How do I grep and show the preceding and following 5 lines surrounding each matched line?
For BSD or GNU grep you can use -B num to set how many lines before the match and -A num for the number of lines after the match.
grep -B 3 -A 2 foo README.txt
If you want the same number of lines before and after you can use -C num.
grep -C 3 foo README.txt
This will show 3 lines before and 3 lines after.
-A and -B will work, as will -C n (for n lines of context), or just -n (for n lines of context... as long as n is 1 to 9).
ack works with similar arguments as grep, and accepts -C. But it's usually better for searching through code.
grep astring myfile -A 5 -B 5
That will grep "myfile" for "astring", and show 5 lines before and after each match
ripgrep
If you care about the performance, use ripgrep which has similar syntax to grep, e.g.
rg -C5 "pattern" .
-C, --context NUM - Show NUM lines before and after each match.
There are also parameters such as -A/--after-context and -B/--before-context.
The tool is built on top of Rust's regex engine which makes it very efficient on the large data.
I normally use
grep searchstring file -C n # n for number of lines of context up and down
Many of the tools like grep also have really great man files too. I find myself referring to grep's man page a lot because there is so much you can do with it.
man grep
Many GNU tools also have an info page that may have more useful information in addition to the man page.
info grep
Use grep
$ grep --help | grep -i context
Context control:
-B, --before-context=NUM print NUM lines of leading context
-A, --after-context=NUM print NUM lines of trailing context
-C, --context=NUM print NUM lines of output context
-NUM same as --context=NUM
If you search code often, AG the silver searcher is much more efficient (ie faster) than grep.
You show context lines by using the -C option.
Eg:
ag -C 3 "foo" myFile
line 1
line 2
line 3
line that has "foo"
line 5
line 6
line 7
Search for "17655" in /some/file.txt showing 10 lines context before and after (using Awk), output preceded with line number followed by a colon. Use this on Solaris when grep does not support the -[ACB] options.
awk '
/17655/ {
for (i = (b + 1) % 10; i != b; i = (i + 1) % 10) {
print before[i]
}
print (NR ":" ($0))
a = 10
}
a-- > 0 {
print (NR ":" ($0))
}
{
before[b] = (NR ":" ($0))
b = (b + 1) % 10
}' /some/file.txt;
Let's understand using an example.
We can use grep with options:
-A 5 # this will give you 5 lines after searched string.
-B 5 # this will give you 5 lines before searched string.
-C 5 # this will give you 5 lines before & after searched string
Example.
File.txt contains 6 lines and following are the operations.
[abc#xyz]~/% cat file.txt # print all file data
this is first line
this is 2nd line
this is 3rd line
this is 4th line
this is 5th line
this is 6th line
[abc#xyz]~% grep "3rd" file.txt # we are searching for keyword '3rd' in the file
this is 3rd line
[abc#xyz]~% grep -A 2 "3rd" file.txt # print 2 lines after finding the searched string
this is 3rd line
this is 4th line
this is 5th line
[abc#xyz]~% grep -B 2 "3rd" file.txt # Print 2 lines before the search string.
this is first line
this is 2nd line
this is 3rd line
[abc#xyz]~% grep -C 2 "3rd" file.txt # print 2 line before and 2 line after the searched string
this is first line
this is 2nd line
this is 3rd line
this is 4th line
this is 5th line
Trick to remember options:
-A → A means "after"
-B → B means "before"
-C → C means "in between"
I do it the compact way:
grep -5 string file
That is the equivalent of:
grep -A 5 -B 5 string file
Here is the #Ygor solution in awk
awk 'c-->0;$0~s{if(b)for(c=b+1;c>1;c--)print r[(NR-c+1)%b];print;c=a}b{r[NR%b]=$0}' b=3 a=3 s="pattern" myfile
Note: Replace a and b variables with number of lines before and after.
It's especially useful for system which doesn't support grep's -A, -B and -C parameters.
Grep has an option called Context Line Control, you can use the --context in that, simply,
| grep -C 5
or
| grep -5
Should do the trick
$ grep thestring thefile -5
-5 gets you 5 lines above and below the match 'thestring' is equivalent to -C 5 or -A 5 -B 5.
Related
I have a file which contains several lines as follows:
>header1
<pattern_1>CGGCGGGCAGATGGCCACCAACAACCAGAGCTCCCTGGCCGGGCCTCTTTTCCTGACGGCCGCCCCCACTGCCCCCACGACCGGCCCGTACAAC<pattern_2>
>header2
<pattern_1>CGGCGGGCAGATGGCCACCAACAACCAGAGCTCCCTGGCCTGCAATCACTACTCGTGTTTTGCCACCACTGCCCCCACGACCGGCACGTACAAC<pattern_2>
>header3
<pattern_1>ATGGCCACCAACAACCAGAGCTCCC
>header4
GACCGGCACGTACAACCTCCAGGAAATCGTGCCCGGCAGCGTGTGGATGGAGAGGGACGTG
>header5
TGCCCCCACGACCGGCACGTACAAC<pattern_2>
I want to extract all lines containing both and including the header lines.
I have tried using grep, but it only extracts the sequence lines but not the header lines.
grep <pattern_1> | grep <pattern_2> input.fasta > output.fasta
How to extract lines containing both the patterns and the headers in Linux? The patterns can be present anywhere in the lines. Not limited to start or end of the lines.
Expected output:
>header1
<pattern_1>CGGCGGGCAGATGGCCACCAACAACCAGAGCTCCCTGGCCGGGCCTCTTTTCCTGACGGCCGCCCCCACTGCCCCCACGACCGGCCCGTACAAC<pattern_2>
>header2
<pattern_1>CGGCGGGCAGATGGCCACCAACAACCAGAGCTCCCTGGCCTGCAATCACTACTCGTGTTTTGCCACCACTGCCCCCACGACCGGCACGTACAAC<pattern_2>
$ grep -A 1 header[12] file
>header1
<pattern_1>CGGCGGGCAGATGGCCACCAACAACCAGAGCTCCCTGGCCGGGCCTCTTTTCCTGACGGCCGCCCCCACTGCCCCCACGACCGGCCCGTACAAC<pattern_2>
>header2
<pattern_1>CGGCGGGCAGATGGCCACCAACAACCAGAGCTCCCTGGCCTGCAATCACTACTCGTGTTTTGCCACCACTGCCCCCACGACCGGCACGTACAAC<pattern_2>
man grep:
-A NUM, --after-context=NUM
Print NUM lines of trailing context after matching lines.
Places a line containing a group separator (--) between
contiguous groups of matches. With the -o or --only-matching
option, this has no effect and a warning is given.
-B NUM, --before-context=NUM
Print NUM lines of leading context before matching lines.
Places a line containing a group separator (--) between
contiguous groups of matches. With the -o or --only-matching
option, this has no effect and a warning is given.
grep -B 1 pattern_[12]could work also, but you have several pattern_1s in the sample data so... not this time.
You can easily do that with awk like this:
awk '/^>/{h=$0;next}
/<pattern_1>/&&/<pattern_2>/{print h;print}' input.fasta > output.fasta
And here is a sed solution which yields the desired output as well:
sed -n '/^>/{N;/<pattern_1>/{/<pattern_2>/p}}' input.fasta > output.fasta
If it is likely that multiline records exist, you can use this:
awk -v pat1='<pattern_1>' -v pat2='<pattern_2>' '
/^>/ {r=$0;p=0;next}
!p {r=r ORS $0;if(chk()){print r;p=1};next}
p
function chk( tmp){
tmp=gensub(/\n/,"","g",r)
return (tmp~pat1&&tmp~pat2)
}' input.fasta > output.fasta
You might be interested in BioAwk, it is an adapted version of awk which is tuned to process fasta files
bioawk -c fastx -v seq1="pattern1" -v seq2="pattern2" \
'($seq ~ seq1) && ($seq ~ seq2) { print ">"$name; print $seq }' file.fasta
If you want seq1 at the beginning and seq2 at the end, you can change it into:
bioawk -c fastx -v seq1="pattern1" -v seq2="pattern2" \
'($seq ~ "^"seq1) && ($seq ~ seq2"$") { print ">"$name; print $seq }' file.fasta
This is really practical for processing fasta files, as often the sequence is spread over multiple lines. The above code handles this very easily as the variable $seq contains the full sequence.
If you do not want to install BioAwk, you can use the following method to process your FASTA file. It will allow multi-line sequences and does the following:
read a single record at a time (this assumes no > in the header, except the first character)
extract the header from the record and store it in name (not really needed)
merge the full sequence in a single string of characters, removing all newlines and spaces. This ensures that searching for pattern1 or pattern2 will not fail if the pattern is split over multiple lines.
if a match is found, print the record.
The following awk does the requested:
awk -v seq1="pattern1" -v seq2="pattern2" \
'BEGIN{RS=">"; ORS=""; FS="\n"}
{ seq="";for(i=2;i<=NF;++i) seq=seq""$i; gsub(/[^a-zA-Z0-9]/,"",seq) }
(seq ~ seq1 && seq ~ seq2){print ">" $0}' file.fasta
If the record header contains other > characters which are not at the beginning of the line, you have to take a slightly different approach (unless you use GNU awk)
awk -v seq1="pattern1" -v seq2="pattern2" \
'/^>/ && (seq ~ seq1 && seq ~ seq2) {
print name
for(i=0;i<n;i++) print aseq[i]
}
/^>/ { seq=""; delete aseq; n=0; name=$0; next }
{ aseq[n++] = $0; seq=seq""$0; sub(/[^a-zA-Z0-9]*$/,"",seq) }
END { if (seq ~ seq1 && seq ~ seq2) {
print name
for(i=0;i<n;i++) print aseq[i]
}
}' file.fasta
note: we make use of sub here in case unexpected characters are introduced in the fasta file (eg. spaces/tabs or CR (\r))
Note: BioAwk is based on Brian Kernighan's awk which is documented in "The AWK Programming Language",
by Al Aho, Brian Kernighan, and Peter Weinberger
(Addison-Wesley, 1988, ISBN 0-201-07981-X)
. I'm not sure if this version is compatible with POSIX.
If you want grep to print lines around the match, use the -B flag for lines before, the -A for lines after, and -C for both before and after the match.
In your case, grep -B 1 seems like it would do the job.
If your input file is exactly as described in your post then you can use:
grep -B1 '^<pattern_1>.*<pattern_2>$' input
>header1
<pattern_1>CGGCGGGCAGATGGCCACCAACAACCAGAGCTCCCTGGCCGGGCCTCTTTTCCTGACGGCCGCCCCCACTGCCCCCACGACCGGCCCGTACAAC<pattern_2>
>header2
<pattern_1>CGGCGGGCAGATGGCCACCAACAACCAGAGCTCCCTGGCCTGCAATCACTACTCGTGTTTTGCCACCACTGCCCCCACGACCGGCACGTACAAC<pattern_2>
Where -B1 will display on top of the matching lines the line before it. The regex used is based on the hypothesis that your 2 patterns are located in the exact order at the beginning and at the end of the line. If this is not the case: use '.*<pattern_1>.*<pattern_2>.*'. Last but not least, if the order of the 2 patterns are not always respected then you can use: '^.*<pattern_1>.*<pattern_2>.*$\|^.*<pattern_2>.*<pattern_1>.*$'
On the following input file:
cat input
>header1
<pattern_1>CGGCGGGCAGATGGCCACCAACAACCAGAGCTCCCTGGCCGGGCCTCTTTTCCTGACGGCCGCCCCCACTGCCCCCACGACCGGCCCGTACAAC<pattern_2>
>header2
<pattern_1>CGGCGGGCAGATGGCCACCAACAACCAGAGCTCCCTGGCCTGCAATCACTACTCGTGTTTTGCCACCACTGCCCCCACGACCGGCACGTACAAC<pattern_2>
>header2b
<pattern_2>CGGCGGGCAGATGGCCACCAACAACCAGAGCTCCCTGGCCTGCAATCACTACTCGTGTTTTGCCACCACTGCCCCCACGACCGGCACGTACAAC<pattern_1>
>header3
<pattern_1>ATGGCCACCAACAACCAGAGCTCCC
>header4
GACCGGCACGTACAACCTCCAGGAAATCGTGCCCGGCAGCGTGTGGATGGAGAGGGACGTG
>header5
TGCCCCCACGACCGGCACGTACAAC<pattern_2>
output:
grep -B1 '^.*<pattern_1>.*<pattern_2>.*$\|^.*<pattern_2>.*<pattern_1>.*$' input
>header1
<pattern_1>CGGCGGGCAGATGGCCACCAACAACCAGAGCTCCCTGGCCGGGCCTCTTTTCCTGACGGCCGCCCCCACTGCCCCCACGACCGGCCCGTACAAC<pattern_2>
>header2
<pattern_1>CGGCGGGCAGATGGCCACCAACAACCAGAGCTCCCTGGCCTGCAATCACTACTCGTGTTTTGCCACCACTGCCCCCACGACCGGCACGTACAAC<pattern_2>
>header2b
<pattern_2>CGGCGGGCAGATGGCCACCAACAACCAGAGCTCCCTGGCCTGCAATCACTACTCGTGTTTTGCCACCACTGCCCCCACGACCGGCACGTACAAC<pattern_1>
i have two text files 'simple' and 'simple1' with following data in them
simple.txt--
hello
hi hi hello
this
is it
simple1.txt--
hello hi
how are you
[]$ tr ' ' '\n' < simple.txt | grep -i -c '\bh\w*'
4
[]$ tr ' ' '\n' < simple1.txt | grep -i -c '\bh\w*'
3
this commands show the number of words that start with "h" for each file but i want to display the total count to be 7 i.e. total of both file. Can i do this in single command/shell script?
P.S.: I had to write two commands as tr does not take two file names.
Try this, the straightforward way :
cat simple.txt simple1.txt | tr ' ' '\n' | grep -i -c '\bh\w*'
This alternative requires no pipelines:
$ awk -v RS='[[:space:]]+' '/^h/{i++} END{print i+0}' simple.txt simple1.txt
7
How it works
-v RS='[[:space:]]+'
This tells awk to treat each word as a record.
/^h/{i++}
For any record (word) that starts with h, we increment variable i by 1.
END{print i+0}
After we have finished reading all the files, we print out the value of i.
It is not the case, that tr accepts only one filename, it does not accept any filename (and always reads from stdin). That's why even in your solution, you didn't provide a filename for tr, but used input redirection.
In your case, I think you can replace tr by fmt, which does accept filenames:
fmt -1 simple.txt simple1.txt | grep -i -c -w 'h.*'
(I also changed the grep a bit, because I personally find it better readable this way, but this is a matter of taste).
Note that both solutions (mine and your original ones) would count a string consisting of letters and one or more non-space characters - for instance the string haaaa.hbbbbbb.hccccc - as a "single block", i.e. it would only add 1 to the count of "h"-words, not 3. Whether or not this is the desired behaviour, it's up to you to decide.
This question already has answers here:
grep - print line before, don't print match
(3 answers)
Closed 7 years ago.
I want to search for a particular string in a certain file. Once found I only want to print the previous line of the greped line but not the line obtained using grep command.
cmd : grep -B 1 line5
Ex:
lin1 with some text
lin2
lin3
lin4
lin5 with some text
Output will be
lin4
lin5 with some text
But is there any solution where I can obtain only lin4 but not lin5.
Thanks for the help in advance.
Display N lines after match
-A is the option which prints the specified N lines after the match as shown below.
Syntax:
grep -A <N> "string" FILENAME
Display N lines before match
-B is the option, which prints the specified N lines before the match.
Syntax:
grep -B <N> "string” FILENAME
Display N lines around match
-C is the option, which prints the specified N lines before the match. In some occasion you might want the match to be appeared with the lines from both the side. This options shows N lines in both the side(before & after) of match.
$ grep -C 2 "Example" FILENAME
Just do this:
grep -B 1 line5 | head -1
to get the first line of your output
I need to get a row based on column value just like querying a database. I have a command output like this,
Name ID Mem VCPUs State
Time(s)
Domain-0 0 15485 16 r-----
1779042.1
prime95-01 512 1
-b---- 61.9
Here I need to list only those rows where state is "r". Something like this,
Domain-0 0 15485 16
r----- 1779042.1
I have tried using "grep" and "awk" but still I am not able to succeed.
Any help me is much appreciated
Regards,
Raaj
There is a variaty of tools available for filtering.
If you only want lines with "r-----" grep is more than enough:
command | grep "r-----"
Or
cat filename | grep "r-----"
grep can handle this for you:
yourcommand | grep -- 'r-----'
It's often useful to save the (full) output to a file to analyse later. For this I use tee.
yourcommand | tee somefile | grep 'r-----'
If you want to find the line containing "-b----" a little later on without re-running yourcommand, you can just use:
grep -- '-b----' somefile
No need for cat here!
I recommend putting -- after your call to grep since your patterns contain minus-signs and if the minus-sign is at the beginning of the pattern, this would look like an option argument to grep rather than a part of the pattern.
try:
awk '$5 ~ /^r.*/ { print }'
Like this:
cat file | awk '$5 ~ /^r.*/ { print }'
grep solution:
command | grep -E "^([^ ]+ ){4}r"
What this does (-E switches on extended regexp):
The first caret (^) matches the beginning of the line.
[^ ] matches exactly one occurence of a non-space character, the following modifier (+) allows it to also match more occurences.
Grouped together with the trailing space in ([^ ]+ ), it matches any sequence of non-space characters followed by a single space. The modifyer {4} requires this construct to be matched exactly four times.
The single "r" is then the literal character you are searching for.
In plain words this could be written like "If the line starts <^> with four strings that are followed by a space <([^ ]+ ){4}> and the next character is , then the line matches."
A very good introduction into regular expressions has been written by Jan Goyvaerts (http://www.regular-expressions.info/quickstart.html).
Filtering by awk cmd in linux:-
Firstly find the column for this cmd and store file2 :-
awk '/Domain-0 0 15485 /' file1 >file2
Output:-
Domain-0 0 15485 16
r----- 1779042.1
after that awk cmd in file2:-
awk '{print $1,$2,$3,$4,"\n",$5,$6}' file2
Final Output:-
Domain-0 0 15485 16
r----- 1779042.1
Does anyone know how to replace line a with line b and line b with line a in a text file using the sed editor?
I can see how to replace a line in the pattern space with a line that is in the hold space (i.e., /^Paco/x or /^Paco/g), but what if I want to take the line starting with Paco and replace it with the line starting with Vinh, and also take the line starting with Vinh and replace it with the line starting with Paco?
Let's assume for starters that there is one line with Paco and one line with Vinh, and that the line Paco occurs before the line Vinh. Then we can move to the general case.
#!/bin/sed -f
/^Paco/ {
:notdone
N
s/^\(Paco[^\n]*\)\(\n\([^\n]*\n\)*\)\(Vinh[^\n]*\)$/\4\2\1/
t
bnotdone
}
After matching /^Paco/ we read into the pattern buffer until s// succeeds (or EOF: the pattern buffer will be printed unchanged). Then we start over searching for /^Paco/.
cat input | tr '\n' 'ç' | sed 's/\(ç__firstline__\)\(ç__secondline__\)/\2\1/g' | tr 'ç' '\n' > output
Replace __firstline__ and __secondline__ with your desired regexps. Be sure to substitute any instances of . in your regexp with [^ç]. If your text actually has ç in it, substitute with something else that your text doesn't have.
try this awk script.
s1="$1"
s2="$2"
awk -vs1="$s1" -vs2="$s2" '
{ a[++d]=$0 }
$0~s1{ h=$0;ind=d}
$0~s2{
a[ind]=$0
for(i=1;i<d;i++ ){ print a[i]}
print h
delete a;d=0;
}
END{ for(i=1;i<=d;i++ ){ print a[i] } }' file
output
$ cat file
1
2
3
4
5
$ bash test.sh 2 3
1
3
2
4
5
$ bash test.sh 1 4
4
2
3
1
5
Use sed (or not at all) for only simple substitution. Anything more complicated, use a programming language
A simple example from the GNU sed texinfo doc:
Note that on implementations other than GNU `sed' this script might
easily overflow internal buffers.
#!/usr/bin/sed -nf
# reverse all lines of input, i.e. first line became last, ...
# from the second line, the buffer (which contains all previous lines)
# is *appended* to current line, so, the order will be reversed
1! G
# on the last line we're done -- print everything
$ p
# store everything on the buffer again
h