How to replace spaces after a certain pattern with commas? - regex

I am new to coding and I'm trying to format some bioinformatics data. I am trying to remove all the spaces after GT:GL:GOF:GQ:NR:NV with commas, but not anything outside of the format xx:xx:xx:xx:xx (like the example). I know I need to use sed with regex option but I'm not very familiar with how to use it. I've never actually used sed before and got confused trying so any help would be appreciated. Sorry if I formatted this poorly (this is my first post).
EDIT 2: I got actual data from the file this time which may help solve the problem. Removed the bad example.
New Example: I pulled this data from my actual file (this is just two samples), and it is surrounded by other data. Essentially the line has a bunch of data followed by "GT:GL:GOF:GQ:NR:NV ", after this there is more data in the format shown below, and finally there is some more random data. Unfortunately I can't post a full line of the data because it is extremely long and will not fit.
Input
0/1:-1,-1,-1:146:28:14,14:4,0 0/1:-1,-1,-1:134:6:2,2:1,0
Output
0/1:-1,-1,-1:146:28:14,14:4,0,0/1:-1,-1,-1:134:6:2,2:1,0

With Basic Regular Expressions, you can use character classes and backreferences to accomplish your task, e.g.
$ sed 's/\([0-9][0-9]*:[0-9][0-9]*\)[ ]\([0-9][0-9]*:[0-9][0-9]*\)/\1,\2/g' file
1/0 ./. 0/1 GT:GL:GOF:GQ:NR:NV 1:12:314,213:132:13:31,14:31:31 AB GT BB
1/0 ./. 0/1 GT:GL:GOF:GQ:NR:NV 10:13:12,41:41:1:13,13:131:1:1 AB GT RT
1/0 ./. 0/1 GT:GL:GOF:GQ:NR:NV 1:12:314,213:132:13:31,14:31:31 AB GT
Which basically says:
find and capture any [0-9][0-9]* one or more digits,
separated by a :, and
followed by [0-9][0-9]* one or more digits -- as capture group 1,
match a space following capture group 1 followed by capture group 2 (which is the same as capture group 1),
then replace the space separating the capture groups with a comma reinserting the capture group text using backreference 1 and 2 (e.g. \1 and \2), finally
make the replacement global (e.g. g) to replace all matching occurrences.
Edit Based On New Input Posted
If you still need all of the original commas added, and you now want to add a comma between ,0 0/ (where there is a comma precedes a single-digit followed by the space to be replaced with a comma, followed by a single-digit and a forward-slash), then all you need to do is make your capture groups conditional (on either capturing the original data as above -or- capturing this new segment. You do that by including an OR (e.g. \| in basic regex terms) between the conditions.
For instance by adding \|,[0-9] at the end of the first capture group and \|[0-9][/] at the end of the second, e.g.
$ sed 's/\([0-9][0-9]*:[0-9][0-9]*\|,[0-9]\)[ ]\([0-9][0-9]*:[0-9][0-9]*\|[0-9][/]\)/\1,\2/g' file
0/1:-1,-1,-1:146:28:14,14:4,0,0/1:-1,-1,-1:134:6:2,2:1,0
If you have other caveats in your file, I suggest you post several complete lines of input, and if they are too long, then create a zip, gzip, bzip or xz file and post it to a site like pastebin and add the link to your question.
If all you really care about now is the space in ,0 0/, then you can shorten the sed command to:
$ sed 's/\(,[0-9]\)[[:space:]]\([0-9][/]\)/\1,\2/g' file
0/1:-1,-1,-1:146:28:14,14:4,0,0/1:-1,-1,-1:134:6:2,2:1,0
(note: I've included [[:space:]] to handle any whitespace (space, tab, ...) instead of just the literal [ ] (space) in the new example)
Let me know if this fixes the issue.

I'm assuming that the xx:xx:xx or xx:xx:xx:xx can have any number of parts, since some have 3, and some have 4.
This is quite difficult to do reliably with sed, as it does not support lookarounds, which seem like they might be needed for this example.
You can try something like:
perl -pe 's/(?<=\d) (?=\d+(:\d+){2,})/,/g' input.txt
If you've got your heart set on sed, you can try this, but it may miss some cases:
sed -r 's/(:[0-9]+) ([0-9]+:)/\1,\2/g' input.txt

Could you please try following. This will take care of printing those values also which are NOT coming in match of regex. Also we would have made regex mentioned in match a bit shorter by doing it as [0-9]+\.{4} etc since this is tested on old awk so couldn't test it.
awk '
BEGIN{
OFS=","
}
match($0,/GT:GL:GOF:GQ:NR:NV [0-9]+:[0-9]+:[0-9]+:[0-9]+:[0-9]+/){
value=substr($0,RSTART!=1?1:RSTART,RSTART+RLENGTH-1)
value1=substr($0,RSTART+RLENGTH+1)
gsub(/[[:space:]]+/,",",value1)
print value,value1
next
}
1
' Input_file

You may also achieve your desired result without regex, using awk:
awk '{printf "%s", $1FS$2FS$3FS$4FS$5","$6","$7; for (i=8;i<=NF;i++) printf "%s", FS$i; print ""}' input.txt
Basically, it outputs from field 1 to 5 with the default field separator ("space"), then from field 5 to 7 with the comma separator, then from field 8 onwards with default separator again.

perl myscript.pl '0/1:-1,-1,-1:146:28:14,14:4,0 0/1:-1,-1,-1:134:6:2,2:1,0'
myscript.pl,
#!/usr/local/ActivePerl-5.20/bin/env perl
my $input = $ARGV[0];
$input =~ s/ /\,/g;
print $input, "\n";
__DATA__
output
0/1:-1,-1,-1:146:28:14,14:4,0,0/1:-1,-1,-1:134:6:2,2:1,0
This will remove all spaces, not just the space in question

Related

Substitute any other character except for a specific pattern in Perl

I have text files with lines like this:
U_town/u_LN0_pk_LN3_bnb_LN155/DD0 U_DESIGN/u_LNxx_pk_LN99_bnb_LN151_LN11_/DD5
U_master/u_LN999_pk_LN767888_bnb_LN9772/Dnn111 u_LN999_pk_LN767888_bnb_LN9772_LN9999_LN11/DD
...
I am trying to substitute any other character except for / to nothing and keep a word with pattern _LN\d+_ with Perl one-liner.
So the edited version would look like:
/_LN0__LN3__LN155/ /_LN99__LN151_LN11_/
/_LN999__LN767888_/ _LN999__LN767888__LN9772_LN9999_/
I tried below which returned empty lines
perl -pe 's/(?! _LN\d+_)[^\/].+//g' file
Below returned only '/'.
perl -pe 's/(?! _LN\d+_)\w+//g' file
Is it perhaps not possible with a one-liner and I should consider writing a code to parse character by character and see if a matching word _LN\d+_ or a character / is there?
To merely remove everything other than these patterns can simply match the patterns and join the matches back
perl -wnE'say join "", m{/ | _LN[0-9]+_ }gx' file
or perhaps, depending on details of the requirements
perl -wnE'say join "", m{/ | _LN[0-9]+(?=_) }gx' file
(See explanation in the last bullet below.)
Prints, for the first line (of the two) of the shown sample input
/_LN0__LN3_//_LN99__LN151_
...
or, in the second version
/_LN0_LN3//_LN99_LN151_LN11/
...
The _LN155 is not there because it is not followed by _. See below.
Questions:
Why are there spaces after some / in the "edited version" shown in the question?
The pattern to keep is shown as _LN\d+_ but _LN155 is shown to be kept even though it is not followed by a _ in the input (but by a /) ...?
Are underscores optional by any chance? If so, append ? to them in the pattern
perl -wnE'say join "", m{/ | _?LN[0-9]+_? }gx' file
with output
/_LN0__LN3__LN155//_LN99__LN151_LN11_/
(It's been clarified that the extra space in the shown desired output is a mistake.)
If the underscores "overlap," like in _LN155_LN11_, in the regex they won't be both matched by the _LN\d+_ pattern, since the first one "takes" the underscore.
But if such overlapping instances nned be kept then replace the trailing _ with a lookahead for it, which doesn't consume it so it's there for the leading _ on the next pattern
perl -wnE'say join "", m{/ | _LN[0-9]+(?=_) }gx' file
(if the underscores are optional and you use _?LN\d+_? pattern then this isn't needed)

Remove anything before primary domain or after forward slash

How can I extract domain names from the text input below? I tried this but it didn't work as expected:
grep -oP '(?<=[.])\w+(?=[.])'
Is there anyway to do this in sed/awk or any other Linux command?
Input:
netgear.com
myapi.arlo.com
https://updates.netgear.com/arlo
https://bugcrowd-pub.bounty.accellion.net
client-api.arkoselabs.com
Output desired:
netgear.com
arlo.com
netgear.com
accellion.net
arkoselabs.com
I found so many solution thanks Google, Tried to craft my own regex ,
^((http[s]?|ftp):\/)?\/?([^:\/\s]+)((\/\w+)*\/)([\w\-\.]+[^#?\s]+)(.*)?(#[\w\-]+)?$
[a-zA-Z0-9-]+\.[a-zA-Z]+($|(?=\/))
awk -F"." '{print $(NF-1)"."$NF}'
It looks like you are not only trying to remove the /, you are actually trying to extract the main domain from those URLs.
If you put the input in a file called input.txt, the following works for me on Ubuntu 20.10:
cat input.txt | sed -e 's;..([a-zA-Z0-9-].[a-zA-Z0-9-]).$;\1;'
As a brief explanation:
The domain name "parts" (the words between the dots) can only use numbers, letters and the dash symbol as characters. That pattern can be represented as:
[a-zA-Z0-9-]*
The regex above will match 2 of those, separated by a dot, proceeded by a dot (and possibly a number of characters), and succeeded by either the end of line or a group of characters that are not part of the previous groups. I believe the greedy nature of .* will make sure that only the main domain is captured.
There is probably more robust solutions available too.

Regex expression to find file extension in a file with multiple periods

How would you write a regular expression to find the file extension of the following files, keeping in mind that what I am looking for is the ".pdf" or ".xls" portion of the string?
REPORTPDF.20130810.pdf.pgp
REPORTXLS.20130810.xls.pgp
EDIT:
The resulting filenames I want to end up with are the following:
REPORT20130810.PDF
REPORT20130810.XLS
I am on a Windows platform. I've played around with this a bit at http://regexpal.com/ but so far I can only figure out how to match the date:
([0-9]{4}[0-9]{2}[0-9]{2})
Using sed:
sed 's/^\(.*[^.]*\)\.[^.]*$/\1/' <<< "REPORTPDF.20130810.pdf.pgp"
REPORTPDF.20130810.pdf
Using grep -P (PCRE regex):
grep -oP '^.+[^.]+(?=\.[^.]+$)' <<< "REPORTPDF.20130810.pdf.pgp"
REPORTPDF.20130810.pdf
.+\.(\w+)\.\w+$ would deliver the last but one extension as group 1, how this is accessed would then be dependent of your host language for the regex.
If you don't need the file extension to be capitalized, this should work
([a-zA-Z]+)\.([0-9]{4}[0-9]{2}[0-9]{2})\.(xls|pdf)\.pgp
Matches:
REPORTXLS.20130810.xls.pgp
And then the groups you'd use are two and three
REPORT\2.\3
Matches:
REPORT20130810.xls
Problem is that you don't provide much context for how you're going about changing these file names.
You don't say what language/library you're using, but this Perl one-liner does the trick:
perl -lpe "s/^([^.]*)(...)\.(\d+)(\.\2)\.pgp/\1\3\4/i; $_=uc"
I think this will work for you :)
^(([A-Z a-z]*)(?:XLS.|PDF.)(\d{8})(.pdf|.xls))
Edit live on Debuggex
^ starts at the beginning of the string
(.*) any character before
\d any number 0-9
{8} only 8 times for that character section (in this case 8 times of
the numbers 0-9)
?: is non capture groups
I wrapped the capture groups into one large one so the thing that you want will be in the first capture group :).
This can be prob be replaced
([A-Z a-z]*)
with
(REPORT)
This (.*?(?:\..*)?)(\..*) will hold things like:
'hello.1a.2bb.3' ---> group(1) == 'hello.1a.2bb', group(2) == '.3'
'yep.1' ---> group(1) == 'yep', group(2) == '.1'
If the format is pretty much fixed you could use
(REPORT)([^.]++)[.]([^.]++)[.]([^.]++)[.](pgp)
and cherry pick replacement based on what you want
Used java here but regex match would still be same
String a = "REPORTPDF.20130810.pdf.pgp".replaceAll(
"(REPORT)([^.]++)[.]([^.]++)[.]([^.]++)[.](pgp)",
"$1--$2--$3--$4--$5");
;
String b = "REPORTXLS.20130810.xls.pgp".replaceAll(
"(REPORT)([^.]++)[.]([^.]++)[.]([^.]++)[.](pgp)",
"$1--$2--$3--$4--$5");
System.out.println(a);
System.out.println(b);
REPORT--PDF--20130810--pdf--pgp
REPORT--XLS--20130810--xls--pgp
in your case "$1$3.$2"
String b = "REPORTXLS.20130810.xls.pgp".replaceAll(
"(REPORT)([^.]++)[.]([^.]++)[.]([^.]++)[.](pgp)",
"$1$3.$2");
which produces intended result
REPORT20130810.XLS

Why doesn't this simple regex match what I think it should?

I have a data file that looks like the following example. I've added '%' in lieu of \t, the tab control character.
1234:56% Alice Worthington
alicew% Jan 1, 2010 10:20:30 AM% Closed% Development
Digg:
Reddit:
Update%% file-one.txt% 1.1% c:/foo/bar/quux
Add%% file-two.txt% 2.5.2% c:/foo/bar/quux
Remove%% file-three.txt% 3.4% c:/bar/quux
Update%% file-four.txt% 4.6.5.3% c:/zzz
... many more records of the above form
The records I'm interested in are the lines beginning with "Update", "Add", "Remove", and so on. I won't know what the lines begin with ahead of time, or how many lines precede them. I do know that they always begin with a string of letters followed by two tabs. So I wrote this regex:
generate-report-for 1234:56 | egrep "^[[:alpha:]]+\t\t.+"
But this matches zero lines. Where did I go wrong?
Edit: I get the same results whether I use '...' or "..." for the egrep expression, so I'm not sure it's a shell thing.
Apparently \t isn't a special character for egrep. You can either use grep -P to enable Perl-compatible regex engine, or insert literal tabs with CtrlvCtrli
Even better, you could use the excellent ack
It looks like the shell is parsing "\t\t" before it is sent to egrep. Try "\\t\\t" or '\t\t' instead. That is 2 slashes in double quotes and one in single quotes.
The file might not be exactly what you see. Maybe there are control characters hidden. It happens, sometimes. My suggestion is that you debug this. First, reduce to the minimum regex pattern that matches, and then keep adding stuff one by one, until you find the problem:
egrep "[[:alpha:]]"
egrep "[[:alpha:]]+"
egrep "[[:alpha:]]+\t"
egrep "[[:alpha:]]+\t\t"
egrep "[[:alpha:]]+\t\t.+"
egrep "^[[:alpha:]]+\t\t.+"
There are variations on that sequence, depending on what you find out at each step. Also, the first step can really be skipped, but this is just for the sake of showing the technique.
you can use awk
awk '/^[[:alpha:]]\t\t/' file

matching text in quotes (newbie)

I'm getting totally lost in shell programming, mainly because every site I use offers different tool to do pattern matching. So my question is what tool to use to do simple pattern matching in piped stream.
context: I have named.conf file, and i need all zones names in a simple file for further processing. So I do ~$ cat named.local | grep zone and get totally lost here. My output is ~hundred or so newlines in form 'zone "domain.tld" {' and I need text in double quotes.
Thanks for showing a way to do this.
J
I think what you're looking for is sed... it's a stream editor which will let you do replacements on a line-by-line basis.
As you're explaining it, the command `cat named.local | grep zone' gives you an output a little like this:
zone "domain1.tld" {
zone "domain2.tld" {
zone "domain3.tld" {
zone "domain4.tld" {
I'm guessing you want the output to be something like this, since you said you need the text in double quotes:
"domain1.tld"
"domain2.tld"
"domain3.tld"
"domain4.tld"
So, in reality, from each line we just want the text between the double-quotes (including the double-quotes themselves.)
I'm not sure you're familiar with Regular Expressions, but they are an invaluable tool for any person writing shell scripts. For example, the regular expression /.o.e/ would match any line where there's a word with the 2nd letter was a lower-case o, and the 4th was e. This would match string containing words like "zone", "tone", or even "I am tone-deaf."
The trick there was to use the . (dot) character to mean "any letter". There's a couple of other special characters, such as * which means "repeat the previous character 0 or more times". Thus a regular expression like a* would match "a", "aaaaaaa", or an empty string: ""
So you can match the string inside the quotes using: /".*"/
There's another thing you would know about sed (and by the comments, you already do!) - it allows backtracking. Once you've told it how to recognize a word, you can have it use that word as part of the replacement. For example, let's say that you wanted to turn this list:
Billy "The Kid" Smith
Jimmy "The Fish" Stuart
Chuck "The Man" Norris
Into this list:
The Kid
The Fish
The Man
First, you'd look for the string inside the quotes. We already saw that, it was /".*"/.
Next, we want to use what's inside the quotes. We can group it using parens: /"(.*)"/
If we wanted to replace the text with the quotes with an underscore, we'd do a replace: s/"(.*)"/_/, and that would leave us with:
Billy _ Smith
Jimmy _ Stuart
Chuck _ Norris
But we have backtracking! That'll let us recall what was inside the parens, using the symbol \1. So if we do now: s/"(.*)"/\1/ we'll get:
Billy The Kid Smith
Jimmy The Fish Stuart
Chuck The Man Norris
Because the quotes weren't in the parens, they weren't part of the contents of \1!
To only leave the stuff inside the double-quotes, we need to match the entire line. To do that we have ^ (which means "beginning of line"), and $ (which means "end of line".)
So now if we use s/^.*"(.*)".*$/\1/, we'll get:
The Kid
The Fish
The Man
Why? Let's read the regular expression s/^.*"(.*)".*$/\1/ from left-to-right:
s/ - Start a substitution regular expression
^ - Look for the beginning of the line. Start from there.
.* - Keep going, reading every character, until...
" - ... until you reach a double-quote.
( - start a group a characters we might want to recall later when backtracking.
.* - Keep going, reading every character, until...
) - (pssst! close the group!)
" - ... until you reach a double-quote.
.* - Keep going, reading every character, until...
$ - The end of the line!
/ - use what's after this to replace what you matched
\1 - paste the contents of the first group (what was in the parens) matched.
/ - end of regular expression
In plain English: "Read the entire line, copying aside the text between the double-quotes. Then replace the entire line with the content between the double qoutes."
You can even add double-quote around the replacing text s/^.*"(.*)".*$/"\1"/, so we'll get:
"The Kid"
"The Fish"
"The Man"
And that can be used by sed to replace the line with the content from within the quotes:
sed -e "s/^.*\"\(.*\)\".*$/\"\1\"/"
(This is just shell-escaped to deal with the double-quotes and slashes and stuff.)
So the whole command would be something like:
cat named.local | grep zone | sed -e "s/^.*\"\(.*\)\".*$/\"\1\"/"
Well, nobody mentioned cut yet, so, to prove that there are many ways to do something with the shell:
% grep '^zone' /etc/bind/named.conf | cut -d' ' -f2
"gennic.net"
"generic-nic.net"
"dyn.generic-nic.net"
"langtag.net"
1.
zoul#naima:etc$ cat named.conf | grep zone
zone "." IN {
zone "localhost" IN {
file "localhost.zone";
zone "0.0.127.in-addr.arpa" IN {
2.
zoul#naima:etc$ cat named.conf | grep ^zone
zone "." IN {
zone "localhost" IN {
zone "0.0.127.in-addr.arpa" IN {
3.
zoul#naima:etc$ cat named.conf | grep ^zone | sed 's/.*"\([^"]*\)".*/\1/'
.
localhost
0.0.127.in-addr.arpa
The regexp is .*"\([^"]*\)".*, which matches:
any number of any characters: .*
a quote: "
starts to remember for later: \(
any characters except quote: [^"]*
ends group to remember: \)
closing quote: "
and any number of characters: .*
When calling sed, the syntax is 's/what_to_match/what_to_replace_it_with/'. The single quotes are there to keep your regexp from being expanded by bash. When you “remember” something in the regexp using parens, you can recall it as \1, \2 etc. Fiddle with it for a while.
You should have a look at awk.
As long as someone is pointing out sed/awk, I'm going to point out that grep is redundant.
sed -ne '/^zone/{s/.*"\([^"]*\)".*/\1/;p}' /etc/bind/named.conf
This gives you what you're looking for without the quotes (move the quotes inside the parenthesis to keep them). In awk, it's even simpler with the quotes:
awk '/^zone/{print $2}' /etc/bind/named.conf
I try to avoid pipelines as much as possible (but not more). Remember, Don't pipe cat. It's not needed. And, insomuch as awk and sed duplicating grep's work, don't pipe grep, either. At least, not into sed or awk.
Personally, I'd probably have used perl. But that's because I probably would have done the rest of whatever you're doing in perl, making it a minor detail (and being able to slurp the whole file in and regex against everything simultaneously, ignoring \n's would be a bonus for cases where I don't control /etc/bind, such as on a shared webhost). But, if I were to do it in shell, one of the above two would be the way I'd approach it.