I would like to get multi-line text in between horizontal delimiter and ignore anything else before and after the delimiter.
An example would be:-
Some text here before any delimiter
----------
Line 1
Line 2
Line 3
Line 4
----------
Line 1
Line 2
Line 3
Line 4
----------
Some text here after last delimiter
And I would like to get
Line 1
Line 2
Line 3
Line 4
Line 1
Line 2
Line 3
Line 4
How do I do this with awk / sed with regex? Thanks.
You can try this.
file: a.awk:
BEGIN { RS = "-+" }
{
if ( NR > 1 && RT != "" )
{
print $0
}
}
run: awk -f a.awk data_file
If you can comfortably fit the entire file into memory, and if Perl is acceptable instead of awk or sed,
perl -0777 -pe 's/\A.*?\n-{10}\n//s;
s/(.*\n)-{10}\n.*?\Z/\1/s;
s/\n-{10}\n/\n\n\n/g' file >newfile
The main FAQs here are the -0777 option (slurp mode) and the /s (dot matches newlines) regex flag.
This might work for you:
sed '1,/^--*$/d;:a;$!{/\(^\|\n\)--*$/!N;//!ba;s///p};d' file
Related
I want to extract lines that have a particular pattern, in a certain column. For example, in my 'input.txt' file, I have many columns. I want to search the 25th column for 'foobar', and extract only those lines that have 'foobar' in the 25th column. I cannot do:
grep foobar input.txt
because other columns may also have 'foobar', and I don't want those lines. Also:
the 25th column will have 'foobar' as part of a string (i.e. it could be 'foobar ; muller' or 'max ; foobar ; john', or 'tom ; foobar35')
I would NOT want 'tom ; foobar35'
The word in column 25 must be an exact match for 'foobar' (and ; so using awk $25=='foobar' is not an option.
In other words, if column 25 had the following lines:
foobar ; muller
max ; foobar ; john
tom ; foobar35
I would want only lines 1 & 2.
How do I use xargs and sed to extract these lines? I am stuck at:
cut -f25 input.txt | grep -nw foobar | xargs -I linenumbers sed ???
thanks!
Do not use xargs and sed, use the other tool common on so many machines and do this:
awk '{if($25=="foobar"){print NR" "$0}}' input.txt
print NR prints the line number of the current match so the first column of the output will be the line number.
print $0 prints the current line. Change it to print $25 if you only want the matching column. If you only want the output, use this:
awk '{if($25=="foobar"){print $0}}' input.txt
EDIT1 to match extended question:
Use what #shellter and #Jotne suggested but add string delimiters.
awk -vFPAT="([^ ]*)|('[^']*')" -vOFS=' ' '$25~/foobar/' input.txt
[^ ]* matches all characters that are not a space.
'[^']*' matches everything inside single quotes.
EDIT2 to exclude everything but foobar:
awk -vFPAT="([^ ]*)|('[^']*')" -vOFS=' ' "\$25~/[;' ]foobar[;' ]/" input.txt
[;' ] only allows ;, ' and in front and after foobar.
Tested with this file:
1 "1 ; 1" 4
2 'kom foobar' 33
3 "ll;3" 3
4 '1; foobar' asd
7 '5 ;foobar' 2
7 '5;foobar' 0
2 'kom foobar35' 33
2 'kom ; foobar' 33
2 'foobar ; john' 33
2 'foobar;paul' 33
2 'foobar1;paul' 33
2 'foobarli;paul' 33
2 'afoobar;paul' 33
and this command awk -vFPAT="([^ ]*)|('[^']*')" -vOFS=' ' "\$2~/[;' ]foobar[;' ]/" input.txt
To get the line with foobar as part of the 25 field.
awk '$25=="foobar"' input.txt
$25 25th filed
== equal to
"foobar"
Since no action spesified, print the complete line will be done, same as {print $0}
Or
awk '$25~/^foobar$/' input.txt
This might work for you (GNU sed):
sed -En 's/\S+/\n&\n/25;s/\n(.*foobar.*)\n/\1/p' file
Surround the 25th field by newlines and pattern match for foobar between newlines.
If you only want to match the word foobar use:
sed -En 's/\S+/\n&\n/25;s/\n(.*\<foobar\>.*)\n/\1/p' file
I have a special file with this kind of format :
title1
_1 texthere
title2
_2 texthere
I would like all newlines starting with "_" to be placed as a second column to the line before
I tried to do that using sed with this command :
sed 's/_\n/ /g' filename
but it is not giving me what I want to do (doing nothing basically)
Can anyone point me to the right way of doing it ?
Thanks
Try following solution:
In sed the loop is done creating a label (:a), and while not match last line ($!) append next one (N) and return to label a:
:a
$! {
N
b a
}
After this we have the whole file into memory, so do a global substitution for each _ preceded by a newline:
s/\n_/ _/g
p
All together is:
sed -ne ':a ; $! { N ; ba }; s/\n_/ _/g ; p' infile
That yields:
title1 _1 texthere
title2 _2 texthere
If your whole file is like your sample (pairs of lines), then the simplest answer is
paste - - < file
Otherwise
awk '
NR > 1 && /^_/ {printf "%s", OFS}
NR > 1 && !/^_/ {print ""}
{printf "%s", $0}
END {print ""}
' file
This might work for you (GNU sed):
sed ':a;N;s/\n_/ /;ta;P;D' file
This avoids slurping the file into memory.
or:
sed -e ':a' -e 'N' -e 's/\n_/ /' -e 'ta' -e 'P' -e 'D' file
A Perl approach:
perl -00pe 's/\n_/ /g' file
Here, the -00 causes perl to read the file in paragraph mode where a "line" is defined by two consecutive newlines. In your example, it will read the entire file into memory and therefore, a simple global substitution of \n_ with a space will work.
That is not very efficient for very large files though. If your data is too large to fit in memory, use this:
perl -ne 'chomp;
s/^_// ? print "$l " : print "$l\n" if $. > 1;
$l=$_;
END{print "$l\n"}' file
Here, the file is read line by line (-n) and the trailing newline removed from all lines (chomp). At the end of each iteration, the current line is saved as $l ($l=$_). At each line, if the substitution is successful and a _ was removed from the beginning of the line (s/^_//), then the previous line is printed with a space in place of a newline print "$l ". If the substitution failed, the previous line is printed with a newline. The END{} block just prints the final line of the file.
I would be happy if anyone can suggest me command (sed or AWK one line command) to divide each line of file in equal number of part. For example divide each line in 4 part.
Input:
ATGCATHLMNPHLNTPLML
Output:
ATGCA THLMN PHLNT PLML
This should work using GNU sed:
sed -r 's/(.{4})/\1 /g'
-r is needed to use extended regular expressions
.{4} captures every four characters
\1 refers to the captured group which is surrounded by the parenthesis ( ) and adds a space behind this group
g makes sure that the replacement is done as many times as possible on each line
A test; this is the input and output in my terminal:
$ echo "ATGCATHLMNPHLNTPLML" | sed -r 's/(.{4})/\1 /g'
ATGC ATHL MNPH LNTP LML
I suspect awk is not the best tool for this, but:
gawk --posix '{ l = sprintf( "%d", 1 + (length()-1)/4);
gsub( ".{"l"}", "& " ) } 1' input-file
If you have a posix compliant awk you can omit the --posix, but --posix is necessary for gnu awk and since that seems to be the most commonly used implementation I've given the solution in terms of gawk.
This might work for you (GNU sed):
sed 'h;s/./X/g;s/^\(.*\)\1\1\1/\1 \1 \1 \1/;G;s/\n/&&/;:a;/^\n/bb;/^ /s/ \(.*\n.*\)\n\(.\)/\1 \n\2/;ta;s/^.\(.*\n.*\)\n\(.\)/\1\2\n/;ta;:b;s/\n//g' file
Explanation:
h copy the pattern space (PS) to the hold space (HS)
s/./X/g replace every character in the HS with the same non-space character (in this case X)
s/^\(.*\)\1\1\1/\1 \1 \1 \1/ split the line into 4 parts (space separated)
G append a newline followed by the contents of the HS to the PS
s/\n/&&/ double the newline (to be later used as markers)
:a introduce a loop namespace
/^\n/bb if we reach a newline we are done and branch to the b namespace
/^ /s/ \(.*\n.*\)\n\(.\)/\1 \n\2/;ta; if the first character is a space add a space to the real line at this point and repeat
s/^.\(.*\n.*\)\n\(.\)/\1\2\n/;ta any other character just bump along and repeat
:b;s/\n//g all done just remove the markers and print out the result
This work for any length of line, however is the line is not exactly divisible by 4 the last portion will contain the remainder as well.
perl
perl might be a better choice here:
export cols=4
perl -ne 'chomp; $fw = 1 + int length()/$ENV{cols}; while(/(.{1,$fw})/gm) { print $1 . " " } print "\n"'
This re-calculates field-width for every line.
coreutils
A GNU coreutils alternative, field-width is chosen based on the first line of infile:
cols=4
len=$(( $(head -n1 infile | wc -c) - 1 ))
fw=$(echo "scale=0; 1 + $len / 4" | bc)
cut_arg=$(paste -d- <(seq 1 $fw 19) <(seq $fw $fw $len) | head -c-1 | tr '\n' ',')
Value of cut_arg is in the above case:
1-5,6-10,11-15,16-
Now cut the line into appropriate chunks:
cut --output-delimiter=' ' -c $cut_arg infile
I need help with using sed to comment a matching lines and 4 lines which follows it.
in a text file.
my text file is like this:
[myprocess-a]
property1=1
property2=2
property3=3
property4=4
[anotherprocess-b]
property1=gffgg
property3=gjdl
property2=red
property4=djfjf
[myprocess-b]
property1=1
property4=4
property2=2
property3=3
I want to prefix # to all the lines having text '[myprocess' and 4 lines that follows it
expected output:
#[myprocess-a]
#property1=1
#property2=2
#property3=3
#property4=4
[anotherprocess-b]
property1=gffgg
property3=gjdl
property2=red
property4=djfjf
#[myprocess-b]
#property1=1
#property4=4
#property2=2
#property3=3
Greatly appreciate your help on this.
You can do this by applying a regular expression to a set of lines:
sed -e '/myprocess/,+4 s/^/#/'
This matches lines with 'myprocess' and the 4 lines after them. For those 4 lines it then inserts a '#' at the beginning of the line.
(I think this might be a GNU extension - it's not in any of the "sed one liner" cheatsheets I know)
sed '/\[myprocess/ { N;N;N;N; s/^/#/gm }' input_file
Using string concatenation and default action in awk.
http://www.gnu.org/software/gawk/manual/html_node/Concatenation.html
awk '/myprocess/{f=1} f>5{f=0} f{f++; $0="#" $0} 1' foo.txt
or if the block always ends with empty line
awk '/myprocess/{f=1} !NF{f=0} f{$0="#" $0} 1' foo.txt
Most command-line programs just operate on one line at a time.
Can I use a common command-line utility (echo, sed, awk, etc) to concatenate every set of two lines, or would I need to write a script/program from scratch to do this?
$ cat myFile
line 1
line 2
line 3
line 4
$ cat myFile | __somecommand__
line 1line 2
line 3line 4
sed 'N;s/\n/ /;'
Grab next line, and substitute newline character with space.
seq 1 6 | sed 'N;s/\n/ /;'
1 2
3 4
5 6
$ awk 'ORS=(NR%2)?" ":"\n"' file
line 1 line 2
line 3 line 4
$ paste - - < file
line 1 line 2
line 3 line 4
Not a particular command, but this snippet of shell should do the trick:
cat myFile | while read line; do echo -n $line; [ "${i}" ] && echo && i= || i=1 ; done
You can also use Perl as:
$ perl -pe 'chomp;$i++;unless($i%2){$_.="\n"};' < file
line 1line 2
line 3line 4
Here's a shell script version that doesn't need to toggle a flag:
while read line1; do read line2; echo $line1$line2; done < inputfile