I have the below lines in a file
Acanthocephala;Palaeacanthocephala;Polymorphida;Polymorphidae;;Profilicollis;Profilicollis_altmani;
Acanthocephala;Eoacanthocephala;Neoechinorhynchida;Neoechinorhynchidae;;;;
Acanthocephala;;;;;;;
Acanthocephala;Palaeacanthocephala;Polymorphida;Polymorphidae;;Polymorphus;;
and I want to remove the repeating semi-colon characters from all lines to look like below (note- there are repeating semi-colons in the middle of some of the above lines too)
Acanthocephala;Palaeacanthocephala;Polymorphida;Polymorphidae;Profilicollis;Profilicollis_altmani;
Acanthocephala;Eoacanthocephala;Neoechinorhynchida;Neoechinorhynchidae;
Acanthocephala;
Acanthocephala;Palaeacanthocephala;Polymorphida;Polymorphidae;Polymorphus;
I would appreciate if someone could kindly share a bash one-liner to accomplish this.
You can use tr with "squeeze":
tr -s ';' < infile
perl -p -e 's/;+/;/g' myfile # writes output to stdout
or
perl -p -i -e 's/;+/;/g' myfile # does an in-place edit
If you want to edit the file itself:
printf "%s\n" 'g/;;/s/;\{2,\}/;/g' w | ed -s foo.txt
If you want to pipe a modified copy of the file to something else and leave the original unchanged:
sed 's/;\{2,\}/;/g' foo.txt | whatever
These replace runs of 2 or more semicolons with single ones.
could be solved easily by substitutions.
I add an awk solution by playing with the FS/OFS variable:
awk -F';+' -v OFS=';' '$1=$1' file
or
awk -F';+' -v OFS=';' '($1=$1)||1' file
Here's a sed version of alaniwi's answer:
sed 's/;\+/;/g' myfile # Write output to stdout
or
sed -i 's/;\+/;/g' myfile # Edit the file in-place
How would I use sed to delete all lines in a text file that contain a specific string?
To remove the line and print the output to standard out:
sed '/pattern to match/d' ./infile
To directly modify the file – does not work with BSD sed:
sed -i '/pattern to match/d' ./infile
Same, but for BSD sed (Mac OS X and FreeBSD) – does not work with GNU sed:
sed -i '' '/pattern to match/d' ./infile
To directly modify the file (and create a backup) – works with BSD and GNU sed:
sed -i.bak '/pattern to match/d' ./infile
There are many other ways to delete lines with specific string besides sed:
AWK
awk '!/pattern/' file > temp && mv temp file
Ruby (1.9+)
ruby -i.bak -ne 'print if not /test/' file
Perl
perl -ni.bak -e "print unless /pattern/" file
Shell (bash 3.2 and later)
while read -r line
do
[[ ! $line =~ pattern ]] && echo "$line"
done <file > o
mv o file
GNU grep
grep -v "pattern" file > temp && mv temp file
And of course sed (printing the inverse is faster than actual deletion):
sed -n '/pattern/!p' file
You can use sed to replace lines in place in a file. However, it seems to be much slower than using grep for the inverse into a second file and then moving the second file over the original.
e.g.
sed -i '/pattern/d' filename
or
grep -v "pattern" filename > filename2; mv filename2 filename
The first command takes 3 times longer on my machine anyway.
The easy way to do it, with GNU sed:
sed --in-place '/some string here/d' yourfile
You may consider using ex (which is a standard Unix command-based editor):
ex +g/match/d -cwq file
where:
+ executes given Ex command (man ex), same as -c which executes wq (write and quit)
g/match/d - Ex command to delete lines with given match, see: Power of g
The above example is a POSIX-compliant method for in-place editing a file as per this post at Unix.SE and POSIX specifications for ex.
The difference with sed is that:
sed is a Stream EDitor, not a file editor.BashFAQ
Unless you enjoy unportable code, I/O overhead and some other bad side effects. So basically some parameters (such as in-place/-i) are non-standard FreeBSD extensions and may not be available on other operating systems.
I was struggling with this on Mac. Plus, I needed to do it using variable replacement.
So I used:
sed -i '' "/$pattern/d" $file
where $file is the file where deletion is needed and $pattern is the pattern to be matched for deletion.
I picked the '' from this comment.
The thing to note here is use of double quotes in "/$pattern/d". Variable won't work when we use single quotes.
You can also use this:
grep -v 'pattern' filename
Here -v will print only other than your pattern (that means invert match).
To get a inplace like result with grep you can do this:
echo "$(grep -v "pattern" filename)" >filename
I have made a small benchmark with a file which contains approximately 345 000 lines. The way with grep seems to be around 15 times faster than the sed method in this case.
I have tried both with and without the setting LC_ALL=C, it does not seem change the timings significantly. The search string (CDGA_00004.pdbqt.gz.tar) is somewhere in the middle of the file.
Here are the commands and the timings:
time sed -i "/CDGA_00004.pdbqt.gz.tar/d" /tmp/input.txt
real 0m0.711s
user 0m0.179s
sys 0m0.530s
time perl -ni -e 'print unless /CDGA_00004.pdbqt.gz.tar/' /tmp/input.txt
real 0m0.105s
user 0m0.088s
sys 0m0.016s
time (grep -v CDGA_00004.pdbqt.gz.tar /tmp/input.txt > /tmp/input.tmp; mv /tmp/input.tmp /tmp/input.txt )
real 0m0.046s
user 0m0.014s
sys 0m0.019s
Delete lines from all files that match the match
grep -rl 'text_to_search' . | xargs sed -i '/text_to_search/d'
SED:
'/James\|John/d'
-n '/James\|John/!p'
AWK:
'!/James|John/'
/James|John/ {next;} {print}
GREP:
-v 'James\|John'
perl -i -nle'/regexp/||print' file1 file2 file3
perl -i.bk -nle'/regexp/||print' file1 file2 file3
The first command edits the file(s) inplace (-i).
The second command does the same thing but keeps a copy or backup of the original file(s) by adding .bk to the file names (.bk can be changed to anything).
You can also delete a range of lines in a file.
For example to delete stored procedures in a SQL file.
sed '/CREATE PROCEDURE.*/,/END ;/d' sqllines.sql
This will remove all lines between CREATE PROCEDURE and END ;.
I have cleaned up many sql files withe this sed command.
echo -e "/thing_to_delete\ndd\033:x\n" | vim file_to_edit.txt
Just in case someone wants to do it for exact matches of strings, you can use the -w flag in grep - w for whole. That is, for example if you want to delete the lines that have number 11, but keep the lines with number 111:
-bash-4.1$ head file
1
11
111
-bash-4.1$ grep -v "11" file
1
-bash-4.1$ grep -w -v "11" file
1
111
It also works with the -f flag if you want to exclude several exact patterns at once. If "blacklist" is a file with several patterns on each line that you want to delete from "file":
grep -w -v -f blacklist file
to show the treated text in console
cat filename | sed '/text to remove/d'
to save treated text into a file
cat filename | sed '/text to remove/d' > newfile
to append treated text info an existing file
cat filename | sed '/text to remove/d' >> newfile
to treat already treated text, in this case remove more lines of what has been removed
cat filename | sed '/text to remove/d' | sed '/remove this too/d' | more
the | more will show text in chunks of one page at a time.
Curiously enough, the accepted answer does not actually answer the question directly. The question asks about using sed to replace a string, but the answer seems to presuppose knowledge of how to convert an arbitrary string into a regex.
Many programming language libraries have a function to perform such a transformation, e.g.
python: re.escape(STRING)
ruby: Regexp.escape(STRING)
java: Pattern.quote(STRING)
But how to do it on the command line?
Since this is a sed-oriented question, one approach would be to use sed itself:
sed 's/\([\[/({.*+^$?]\)/\\\1/g'
So given an arbitrary string $STRING we could write something like:
re=$(sed 's/\([\[({.*+^$?]\)/\\\1/g' <<< "$STRING")
sed "/$re/d" FILE
or as a one-liner:
sed "/$(sed 's/\([\[/({.*+^$?]\)/\\\1/g' <<< "$STRING")/d"
with variations as described elsewhere on this page.
cat filename | grep -v "pattern" > filename.1
mv filename.1 filename
You can use good old ed to edit a file in a similar fashion to the answer that uses ex. The big difference in this case is that ed takes its commands via standard input, not as command line arguments like ex can. When using it in a script, the usual way to accomodate this is to use printf to pipe commands to it:
printf "%s\n" "g/pattern/d" w | ed -s filename
or with a heredoc:
ed -s filename <<EOF
g/pattern/d
w
EOF
This solution is for doing the same operation on multiple file.
for file in *.txt; do grep -v "Matching Text" $file > temp_file.txt; mv temp_file.txt $file; done
I found most of the answers not useful for me, If you use vim I found this very easy and straightforward:
:g/<pattern>/d
Source
I can remove the first line of csv file's starting with myfile and merge them using:
sed 1d myfile*.csv > myfile_merged.csv
I'd like to also remove the last line of the csv files.
I've tried:
sed 1d -i '$d' myfile*.csv > myfile_merged.csv
But get the error:
sed: can't read $d: No such file or directory
Problem is this command:
sed 1d -i '$d' myfile*.csv > myfile_merged.csv
You need not have an argument after -i (inline replacement) in sed otherwise it is treated as a SUFFIX to create a backup for inline replacement.
What you need is this gnu sed command:
sed -i '1d;$d' myfile*.csv
This will remove 1st and last line in each of the matched file and save it in place.
What you're probably trying to do is:
sed -e '1d' -e '$d' myfile*.csv > merged.csv
But this won't work, because it tells sed to remove the first and last line of ALL files, rather than EACH file. In other words, you'll strip the first line of the first file, and the last line of the last file ... and that's it.
To process each file individually, you probably need to process each file .. individually. :)
for f in myfile*.csv
sed -e '1d;$d' "$f"
done > merged.csv
Note that while this will run in bash, it's also POSIX compatible (both the shell and sed parts). And it does not care whether your input is CSV or any other format, as long as it can be parsed line by line using sed.
I've writted a sed script to replace all ^^ with NULL. It seems though that sed is only catching a pair, but not including the second in that pair as it continues to search.
echo "^^^^" | sed 's/\^\^/\^NULL\^/g'
produces
^NULL^^NULL^
when it should produce
^NULL^NULL^NULL^
Try with a loop to apply your command again to modified pattern space:
echo "^^^^" | sed ':a;s/\^\^/\^NULL\^/;t a;'
To edit a file in place on OSX, try the -i flag and multiline command:
sed -i '' ':a
s/\^\^/\^NULL\^/
t a' file
With GNU sed:
sed -i ':a;s/\^\^/\^NULL\^/;t a;' file
or simply redirect the command to a temporary file before renaming it:
sed ':a;s/\^\^/\^NULL\^/;t a;' file > tmp && mv tmp file
I really like SLePort solution, but since it is not working for you, you can try with (tested on Linux, not Mac):
echo "^^^^" | sed 's/\^\^/\^NULL\^/g; s//\^NULL\^/g'
It is doing the same as the former solution, but explicitly, not looping with tags.
You can omit the pattern in the second command and sed will use the previous pattern.
I have the following (example.txt) file:
blue(4) red(8) green(5) yellow(19) brown(60) black(5)
how can I achieve in unix the following result?
blue(4)
red(8)
green(5)
yellow(19)
brown(60)
black(5)
If you need to insert newline after closing brackets, try
sed 's/) \?/)\n/g' example.txt
The following in-line sed script will replace a space with a newline, and should solve your problem.
sed -i 's/ /\n/g' example.txt > example_out.txt
xargs -n 1 < example.txt
By passing example.txt into xargs taking one argument at a time -n 1, xargs will place each entry on a separate line.
E.g., to put two entries per line one would simply change the -n 1 to -n 2
The option -n is also referred to as max-args on the man page.
Pass your data to this sed command, like so:
sed 's/ /\n/g' example.txt
I needed to achieve something like this and used sed command. It can be used to perform functions on streams.
For your requirement, you can use it like this:
sed -i 's/ /\n/g' example.txt
You can read more about this in the sed man page.