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I'm trying to format cut, paste output but sed not working.
file.txt
Apple
Banana
Apple
Banana
Orange
Apple
Orange
code.sh
cut -f2 file.txt | sort | uniq | sed 's/^\|$/#/g'| paste -sd,\& -
expected output / output on ubuntu
#Apple#,#Banana#&#Orange#
getting output / output on macos
Apple,Banana&Orange
Note: The code works on Ubuntu, but on MacOS it doesn't.
This can be done in a single gnu-awk:
awk '!seen[$1]++{} END {
PROCINFO["sorted_in"]="#ind_str_asc"
for (i in seen)
s = s (s == "" ? "" : (++j==1?",":"&")) "#" i "#"
print s
}' file
#Apple#,#Banana#&#Orange#
On OSX I have gnu awk installed via home brew.
As mentioned elsewhere, BSD sed doesn't support \|. Instead of replacing ^ and $, you can substitute # around the whole line.
sort -u file.txt | sed 's/.*/#&#/' | paste -sd,'&' -
As far as I know, BSD/Mac sed doesn't support \|. See sed not giving me correct substitute operation for newline with Mac - differences between GNU sed and BSD / OSX sed for details.
As an alternate, you can use ERE instead of BRE. I checked it on Linux, apparently this still doesn't seem to work on Mac (See also: MacOS sed: match either beginning or end).
$ echo 'Apple' | sed -E 's/^|$/#/g'
#Apple#
# workaround for Mac
$ echo 'Apple' | sed -e 's/^/#/' -e 's/$/#/'
#Apple#
Instead of sort+uniq+sed, you can also use awk (but note that awk solution shown here removes duplicates while preserving original order, doesn't sort the input):
$ awk '!seen[$0]++{print "#" $0 "#"}' ip.txt
#Apple#
#Banana#
#Orange#
Change $0 to $2 if you want only the second field, based on your use of cut
A simple way to do it using the sed command:
sed -E 's/[[:alnum:]]+/#&#/'
the -E option for enabling the POSIX ERE (extended regular
expression)
[[:alnum:]]+ The alphanumeric characters; in ASCII, equivalent to [A-Za-z0-9] with the plus (+) to refer to one or more.
the & symbol, does bring or refer to the content of the pattern we found. (on which we surrounded it with #)
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 $file; done
I found most of the answers not useful for me, If you use vim I found this very easy and straightforward:
:g/<pattern>/d
Source
I have a file parse.txt
parse.txt contains the following
remo/hello/1.0,remo/hello2/2.0,remo/hello3/3.0,whitney/hello/1.0,julie/hello/2.0,julie/hello/3.0
and I want the output.txt file as (to reverse the order from last to first)using parse.txt
julie/hello/3.0,julie/hello/2.0,whitney/hello/1.0,remo/hello3/3.0,remo/hello2/2.0,remo/hello/1.0
I have tried the following code:
tail -r parse.txt
You can use the surprisingly helpful tac from GNU Coreutils.
tac -s "," parse.txt > newparse.txt
tac by default will "cat" the file to standard out, reversing the lines. By specifying the separator using the -s flag, you can simply reverse your fields as desired.
(You may need to do a post-processing step to get the commas to work out correctly, which can be another step in your pipeline.)
I like the tac solution; it's tight and elegant, but as Micah pointed out, tac is part of GNU Coreutils, which means that it's not available by default in FreeBSD, OSX, Solaris, etc.
This can be done in pure bash, no external tools required.
#!/usr/bin/env bash
unset comma
read foo < parse.txt
bar=(${foo//,/ })
for (( count="${#bar[#]}"; --count >= 0; )); do
printf "%s%s" "$comma" "${bar[$count]}"
comma=","
done
This obviously only handles one line, per your sample input. You can wrap it in something if you need to handle multiple lines of input.
The logic here is that we can convert the input into an array by replacing commas with spaces. Of course, if our input data included spaces, this would have to be adjusted. Once we have the array, we simply step backwards through it, printing each record.
Note that this does not include a terminating newline. If you want one, you can add it with:
printf '\n'
as a final line.
perl -F, -lane 'print join ",", reverse #F' parse.txt > output.txt
You can use this awk command:
awk -v RS=, '{a[++i]=$1} END{for (k=i; k>=1; k--) printf a[k] (k>1?RS:ORS)}' parse.txt
julie/hello/3.0,julie/hello/2.0,whitney/hello/1.0,remo/hello3/3.0,remo/hello2/2.0,remo/hello/1.0
The question is tagged unix and you have mentioned tail -r which suggests you might not be using Linux (with full GNU toolchain), but instead some "real" Unix (BSD variant), e.g. osx.
As such, the tac command is not available, but as mentioned in the question, tail -r is. So you can use the following:
$ tr ',' '\n' < parse.txt | tail -r | tr '\n' ',' | sed 's/,$//'
julie/hello/3.0,julie/hello/2.0,whitney/hello/1.0,remo/hello3/3.0,remo/hello2/2.0,remo/hello/1.0
$
Notes:
This only works for files that have one line, as we are relying on converting commas to newlines and back. If there is more than one line, then the newlines in between will get converted to commas by the second tr.
The final sed is to remove a trailing comma, that was converted from a trailing newline inserted by tail
Emulating tac with sed:
tr , '\n' <parse.txt | sed '1!G; h; $!d' | paste -sd ,
Alternatively, if you don't have paste:
tr , '\n' <parse.txt | sed '1!G; h; $!d' | tr '\n' , | sed 's/,$//'
Output:
julie/hello/3.0,julie/hello/2.0,whitney/hello/1.0,remo/hello3/3.0,remo/hello2/2.0,remo/hello/1.0
You can use any language to do that
xargs ruby -e "puts ARGV[0].split(',').reverse.join(',')" < parse.txt
Reverse can be done by tac (from cat). As commented this will reverse the lines not what the OP asked for.
tac filename
You can still you tac if you provide line by line and reverse by not linefeed delimiter but the field separator, here ,.
echo "a,b,c" | tr '\n' ',' | tac -s "," | sed 's/,$/\n/'
I have a text file that's about 300KB in size. I want to remove all lines from this file that begin with the letter "P". This is what I've been using:
> cat file.txt | egrep -v P*
That isn't outputting to console. I can use cat on the file without another other commands and it prints out fine. My final intention being to:
> cat file.txt | egrep -v P* > new.txt
No error appears, it just doesn't print anything out and if I run the 2nd command, new.txt is empty.
I should say I'm running Windows 7 with Cygwin installed.
Explanation
use ^ to anchor your pattern to the beginning of the line ;
delete lines matching the pattern using sed and the d flag.
Solution #1
cat file.txt | sed '/^P/d'
Better solution
Use sed-only:
sed '/^P/d' file.txt > new.txt
With awk:
awk '!/^P/' file.txt
Explanation
The condition starts with an ! (negation), that negates the following pattern ;
/^P/ means "match all lines starting with a capital P",
So, the pattern is negated to "ignore lines starting with a capital P".
Finally, it leverage awk's behavior when { … } (action block) is missing, that is to print the record validating the condition.
So, to rephrase, it ignores lines starting with a capital P and print everything else.
Note
sed is line oriented and awk column oriented. For your case you should use the first one, see Edouard Lopez's reponse.
Use sed with inplace substitution (for GNU sed, will also for your cygwin)
sed -i '/^P/d' file.txt
BSD (Mac) sed
sed -i '' '/^P/d' file.txt
Use start of line mark and quotes:
cat file.txt | egrep -v '^P.*'
P* means P zero or more times so together with -v gives you no lines
^P.* means start of line, then P, and any char zero or more times
Quoting is needed to prevent shell expansion.
This can be shortened to
egrep -v ^P file.txt
because .* is not needed, therefore quoting is not needed and egrep can read data from file.
As we don't use extended regular expressions grep will also work fine
grep -v ^P file.txt
Finally
grep -v ^P file.txt > new.txt
This works:
cat file.txt | egrep -v -e '^P'
-e indicates expression.
I have input.txt
1
2
3
4
5
I need to get such output.txt
1,2,3,4,5
How to do it?
Try this:
tr '\n' ',' < input.txt > output.txt
With sed, you could use:
sed -e 'H;${x;s/\n/,/g;s/^,//;p;};d'
The H appends the pattern space to the hold space (saving the current line in the hold space). The ${...} surrounds actions that apply to the last line only. Those actions are: x swap hold and pattern space; s/\n/,/g substitute embedded newlines with commas; s/^,// delete the leading comma (there's a newline at the start of the hold space); and p print. The d deletes the pattern space - no printing.
You could also use, therefore:
sed -n -e 'H;${x;s/\n/,/g;s/^,//;p;}'
The -n suppresses default printing so the final d is no longer needed.
This solution assumes that the CRLF line endings are the local native line ending (so you are working on DOS) and that sed will therefore generate the local native line ending in the print operation. If you have DOS-format input but want Unix-format (LF only) output, then you have to work a bit harder - but you also need to stipulate this explicitly in the question.
It worked OK for me on MacOS X 10.6.5 with the numbers 1..5, and 1..50, and 1..5000 (23,893 characters in the single line of output); I'm not sure that I'd want to push it any harder than that.
In response to #Jonathan's comment to #eumiro's answer:
tr -s '\r\n' ',' < input.txt | sed -e 's/,$/\n/' > output.txt
tr and sed used be very good but when it comes to file parsing and regex you can't beat perl
(Not sure why people think that sed and tr are closer to shell than perl... )
perl -pe 's/\n/$1,/' your_file
if you want pure shell to do it then look at string matching
${string/#substring/replacement}
Use paste command. Here is using pipes:
echo "1\n2\n3\n4\n5" | paste -s -d, /dev/stdin
Here is using a file:
echo "1\n2\n3\n4\n5" > /tmp/input.txt
paste -s -d, /tmp/input.txt
Per man pages the s concatenates all lines and d allows to define the delimiter character.
Awk versions:
awk '{printf("%s,",$0)}' input.txt
awk 'BEGIN{ORS=","} {print $0}' input.txt
Output - 1,2,3,4,5,
Since you asked for 1,2,3,4,5, as compared to 1,2,3,4,5, (note the comma after 5, most of the solutions above also include the trailing comma), here are two more versions with Awk (with wc and sed) to get rid of the last comma:
i='input.txt'; awk -v c=$(wc -l $i | cut -d' ' -f1) '{printf("%s",$0);if(NR<c){printf(",")}}' $i
awk '{printf("%s,",$0)}' input.txt | sed 's/,\s*$//'
printf "1\n2\n3" | tr '\n' ','
if you want to output that to a file just do
printf "1\n2\n3" | tr '\n' ',' > myFile
if you have the content in a file do
cat myInput.txt | tr '\n' ',' > myOutput.txt
python version:
python -c 'import sys; print(",".join(sys.stdin.read().splitlines()))'
Doesn't have the trailing comma problem (because join works that way), and splitlines splits data on native line endings (and removes them).
cat input.txt | sed -e 's|$|,|' | xargs -i echo "{}"