please provide me a sed oneliner which provides this output:
sdc3 sdc2
for Input :
sdc3[1] sdc2[0]
I mean remove all subscript value from the string ..
sed 's/\[[^]]*\]//g'
reads: substitute any string with literal "[" followed by zero or more characters that aren't a "]", and then the closing "]", with an empty string.
You need the [^]] bit to prevent greedy matching treating "[1] sdc2[0]" as a single match in your sample string.
As for your comment:
sed 's#\([^[ ]*\)\[[^]]*\]#/dev/\1#g'
I switch the seperator from the usual '/' to '#', just to avoid escaping the /dev/ bit you asked for (I won't say "for clarity")
the \(...\) bit matches a subgroup, here sdc2 or whatever, so we can refer to it in the replacement
the subgroup uses a similar character class to the one we used discarding the index: [^[ ] means any character except an "[" (again, to avoid greedily matching the index) or a space (assuming your values are space-delimited as per your post)
the replacement is now the literal "/dev/" followed by the first (and only) subgroup match
the g flag at the end tells it to perform multiple matches per line, instead of stopping at the first one
Related
I have a file in which I want to replace the "_" string with "-" in cases where it makes up a part of my gene name. Examples of the gene names and my intended output are:
aa1c1_123 -> aa1c1-123
aa1c2_456 -> aa1c1-456
aa1c10_789 -> aa1c1-789
In essence, the first four characters are fixed, followed by 1 or 2 characters depending on the chromosome, an underscore and then the remainder of the gene ID which could vary in length and character. Important is that there are other strings in this gene information column contains other strings with underscores (e.g. "gene_id", "transcript_id", "five_prime_utr") so using sed -i.bak s/_/-/g' file.gtf
can't be done.
Perhaps not the most elegant way, but this should work:
sed -i.bak 's/\([0-9a-z]\{4\}[0-9][0-9]\?\)_/\1-/g' file.gtf
i.e. capture a group (referenced by \1 in the substitution) of 4 characters consisting of lower case letters and digits followed by exactly one digit and perhaps another digit, which is followed by an underscore; if found, replace it by the group's content and a dash. This should exclude your other occurrences consisting of only characters and an underscore.
I'm using the vscode vimplugin. I have a bunch of lines that look like:
Terry,169,80,,,47,,,22,,,6,,
I want to remove all the alphanumeric characters after the first comma so I get:
Terry,,,,,,,,,,,,,
In command mode I tried:
s/^.+\,[a-zA-Z0-9-]\+//g
But this does not appear to do anything. How can I get this working?
edit:
s/^[^,]\+,[a-zA-Z0-9-]\+//g
\+ is greedy; ^.\+, eats the entire line up to the last ,.
Instead of the dot (which means "any character") use [^,] which means "any but a comma". Then ^[^,]\+, means "any characters up to the first comma".
The problem with your requirement is that you want to anchor at the beginning using ^ so you cannot use flag g — with the anchor any substitution will be done once. The only way I can solve the puzzle is to use expressions: match and preserve the anchored text and then use function substitute() with flag g.
I managed with the following expression:
:s/\(^[^,]\+\)\(,\+\)\(.\+\)$/\=submatch(1) . submatch(2) . substitute(submatch(3), '[^,]', '', 'g')/
Let me split it in parts. Searching:
\(^[^,]\+\) — first, match any non-commas
\(,\+\) — any number of commas
\(.\+\)$ — all chars to the end of the string
Substituting:
\= — the substitution is an expression
See http://vimdoc.sourceforge.net/htmldoc/change.html#sub-replace-expression
submatch(1) — replace with the first match (non-commas anchored with ^)
submatch(2) — replace with the second match (commas)
substitute(submatch(3), '[^,]', '', 'g') — replace in the rest of the string
The last call to substitute() is simple, it replaces all non-commas with empty strings.
PS. Tested in real vim, not vscode.
I've got an application that has no useful api implemented, and the only way to get certain information is to parse string output. This is proving to be very painful...
I'm trying to achieve this in bash on SLES12.
Given I have the following strings:
QMNAME(QMTKGW01) STATUS(Running)
QMNAME(QMTKGW01) STATUS(Ended normally)
I want to extract the STATUS value, ie "Ended normally" or "Running".
Note that the line structure can move around, so I can't count on the "STATUS" being the second field.
The closest I have managed to get so far is to extract a single word from inside STATUS like so
echo "QMNAME(QMTKGW01) STATUS(Running)" | sed "s/^.*STATUS(\(\S*\)).*/\1/"
This works for "Running" but not for "Ended normally"
I've tried switching the \S* for [\S\s]* in both "grep -o" and "sed" but it seems to corrupt the entire regex.
This is purely a regex issue, by doing \S you requested to match non-white space characters within (..) but the failing case has a space between which does not comply with the grammar defined. Make it simple by explicitly calling out the characters to match inside (..) as [a-zA-Z ]* i.e. zero or more upper & lower case characters and spaces.
sed 's/^.*STATUS(\([a-zA-Z ]*\)).*/\1/'
Or use character classes [:alnum:] if you want numbers too
sed 's/^.*STATUS(\([[:alnum:] ]*\)).*/\1/'
sed 's/.*STATUS(\([^)]*\)).*/\1/' file
Output:
Running
Ended normally
Extracting a substring matching a given pattern is a job for grep, not sed. We should use sed when we must edit the input string. (A lot of people use sed and even awk just to extract substrings, but that's wasteful in my opinion.)
So, here is a grep solution. We need to make some assumptions (in any solution) about your input - some are easy to relax, others are not. In your example the word STATUS is always capitalized, and it is immediately followed by the opening parenthesis (no space, no colon etc.). These assumptions can be relaxed easily. More importantly, and not easy to work around: there are no nested parentheses. You will want the longest substring of non-closing-parenthesis characters following the opening parenthesis, no mater what they are.
With these assumptions:
$ grep -oP '\bSTATUS\(\K[^)]*(?=\))' << EOF
> QMNAME(QMTKGW01) STATUS(Running)
> QMNAME(QMTKGW01) STATUS(Ended normally)
> EOF
Running
Ended normally
Explanation:
Command options: o to return only the matched substring; P to use Perl extensions (the \K marker and the lookahead). The regexp: we look for a word boundary (\b) - so the word STATUS is a complete word, not part of a longer word like SUBSTATUS; then the word STATUS and opening parenthesis. This is required for a match, but \K instructs that this part of the matched string will not be returned in the output. Then we seek zero or more non-closing-parenthesis characters ([^)]*) and we require that this be followed by a closing parenthesis - but the closing parenthesis is also not included in the returned string. That's a "lookahead" (the (?= ... ) construct).
Here is a piece of code
while($l=~/(\\\s*)$/) {
statements;
}
$l contains a line of text taken form file, in effect this code is for go through lines in file.
Questions:
I don't clearly understand what the condition in while is doing. I think it is trying to match group of \ followed by some number of white spaces at the end of line and loop should stop whenever a line ends with \ and may be some white spaces. I am not sure of it.
I came across statement $a ~= s/^(.*$)/$1/ . What I understand that ^ will force matching at the beginning of string, but in (.*$) would mean match all the characters at the end of string . Dose it mean that the statement is trying to find if any group of character at the end is same as group of character in the beginning of text ?
It is interesting to note that this statement:
while ( $l =~ /(\\\s*)$/ ) {
Is an infinite loop unless $l is altered inside the loop so that the regex no longer matches. As has already been mentioned by others, this is what it matches:
( ... ) a capture group, captures string to $1 (that's the number one, not lower case L)
\\ matches a literal backslash
\s* matches 0 or more whitespace characters.
$ matches end of line with optional newline.
Since you do not have the /g modifier, this regex will not iterate through matches, it will simply check if there is a match, resetting the regex each iteration, thereby causing an endless loop.
The statement
$a ~= s/^(.*$)/$1/
Looks rather pointless. It captures a string of characters up until end of string, then replaces it with itself. The captured text is stored in $1 and is simply replaced. The only marginally useful thing about this regex is that:
It matches up until newline \n, and nothing further, which may be of some use to a parser. A period . matches any character except newline, unless the /s modifier is present on the regex.
It captures the line in $1 for future use. However, a simple /^(.*$)/ would do the same.
1. the while
Usually while (regex) is used with the /g modifier, otherwise, if it matches, you get an infinite loop (unless you exit the loop, like using last).
statements would be executed continuously in an infinite loop.
In your case, adding the g
while($l=~/(\\\s*)$/g)
will have the while make only one loop, due to the $ - making a match unique (whatever matches up to the end of string is unique, as $ marks the end, and there is nothing after...).
2. $a ~= s/^(.*$)/$1/
This is a substitution. If the string ^.*$ matches (and it will, since ^.*$ matches (almost, see comment) anything) it is replaced with... $1 or what's inside the (), ie itself, since the match occurs from 1st char to the end of string
^ means beginning of string
(.*) means all chars
$ end of string
so that will replace $a with itself - probably not what you want.
it matches a literal backslash followed by 0 or more spaces followed by the end of the line.
it executes statements for all the lines in that text file that contain a \, followed by zero or more spaces ( \s* ), at the end of the line ($).
It matches lines that end with a backslash character, ignoring any trailing whitespace characters.
Ending a line with a backslash is used in some languages and data files to indicate that the line is being continued on the next line. So I suspect this is part of a parser that merges these continuation lines.
If you enter a regular expression at RegExr and hover your mouse over the pieces, it displays the meaning of each piece in a tooltip.
(\\\s*)$ this regex means --- a \ followed by zero or more number of white space characters which is followed by end of the line. Since you have your regex in (...), you can extract what you matched using $1, if you need.
http://rubular.com/r/dtHtEPh5DX
EDIT -- based on your update
$a ~= s/^(.$)/$1/ --- this is search and replace. So your regex matches a line which contains exactly one character (since you use . http://www.regular-expressions.info/dot.html), except a new-line character. Since you use (...), the character which matched the regex is extracted and stored in variable a
EDIT -- you changed your regex so here is the updated answer
$a ~= s/^(.*$)/$1/ -- same as above except now it matches zero or more characters (except new-line)
I have a file of string records where one of the fields - delimited by "," - can contain one or more "-" inside it.
The goal is to delete the field value if it contains more than two "-".
i am trying to recoup my past knowledge of sed/awk but can't make much headway
==========
info,whitepaper,Data-Centers,yes-the-6-top-problems-in-your-data-center-lane
info,whitepaper,Data-Centers,the-evolution-center
info,whitepaper,Data-Centers,the-evolution-of-lan-technology-lanner
==========
expected outcome:
info,whitepaper,Data-Centers
info,whitepaper,Data-Centers,the-evolution-center
info,whitepaper,Data-Centers
thanks
Try
sed -r 's/(^|,)([^,-]+-){3,}[^,]+(,|$)/\3/g'
or if you're into slashes
sed 's/\(^\|,\)\([^,-]\+-\)\{3,\}[^,]\+\(,\|$\)/\3/g'
Explanation:
I'm using the most basic sed command: substitution. The syntax is: s/pattern/replacement/flags.
Here pattern is (^|,)([^,-]+-){3,}[^,]+(,|$), replacement is \3, flags is g.
The g flag means global replacement (all matching parts are replaced, not only the first in line).
In pattern:
brackets () create a group. Somewhat like in math. They also allow to refer to a group with a number later.
^ and $ mean beginning and end of the string.
| means "or", so (^|,) means "comma or beginning of the string".
square brackets [] mean a character class, ^ inside means negation. So [^,-] means "anything but comma or hyphen". Not that usually the hyphen has a special meaning in character classes: [a-z] means all lowercase letters. But here it's just a hyphen because it's not in the middle.
+ after an expression means "match it 1 or more times" (like * means match it 0 or more times).
{N} means "match it exactly N times. {N,M} is "from N to M times". {3,} means "three times or more". + is equivalent to {1,}.
So this is it. The replacement is just \3. This refers to the third group in (), in this case (,|$). This will be the only thing left after the substitution.
P.S. the -r option just changes what characters need to be escaped: without it all of ()-{}| are treated as regular chars unless you escape them with \. Conversely, to match literal ( with -r option you'll need to escape it.
P.P.S. Here's a reference for sed. man sed is your friend as well.
Let me know if you have further questions.
You could try perl instead of sed or awk:
perl -F, -lane 'print join ",", grep { !/-.*-.*-/ } #F' < file.txt
This might work for you:
sed 's/,\{,1\}[^,-]*\(-[^,]*\)\{3,\}//g file
sed 's/\(^\|,\)\([^,]*-\)\{3\}[^,]*\(,\|$\)//g'
This should work in more cases:
sed 's/,$/\n/g;s/\(^\|,\|\n\)\([^,\n]*-\)\{3\}[^,\n]*\(,\|\n\|$\)/\3/g;s/,$//;s/\n/,/g'