
The direct analog to that answer, however, would be:
awk '/^#/ {next; next; next; print}' inputfile
this can be done using grep easily
grep -A 1 '^#' ./infile
This might work for you (GNU sed):
sed '/^#/,+2d' file

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