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
I am trying to write a regex that matches and excludes all strings in a file that contain ${ followed by } with any characters between or around it. In between could be any characters/numbers/underscores/dashes/etc (there won't be another parenthesis inside).
Example matches:
hello ${VAR}
${HELLO_VAR} world
https://${WEB_VAR}
I came up with this: egrep -v '^\${[a-zA-Z?]', though it seems to be working partially and I am not too sure if its right. How can I do this?
The input file has strings separated by a newline, very similar to simple java properties.
You can trying using sed command.
sed 's/\$\{[^}]*\}//g' <input_file> > <output_file>
Sed here excludes all the characters between '{' and '}' and writes the new content in a new output file.
You can give this one a shot:
\$\{[^}]*\}
Match ${ literally, followed by everything except }, followed by }
You say you're trying to exclude all strings in a file, so it sounds like you need something a bit more advanced than just a regex with grep. I'd do this with an awk script:
awk '{while(match($0,/\$\{[^}]*\}/)){$0=substr($0,0,RSTART-1) substr($0,RSTART+RLENGTH)}} 1' input.txt
Or, split for easier reading and commenting:
{
while (match($0,/\$\{[^}]*\}/)) {
$0=substr($0,0,RSTART-1) substr($0,RSTART+RLENGTH)
}
}
1
The idea here is that for each line, we'll check to see whether the regex matches anything on the line. If it does, we'll replace the line with the parts around the matched regex. (We could alternate sub(/RE/,""), but that would require applying the regex twice per match rather than once.)
The final 1 is shorthand that says "print the current line". It runs whether or not the loop processed any matches.
Just use the global wilcard .* around the two sequences, as in:
.*\$\{.*\}.*
As you want to match entire lines, you have to use wilcard at both sides, to extend the regexp to both ends (it doesn't matter if you anchor it with ^ and $ as the greedy algorithm will try to extend as much as possible) Note that the $, { and } must be escaped, as they are reserved by the regexp language.
This can be seen in action here.
note
the title of this question doesn't specify that the substring between the two curly braces should not have a }, and as you want only to match the whole line, then it is not necessary to check for something except a }, the only requirement is that } must be after the ${ in the line. Anyway, this has no drawback in efficiency, as the NFA that parses this regexp has the same number of states as the other.
In a code file, I want to remove any (one or more) consecutive white lines (lines that may include only zero or more spaces/tabs and then a newline) that go between a code text and the concluding } of a block. This concluding } may have spaces for indentation before it, so I want to keep them.
Here is what I try to do:
perl -i -0777 -pe 's/\s+\n([ ]*)\}/\n($1)\}/g' file
For example, if my code file looks like (□ is the space character):
□□□□while (true) {\n
□□□□□□□□print("Yay!");□□□□□□\n
□□□□□□□□□□□□□□□□\n
□□□□}\n
Then I want it to become:
□□□□while (true) {\n
□□□□□□□□print("Yay!");\n
□□□□}\n
However it does not do the change I expected. Any idea what I am doing wrong here?
The only issues I can see with your regex are
you don't need the parenthesis around the matching variable,
and
the use of a character class when extracting the match is
redundant (unless you want to match tabs as well as spaces).
So, you could try
s/\s+\n( *)\}/\n$1\}/g
instead.
This works as expected when run on your test input.
To tidy it up even more, you could try the following.
s/\s+(\n *\})/$1/g
If there might be tabs as well as spaces, you can use a character class. (You do not need to include '|' inside the character class).
s/\s+(\n[ \t]*\})/$1/g
perl -pi -0777 -e's/^\s*\n(?=\s*})//mg' yourfile
(Remove whitespace from the beginning of a line through a newline that precedes a line with } as the first non-whitespace.)
Try using this regex instead, which uses a positive look-ahead assertion. This way you only capture the part that you want to remove, and then replace it with nothing:
s/\s+(?=\n[ ]*\})//g
You can try the following one liner
perl -0777 -pe 's/\s*\n*(\s*\n)/$1/g' test
I am trying to match a paragraph pattern and I am having trouble.
The pattern is:
[image.gif]
some words, usually a few lines
name
emailaddress<mailto:theemailaddress#mail.com>
I tried matching everything between the gif image and the <mailto: but this happens multiple times in the file meaning I get a bad result.
I tried it with this
(?<=\[image.gif\].*?(\[image.gif\])).*?(?=<mailto:)
Is there a way to use Regex to match the general layout of a paragraph?
"the general layout of a paragraph" needs a better definition. Given the lack of an input plus expected output, I'm having to guess what you want here. I'm also guessing that you will accept any language. Here's perl, almost certainly not a language you're familiar with.
Assumed input:
do not match this line
[image.gif]
some words, usually a few lines
Bobert McBobson
emailaddress<mailto:bobertmb#example.com>
don't match this line either
[image.gif]
another few words
on another few lines
Bobina Robertsdaughter
emailaddress<mailto:bobinard#example.info>
this line is also not for matching
Expected output:
[image.gif]
some words, usually a few lines
Bobert McBobson
emailaddress<mailto:bobertmb#example.com>
---
[image.gif]
another few words
on another few lines
Bobina Robertsdaughter
emailaddress<mailto:bobinard#example.info>
Solution using perl:
#!/usr/bin/perl -n007
my $sep = "";
while (/(\[image\.gif\].*?<mailto:[^>]*>(\r)?\n)/gms) {
print $sep . $1;
$sep = "---$2\n";
}
perl is the king of regex languages; many would say that's all it is good for. Here, we use the -n007 option to tell it to read the entire contents of each file and run the code on it as the default variable.
$sep starts blank because there's nothing to separate until the second match.
Then we loop over each block of text that matches the regex:
matches a literal [image.gif]
then matches as little content following that as possible
then matches a literal <mailto: and continues until the next >
then captures the line break (including optional support for DOS line endings)
(see full regex explanation and example at regex101)
We then print the match and finally set the separator to three dashes and a line break (DOS line endings added when needed).
Now you can run it:
$ perl answer.pl input.txt
[image.gif]
some words, usually a few lines
Bobert McBobson
emailaddress<mailto:bobertmb#example.com>
---
[image.gif]
another few words
on another few lines
Bobina Robertsdaughter
emailaddress<mailto:bobinard#example.info>
I have some lines of code I am trying to remove some leading text from which appears like so:
Line 1: myApp.name;
Line 2: myApp.version
Line 3: myApp.defaults, myApp.numbers;
I am trying and trying to find a regex that will remove anything up to (but excluding) myApp.
I have tried various regular expressions, but they all seem to fail when it comes to line 3 (because myApp appears twice).
The closest I have come so far is:
.*?myApp
Pretty simple - but that matches both instances of myApp occurrences in Line 3 - whereas I'd like it to match only the first.
There's a few hundred lines - otherwise I'd have deleted them all manually by now.
Can somebody help me? Thanks.
You need to add an anchor ^ which matches the starting point of a line ,
^.*?(myApp)
DEMO
Use the above regex and replace the matched characters with $1 or \1. So that you could get the string myApp in the final result after replacement.
Pattern explanation:
^ Start of a line.
.*?(myApp) Shortest possible match upto the first myApp. The string myApp was captured and stored into a group.(group 1)
All matched characters are replaced with the chars present inside the group 1.
Your regular expression works in Perl if you add the ^ to ensure that you only match the beginnings of lines:
cat /tmp/test.txt | perl -pe 's/^.*?myApp/myApp/g'
myApp.name;
myApp.version
myApp.defaults, myApp.numbers;
If you wanted to get fancy, you could put the "myApp" into a group that doesn't get captured as part of the expression using (?=) syntax. That way it doesn't have to be replaced back in.
cat /tmp/test.txt | perl -pe 's/^.*?(?=myApp)//g'
myApp.name;
myApp.version
myApp.defaults, myApp.numbers;