I have text file sample.txt like following
ID=Sam-S-PA.path1;Name=Sam-S-PA 23 Hz42
ID=GlcAT-S-PA.path1;Name=GlcAT-S-PA 45 iu7s
ID=TfIIA-S-PA.path1;Name=TfIIA-S-PA 76 5ghz
ID=S-PA.path1;Name=S-PA 69 ivcs
ID=TyrRS-PA.path1;Name=TyrRS-PA 51 Pqas
ID=HisRS-PA.path1;Name=HisRS-PA 32 Majs
I would like to extract row containing only S-PA using grep. I tried following command:
grep -w "S-PA" sample.txt
But it gave a output that included all the entries which I dont want. I want the following output
ID=S-PA.path1;Name=S-PA 69 ivcs
Kindly guide me. Thanks in advance.
Using negative look-ahead and look-behind.
$ grep -P '(?<![\w-])S-PA(?![\w-])' sample.txt
ID=S-PA.path1;Name=S-PA 69 ivcs
Effectively you include - into the "word" for word boundary considerations.
(?<![\w-]) ensures that S-PA is not preceded with a word character or -.
Similarly (?![\w-]) ensures the same for the following characters.
Using regex.
grep -oE "S-PA (.+)" sample.txt
or
egrep -o "S-PA (.+)" sample.txt
It seems you want to match =S-PA followed with a space. Use
grep '=S-PA ' sample.txt
or
grep '=S-PA[[:blank:]]' sample.txt
where [[:blank:]] matches either a regular space or a tab char.
See this regex demo showing how this regex works.
Related
I would like grep to print out all complete words that include the match.
Google did not help me. Here what I tried:
cat file.txt
21676 Mm.24685 NM_009346 ENSMUSG00000055320
20349 Mm.134093 NM_011348 ENSMUSG00000063531
12456 Mm.134000 NM_011228 GM415666
grep -o "ENSMUS" file.txt
ENSMUS
ENSMUS
Desired output:
ENSMUSG00000055320
ENSMUSG00000063531
Thanks for your help!
You may use:
grep -wo "ENSMUS[^[:blank:]]*" file.txt
ENSMUSG00000055320
ENSMUSG00000063531
Here [^[:blank:]]* will match 0 or more characters that are not whitespaces. -w will ensure full word matches.
To extract ENSEMBL mouse accession numbers without the version number:
grep -Po 'ENSMUS\w+' in_file
With the version number:
grep -Po 'ENSMUS\S+' in_file
Here,
\w+ : 1 or more word characters ([A-Za-z0-9_]).
\S+ : 1 or more non-whitespace characters (you can also be more restrictive and use [\w.]+, which is 1 or more word character or literal dot).
Here, GNU grep uses the following options:
-P : Use Perl regexes.
-o : Print the matches only (1 match per line), not the entire lines.
SEE ALSO:
grep manual
perlre - Perl regular expressions
File1:
Btr_0449a 447
Btr_0449 447
Desired output:
Btr_0449 447
I want grep to find 'Btr_0449', not 'Btr_0449a'. Seems I'm doing something wrong since:
grep -F "Btr_0449"
Btr_0449a 447
Btr_0449 447
This should do it:
grep -Fw "Btr_0449"
From the grep manpage:
"-w Select only those lines containing matches that form whole words. "
If you insist to use '-F' flag, then adding a space after your string will do.
grep -F "Btr_0449 "
For the future, you will get much better results if you'll use regex patterns, so for the above query, you could do:
grep -e "Btr_0449\s"
...which will match your string followed by any whitespace character (space, tab, new line, carriage return...)
I have a very large file 100Mb+ where all the content is on one line.
I wish to find a pattern in that file and a number of characters around that pattern.
For example I would like to call a command like the one below but where -A and -B are number of bytes not lines:
cat very_large_file | grep -A 100 -B 100 somepattern
So for a file containing content like this:
1234567890abcdefghijklmnopqrstuvwxyz
With a pattern of
890abc
and a before size of -B 3
and an after size of -A 3
I want it to return:
567890abcdef
Any tips would be great.
Many thanks.
You could try the -o option:
-o, --only-matching
Show only the part of a matching line that matches PATTERN.
and use a regular expression to match your pattern and the 3 preceding/following characters i.e.
grep -o -P ".{3}pattern.{3}" very_large_file
In the example you gave, it would be
echo "1234567890abcdefghijklmnopqrstuvwxyz" > tmp.txt
grep -o -P ".{3}890abc.{3}" tmp.txt
Another one with sed (you may need it on systems where GNU grep is not available):
sed -n '
s/.*\(...890abc...\).*/\1/p
' infile
Best way I can think of doing this is with a tiny Perl script.
#!/usr/bin/perl
$pattern = $ARGV[0];
$before = $ARGV[1];
$after = $ARGV[2];
while(<>) {
print $& if( /.{$before}$pattern.{$after}/ );
}
You would then execute it thusly:
cat very_large_file | ./myPerlScript.pl 890abc 3 3
EDIT: Dang, Paolo's solution is much easier. Oh well, viva la Perl!
I want to run ack or grep on HTML files that often have very long lines. I don't want to see very long lines that wrap repeatedly. But I do want to see just that portion of a long line that surrounds a string that matches the regular expression. How can I get this using any combination of Unix tools?
You could use the grep options -oE, possibly in combination with changing your pattern to ".{0,10}<original pattern>.{0,10}" in order to see some context around it:
-o, --only-matching
Show only the part of a matching line that matches PATTERN.
-E, --extended-regexp
Interpret pattern as an extended regular expression (i.e., force grep to behave as egrep).
For example (from #Renaud's comment):
grep -oE ".{0,10}mysearchstring.{0,10}" myfile.txt
Alternatively, you could try -c:
-c, --count
Suppress normal output; instead print a count of matching lines
for each input file. With the -v, --invert-match option (see
below), count non-matching lines.
Pipe your results thru cut. I'm also considering adding a --cut switch so you could say --cut=80 and only get 80 columns.
You could use less as a pager for ack and chop long lines: ack --pager="less -S" This retains the long line but leaves it on one line instead of wrapping. To see more of the line, scroll left/right in less with the arrow keys.
I have the following alias setup for ack to do this:
alias ick='ack -i --pager="less -R -S"'
grep -oE ".\{0,10\}error.\{0,10\}" mylogfile.txt
In the unusual situation where you cannot use -E, use lowercase -e instead.
Explanation:
cut -c 1-100
gets characters from 1 to 100.
The Silver Searcher (ag) supports its natively via the --width NUM option. It will replace the rest of longer lines by [...].
Example (truncate after 120 characters):
$ ag --width 120 '#patternfly'
...
1:{"version":3,"file":"react-icons.js","sources":["../../node_modules/#patternfly/ [...]
In ack3, a similar feature is planned but currently not implemented.
Taken from: http://www.topbug.net/blog/2016/08/18/truncate-long-matching-lines-of-grep-a-solution-that-preserves-color/
The suggested approach ".{0,10}<original pattern>.{0,10}" is perfectly good except for that the highlighting color is often messed up. I've created a script with a similar output but the color is also preserved:
#!/bin/bash
# Usage:
# grepl PATTERN [FILE]
# how many characters around the searching keyword should be shown?
context_length=10
# What is the length of the control character for the color before and after the
# matching string?
# This is mostly determined by the environmental variable GREP_COLORS.
control_length_before=$(($(echo a | grep --color=always a | cut -d a -f '1' | wc -c)-1))
control_length_after=$(($(echo a | grep --color=always a | cut -d a -f '2' | wc -c)-1))
grep -E --color=always "$1" $2 |
grep --color=none -oE \
".{0,$(($control_length_before + $context_length))}$1.{0,$(($control_length_after + $context_length))}"
Assuming the script is saved as grepl, then grepl pattern file_with_long_lines should display the matching lines but with only 10 characters around the matching string.
I put the following into my .bashrc:
grepl() {
$(which grep) --color=always $# | less -RS
}
You can then use grepl on the command line with any arguments that are available for grep. Use the arrow keys to see the tail of longer lines. Use q to quit.
Explanation:
grepl() {: Define a new function that will be available in every (new) bash console.
$(which grep): Get the full path of grep. (Ubuntu defines an alias for grep that is equivalent to grep --color=auto. We don't want that alias but the original grep.)
--color=always: Colorize the output. (--color=auto from the alias won't work since grep detects that the output is put into a pipe and won't color it then.)
$#: Put all arguments given to the grepl function here.
less: Display the lines using less
-R: Show colors
S: Don't break long lines
Here's what I do:
function grep () {
tput rmam;
command grep "$#";
tput smam;
}
In my .bash_profile, I override grep so that it automatically runs tput rmam before and tput smam after, which disabled wrapping and then re-enables it.
ag can also take the regex trick, if you prefer it:
ag --column -o ".{0,20}error.{0,20}"
Is there a way to make grep output "words" from files that match the search expression?
If I want to find all the instances of, say, "th" in a number of files, I can do:
grep "th" *
but the output will be something like (bold is by me);
some-text-file : the cat sat on the mat
some-other-text-file : the quick brown fox
yet-another-text-file : i hope this explains it thoroughly
What I want it to output, using the same search, is:
the
the
the
this
thoroughly
Is this possible using grep? Or using another combination of tools?
Try grep -o:
grep -oh "\w*th\w*" *
Edit: matching from Phil's comment.
From the docs:
-h, --no-filename
Suppress the prefixing of file names on output. This is the default
when there is only one file (or only standard input) to search.
-o, --only-matching
Print only the matched (non-empty) parts of a matching line,
with each such part on a separate output line.
Cross distribution safe answer (including windows minGW?)
grep -h "[[:alpha:]]*th[[:alpha:]]*" 'filename' | tr ' ' '\n' | grep -h "[[:alpha:]]*th[[:alpha:]]*"
If you're using older versions of grep (like 2.4.2) which do not include the -o option, then use the above. Else use the simpler to maintain version below.
Linux cross distribution safe answer
grep -oh "[[:alpha:]]*th[[:alpha:]]*" 'filename'
To summarize: -oh outputs the regular expression matches to the file content (and not its filename), just like how you would expect a regular expression to work in vim/etc... What word or regular expression you would be searching for then, is up to you! As long as you remain with POSIX and not perl syntax (refer below)
More from the manual for grep
-o Print each match, but only the match, not the entire line.
-h Never print filename headers (i.e. filenames) with output lines.
-w The expression is searched for as a word (as if surrounded by
`[[:<:]]' and `[[:>:]]';
The reason why the original answer does not work for everyone
The usage of \w varies from platform to platform, as it's an extended "perl" syntax. As such, those grep installations that are limited to work with POSIX character classes use [[:alpha:]] and not its perl equivalent of \w. See the Wikipedia page on regular expression for more
Ultimately, the POSIX answer above will be a lot more reliable regardless of platform (being the original) for grep
As for support of grep without -o option, the first grep outputs the relevant lines, the tr splits the spaces to new lines, the final grep filters only for the respective lines.
(PS: I know most platforms by now would have been patched for \w.... but there are always those that lag behind)
Credit for the "-o" workaround from #AdamRosenfield answer
It's more simple than you think. Try this:
egrep -wo 'th.[a-z]*' filename.txt #### (Case Sensitive)
egrep -iwo 'th.[a-z]*' filename.txt ### (Case Insensitive)
Where,
egrep: Grep will work with extended regular expression.
w : Matches only word/words instead of substring.
o : Display only matched pattern instead of whole line.
i : If u want to ignore case sensitivity.
You could translate spaces to newlines and then grep, e.g.:
cat * | tr ' ' '\n' | grep th
Just awk, no need combination of tools.
# awk '{for(i=1;i<=NF;i++){if($i~/^th/){print $i}}}' file
the
the
the
this
thoroughly
grep command for only matching and perl
grep -o -P 'th.*? ' filename
I was unsatisfied with awk's hard to remember syntax but I liked the idea of using one utility to do this.
It seems like ack (or ack-grep if you use Ubuntu) can do this easily:
# ack-grep -ho "\bth.*?\b" *
the
the
the
this
thoroughly
If you omit the -h flag you get:
# ack-grep -o "\bth.*?\b" *
some-other-text-file
1:the
some-text-file
1:the
the
yet-another-text-file
1:this
thoroughly
As a bonus, you can use the --output flag to do this for more complex searches with just about the easiest syntax I've found:
# echo "bug: 1, id: 5, time: 12/27/2010" > test-file
# ack-grep -ho "bug: (\d*), id: (\d*), time: (.*)" --output '$1, $2, $3' test-file
1, 5, 12/27/2010
cat *-text-file | grep -Eio "th[a-z]+"
You can also try pcregrep. There is also a -w option in grep, but in some cases it doesn't work as expected.
From Wikipedia:
cat fruitlist.txt
apple
apples
pineapple
apple-
apple-fruit
fruit-apple
grep -w apple fruitlist.txt
apple
apple-
apple-fruit
fruit-apple
I had a similar problem, looking for grep/pattern regex and the "matched pattern found" as output.
At the end I used egrep (same regex on grep -e or -G didn't give me the same result of egrep) with the option -o
so, I think that could be something similar to (I'm NOT a regex Master) :
egrep -o "the*|this{1}|thoroughly{1}" filename
To search all the words with start with "icon-" the following command works perfect. I am using Ack here which is similar to grep but with better options and nice formatting.
ack -oh --type=html "\w*icon-\w*" | sort | uniq
You could pipe your grep output into Perl like this:
grep "th" * | perl -n -e'while(/(\w*th\w*)/g) {print "$1\n"}'
grep --color -o -E "Begin.{0,}?End" file.txt
? - Match as few as possible until the End
Tested on macos terminal
$ grep -w
Excerpt from grep man page:
-w: Select only those lines containing matches that form whole words. The test is that the matching substring must either be at the beginning of the line, or preceded by a non-word constituent character.
ripgrep
Here are the example using ripgrep:
rg -o "(\w+)?th(\w+)?"
It'll match all words matching th.