Trying to use awk to remove the IonCode_4 digits (always 4 may be different) and leave the file extension. Is the below the best way? Thank you :).
file
1112233 ID_1234_000000-Control_z_zzzz_zz_zz_zz_zz_zz_zzz_zz-zzzz-zzz-zzz_zzzz_zzzz_zzz_zzz_zzz_zzz_zzz.txt
1112231 ID_1234_000000-Control_z_zzzz_zz_zz_zz_zz_zz_zzz_zz-zzzz-zzz-zzz_zzzz_zzzz_zzz_zzz_zzz_zzz_zzz.txt
awk
awk '/_tn_/ {next} gsub ("^.*/|_.*$|IonCode_...._", "", $2)'f
current
1112233 000000-Control
1112231 000000-Control
desired
1112233 000000-Control.txt
1112231 000000-Control.txt
Split records by 1+ spaces or underscore, so the 4th field will be the part you're interested in.
awk -F '[[:space:]]+|_' '!/_tn_/{print $1,$4".txt"}' file
Could you please try following. This is simplest I could think, though we could do it with number of fields mentioning too but that will be more like hard-coding of numbers, so I went with this approach here.
awk '
{
sub(/[^_]*_/,"",$2)
sub(/[^_]*_/,"",$2)
sub(/_.*/,".txt")
}
1
' Input_file
with sed
$ sed -E 's/ID_[0-9]{4}_([^_]+).*(\..*)/\1\2/' file
1112233 000000-Control.txt
1112231 000000-Control.txt
Related
I made a mistake, adding Ini file (Yes we're in 2022 :D) a section with errors
I added a line [End[edit=true]
How could remove this entire line using awk (I don't have any others choice đŸ˜•)
I don't understand how escape the [ in the AWK command line.
Could you please help me?
Thanks
I don't understand how escape the [ in the AWK command line.
If line is always literal [End[edit=true] then you do not need to, just request lines which are not that one following way, let file.ini content be
[someline=true]
[End[edit=true]
[another=true]
then
awk '$0!="[End[edit=true]"' file.ini
gives output
[someline=true]
[another=true]
Explanation: $0 denotes whole line, if it is not [End[edit=true] then it is printed.
(tested in GNU Awk 5.0.1)
A couple ideas where you escape the leading (left) brackets:
awk '/\[End\[edit=true]/ {next} 1' file
# or
awk '!/\[End\[edit=true]/' file
Once you've confirmed the results, and assuming you're using GNU awk, you can add -i inplace to update the file:
awk -i inplace '!/\[End\[edit=true]/' file
Having a very large file where two lines shown below and having two fields name and revision having colon delimiter. I need to print only the second column.
sam:7.[0:6]
Ram:8.[6:6]_rev[2:4] h_ack[2:6]
vincent:58
I tried this code:
#!/bin/bash
awk -F: '{print $2}'
7.[0
8.[6
58
Output should be:
7.[0:6]
8.[6:6]_rev[2:4] h_ack[2:6]
58
What went wrong in my code.
The problem in your awk expression is that you are splitting on all :.
Instead, you want to split only on the first : from the start.
$ awk -F'^[^:]+:' '{print $2}' file
The regex pattern matches the start of the string ^, any character other than a :, and finally a :.
If you specify field separator as :, it's normal behavior of awk to output this, ex:
7.[0, because you need the other columns after $2.
cut here, better suits the requirement:
cut -d: -f2- file
Could you please try following.
awk '
match($0,/:.*/){
print substr($0,RSTART+1,RLENGTH-1)
}
' Input_file
I am trying to match two different Regexp to long strings with awk, removing the part of the string that matches in a 35 characters window.
The problem is that the same bunch of code works when I am looking for the first (which matches at the beginnng) whereas fails to match with the second one (end of string).
Input:
Regexp1(1)(2)(3)(4)(5)xxxxxxxxxxxxxxx(20)(21)(22)(23)Regexp2
Desired output
(1)(2)(3)(4)(5)xxxxxxxxxxxxxxx(20)(21)(22)(23)
So far I used this code that extracts correctly Regexp1, but, unfortunately, is not able to extract also Regexp2 since indexed of RSTART and RLENGTH for Regexp2 are incorrect.
Code for extracting Regexp1 (correct output):
awk -v F="Regexp1" '{if (match(substr($1,1,35),F)) print substr($1,RSTART,RLENGTH)}' file
Code for extracting Regexp2 (wrong output)
awk -v F="Regexp2" '{if (match(substr($1,length($1)-35,35),F)) print substr($1,RSTART,RLENGTH)}' file
Despite the indexes for Regexp1 are correct, for Regexp2 indexes are wrond (RSTART=13). I cannot figure out how to extract the second Regexp.
Considering that your actual Input_file is same as shown samples, if this is the case could you please try following then(good to have new version of awk since old versions may not support number of times logic for regex).
awk '
match($0,/\([0-9]+\){5}.*\([0-9]\){4}/){
print substr($0,RSTART,RLENGTH)
}' Input_file
In case your number of parenthesis values are not fixed then you could do like as follows:
awk '
match($0,/\([0-9]+\){1,}.*\([0-9]\){1,}/){
print substr($0,RSTART,RLENGTH)
}' Input_file
If this isn't all you need:
$ sed 's/Regexp1\(.*\)Regexp2/\1/' file
(1)(2)(3)(4)(5)xxxxxxxxxxxxxxx(20)(21)(22)(23)
or using GNU awk for gensub():
$ awk '{print gensub(/Regexp1(.*)Regexp2/,"\\1",1)}' file
(1)(2)(3)(4)(5)xxxxxxxxxxxxxxx(20)(21)(22)(23)
then edit your question to be far clearer with your requirements and example.
I would like to take the number after the - sign and put is as column 2 in my matrix. I know how to grep the string but not how to print it after the text string.
in:
1-967764 GGCTGGTCCGATGGTAGTGGGTTATCAGAACT
3-425354 GCATTGGTGGTTCAGTGGTAGAATTCTCGCC
4-376323 GGCTGGTCCGATGGTAGTGGGTTATCAGAAC
5-221398 GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT
6-180339 TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT
out:
GGCTGGTCCGATGGTAGTGGGTTATCAGAACT 967764
GCATTGGTGGTTCAGTGGTAGAATTCTCGCC 425354
GGCTGGTCCGATGGTAGTGGGTTATCAGAAC 376323
GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT 221398
TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT 180339
awk -F'[[:space:]-]+' '{print $3,$2}' file
Seems like a simple substitution should do the job:
sed -E 's/[0-9]+-([0-9]+)[[:space:]]*(.*)/\2 \1/' file
Capture the parts you're interested in and use them in the replacement.
Alternatively, using awk:
awk 'sub(/^[0-9]+-/, "") { print $2, $1 }' file
Remove the leading digits and - from the start of the line. When this is successful, sub returns true, so the action is performed, printing the second field, followed by the first.
Using regex ( +|-) as field separator:
$ awk -F"( +|-)" '{print $3,$2}' file
GGCTGGTCCGATGGTAGTGGGTTATCAGAACT 967764
GCATTGGTGGTTCAGTGGTAGAATTCTCGCC 425354
GGCTGGTCCGATGGTAGTGGGTTATCAGAAC 376323
GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT 221398
TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT 180339
here is another awk
$ awk 'split($1,a,"-") {print $2,a[2]}' file
awk '{sub(/.-/,"");print $2,$1}' file
GGCTGGTCCGATGGTAGTGGGTTATCAGAACT 967764
GCATTGGTGGTTCAGTGGTAGAATTCTCGCC 425354
GGCTGGTCCGATGGTAGTGGGTTATCAGAAC 376323
GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT 221398
TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT 180339
I am trying to fetch the lines in which the second part of the line contains a pattern from the first part of the line.
$ cat file.txt
String1 is a big string|big
$ awk -F'|' ' { if ($2 ~ /$1/) { print $0 } } ' file.txt
But it is not working.
I am not able to find out what is the mistake here.
Can someone please help?
Two things: No slashes, and your numbers are backwards.
awk -F\| '$1~$2' file.txt
I guess what you meant is part of the string in the first part should be a part of the 2nd part.if this is what you want! then,
awk -F'|' '{n=split($1,a,' ');for(i=1,i<=n;i++){if($2~/a[i]/)print $0}}' your_file
There are surprisingly many things wrong with your command line:
1) You aren't using the awk condition/action syntax but instead needlessly embedding a condition within an action,
2) You aren't using the default awk action but instead needlessly hand-coding a print $0.
3) You have your RE operands reversed.
4) You are using RE comparison but it looks like you really want to match strings.
You can fix the first 3 of the above by modifying your command to:
awk -F'|' '$1~$2' file.txt
but I think what you really want is "4" which would mean you need to do this instead:
awk -F'|' 'index($1,$2)' file.txt