I have a directory of 40 or so csv's. Each csv file has an extra 10 lines at the top that I don't need. I'm new to bash commands, but I have found that I can use
tail -n +10 oldfile.csv > newfile.csv
to cut 10 lines from a file one at a time. How can I do this across all csv's in the directory? I have tried doing this:
for filename in *foo*; do echo tail -n +10 \"$filename\" > \"${filename}\"; done
From what I've read, I thought this would pass in every csv containing foo in its name, run the formula, and leave the filename alone. Where am I going wrong?
You cannot use the same file as input and ouput.
With sed, your can edit the file in place with the -i flag:
for f in *.csv; do
sed -i '1,10d' "$f"
done
or as one-liner for the command line:
for f in *.csv; do sed -i '1,10d' "$f"; done
As a side note, your tail should be tail -n +11 to output 11th line to end of file.
Use a proper loop as below. Am using the native ex editor which Vim uses internally to in-place replacement, so you don't have to move the files back again using mv or any other command.
for file in *.csv
do
ex -sc '1d10|x' "$file"
done
The command moves to first line, selects 10 lines from first, deletes it and saves & closes the file.
Use a command-line friendly version in a single line as
for file in *.csv; do ex -sc '1d10|x' "$file"; done
The ex command is POSIX compatible and can work on all major platforms and distros.
In awk:
$ awk 'FNR>10{ print > "new-" FILENAME }' files*
Explained:
FNR>10 if current record number in current file is greater than 10, condition is true.
print well, output
> "new-" FILENAME redirect output to a new file named new-file, for example.
Edited to writing output to multiple files. Original which just outputed to screen was awk 'FNR>10' files*
Related
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 $file; done
I found most of the answers not useful for me, If you use vim I found this very easy and straightforward:
:g/<pattern>/d
Source
I am trying to remove 500+ non-consecutive lines from a very large file with sed.
I have the lines stored in a list.txt file but I cant't use it in a for loop
for i in `cat list`; do echo 'sed -i -e ' \'"$i"d\'' huge_file.txt' ; done
because line numbers in the original file would change every time sed removes one and exits.
I should do:
sed -i -e '1d;2d;93572277d;93572278d; ......;nth ' huge_file.txt
Is there a way to pass that list to sed in a file?
you can try with awk:
awk -v s="2,3,..,n" 'BEGIN{n=split(s,t,",");for(i=1;i<=n;i++)d[t[i]]=1}
!d[NR]' huge.txt
You pass the comma-separated line numbers to awk by -v, in awk split it in array, and check each line, if the line number in the array, skip.
Test it with small file, if it worked as you expected, you can do:
awk -v '....' '....' huge.txt > tmp.txt && mv tmp.txt huge.txt
to write the change back to your original input file.
update
If you have 500 line numbers in another file, say, each number in a line, you can:
awk 'NR==FNR{a[$0]=1;next}!a[FNR]' ln.txt huge.txt
If it's just for a single particular task (not frequent) you may use the following GNU sed approach (assuming that numbers in list.txt are separated with newline \n):
sed -i "$(sed -z 's/\n/d;/g' list.txt)" huge_file.txt
So, disclaimer: I am pretty new to using bash and zsh, so there is a chance the answer is really simple. Nonetheless. I checked previous postings and couldn't find anything. (edit: I have tried this in both bash and zsh shells- same problem.)
I have a directory with many files and am trying to remove the first line from each file.
So say the directory contains: file1.txt file2.txt file3.txt ... etc.
I am using the sed command (non-GNU):
sed -i -e "1d" *.txt
For some reason, this is only removing the first line of the first file. I thought that the *.txt would affect all files matching the pattern in directory. Strangely, it is creating the file duplicates with -e appended, but both the duplicate and original are the same.
I tried this with other commands (e.g. ls *.txt) and it works fine. Is there something about sed I am missing?
Thank you in advance.
Different versions of sed in differing operating systems support various parameters.
OpenBSD (5.4) sed
The -i flag is unavailable. You can use the following /bin/sh syntax:
for i in *.txt
do
f=`mktemp -p .`
sed -e "1d" "${i}" > "${f}" && mv -- "${f}" "${i}"
done
FreeBSD (11-CURRENT) sed
The -i flag requires an extension, even if it's empty. Thus must be written as sed -i "" -e "1d" *.txt
GNU sed
This looks to see if the argument following -i is another option (or possibly a command). If so, it assumes an in-place modification. If it appears to be a file extension such as ".bak", it will rename the original with the ".bak" and then modify it into the original file's name.
There might be other variations on other platforms, but those are the three I have at hand.
use it without -e !
for one file use:
sed -i '1d' filename
for all files use :
sed -i '1d' *.txt
or
files=/path/to/files/*.extension ; for var in $files ; do sed -i '1d' $var ; done
.for me i use ubuntu and debian based systems , this method is working for me 100% , but for other platformes i'm not sure , so this is other method :
replace first line with emty pattern , and remove empty lines , (double commands):
for files in $(ls /path/to/files/*.txt); do sed -i "s/$(head -1 "$files")//g" "$files" ; sed -i '/^$/d' "$files" ; done
Note: if your files contain splash '/' , then it will give error , so in this case sed command should look like this ( sed -i "s[$(head -1 "$files")[[g" )
hope that's what you're looking for :)
The issue here is that the line number isn't reset when sed opens a new file, so 1 only matches the first line of the first file.
One solution is to use a shell loop, calling sed once for each file. Gumnos' answer shows how to do this in the most widely compatible way, although if you have a version of sed supporting the -i flag, you could do this instead:
for i in *.txt; do
sed -i.bak '1d' "$i"
done
It is possible to avoid creating the backup file by passing an empty suffix but personally, I don't think it's such a bad thing. One day you'll be grateful for it!
It appears that you're not working with GNU tools but if you were, I would recommend using GNU awk for this task. The variable FNR is useful here, as it keeps track of the record number for each file individually, allowing you to do this:
gawk -i inplace 'FNR>1' *.txt
Using the inplace extension, this allows you to remove the first line from each of your files, by only printing the lines where FNR is greater than 1.
Testing it out:
$ seq 5 > file1
$ seq 5 > file2
$ gawk -i inplace 'FNR>1' file1 file2
$ cat file1
2
3
4
5
$ cat file2
2
3
4
5
The last argument you are passing to the Sed is the problem
try something like this.
var=(`find *txt`)
for file in "${var[#]}"
do
sed -i -e 1d $file
done
This did the trick for me.
I have hundreds of text files in one directory. For all files, I want to delete all the lines that begin with HETATM. I would need a csh or bash code.
I would think you would use grep, but I'm not sure.
Use sed like this:
sed -i -e '/^HETATM/d' *.txt
to process all files in place.
-i means "in place".
-e means to execute the command that follows.
/^HETATM/ means "find lines starting with HETATM", and the following d means "delete".
Make a backup first!
If you really want to do it with grep, you could do this:
#!/bin/bash
for f in *.txt
do
grep -v "^HETATM" "%f" > $$.tmp && mv $$.tmp "$f"
done
It makes a temporary file of the output from grep (in file $$.tmp) and only overwrites your original file if the command executes successfully.
Using the -v option of grep to get all the lines that do not match:
grep -v '^HETATM' input.txt > output.txt
I need to repeatedly remove the first line from a huge text file using a bash script.
Right now I am using sed -i -e "1d" $FILE - but it takes around a minute to do the deletion.
Is there a more efficient way to accomplish this?
Try tail:
tail -n +2 "$FILE"
-n x: Just print the last x lines. tail -n 5 would give you the last 5 lines of the input. The + sign kind of inverts the argument and make tail print anything but the first x-1 lines. tail -n +1 would print the whole file, tail -n +2 everything but the first line, etc.
GNU tail is much faster than sed. tail is also available on BSD and the -n +2 flag is consistent across both tools. Check the FreeBSD or OS X man pages for more.
The BSD version can be much slower than sed, though. I wonder how they managed that; tail should just read a file line by line while sed does pretty complex operations involving interpreting a script, applying regular expressions and the like.
Note: You may be tempted to use
# THIS WILL GIVE YOU AN EMPTY FILE!
tail -n +2 "$FILE" > "$FILE"
but this will give you an empty file. The reason is that the redirection (>) happens before tail is invoked by the shell:
Shell truncates file $FILE
Shell creates a new process for tail
Shell redirects stdout of the tail process to $FILE
tail reads from the now empty $FILE
If you want to remove the first line inside the file, you should use:
tail -n +2 "$FILE" > "$FILE.tmp" && mv "$FILE.tmp" "$FILE"
The && will make sure that the file doesn't get overwritten when there is a problem.
You can use -i to update the file without using '>' operator. The following command will delete the first line from the file and save it to the file (uses a temp file behind the scenes).
sed -i '1d' filename
For those who are on SunOS which is non-GNU, the following code will help:
sed '1d' test.dat > tmp.dat
You can easily do this with:
cat filename | sed 1d > filename_without_first_line
on the command line; or to remove the first line of a file permanently, use the in-place mode of sed with the -i flag:
sed -i 1d <filename>
No, that's about as efficient as you're going to get. You could write a C program which could do the job a little faster (less startup time and processing arguments) but it will probably tend towards the same speed as sed as files get large (and I assume they're large if it's taking a minute).
But your question suffers from the same problem as so many others in that it pre-supposes the solution. If you were to tell us in detail what you're trying to do rather then how, we may be able to suggest a better option.
For example, if this is a file A that some other program B processes, one solution would be to not strip off the first line, but modify program B to process it differently.
Let's say all your programs append to this file A and program B currently reads and processes the first line before deleting it.
You could re-engineer program B so that it didn't try to delete the first line but maintains a persistent (probably file-based) offset into the file A so that, next time it runs, it could seek to that offset, process the line there, and update the offset.
Then, at a quiet time (midnight?), it could do special processing of file A to delete all lines currently processed and set the offset back to 0.
It will certainly be faster for a program to open and seek a file rather than open and rewrite. This discussion assumes you have control over program B, of course. I don't know if that's the case but there may be other possible solutions if you provide further information.
The sponge util avoids the need for juggling a temp file:
tail -n +2 "$FILE" | sponge "$FILE"
If you want to modify the file in place, you could always use the original ed instead of its streaming successor sed:
ed "$FILE" <<<$'1d\nwq\n'
The ed command was the original UNIX text editor, before there were even full-screen terminals, much less graphical workstations. The ex editor, best known as what you're using when typing at the colon prompt in vi, is an extended version of ed, so many of the same commands work. While ed is meant to be used interactively, it can also be used in batch mode by sending a string of commands to it, which is what this solution does.
The sequence <<<$'1d\nwq\n' takes advantage of modern shells' support for here-strings (<<<) and ANSI quotes ($'...') to feed input to the ed command consisting of two lines: 1d, which deletes line 1, and then wq, which writes the file back out to disk and then quits the editing session.
As Pax said, you probably aren't going to get any faster than this. The reason is that there are almost no filesystems that support truncating from the beginning of the file so this is going to be an O(n) operation where n is the size of the file. What you can do much faster though is overwrite the first line with the same number of bytes (maybe with spaces or a comment) which might work for you depending on exactly what you are trying to do (what is that by the way?).
You can edit the files in place: Just use perl's -i flag, like this:
perl -ni -e 'print unless $. == 1' filename.txt
This makes the first line disappear, as you ask. Perl will need to read and copy the entire file, but it arranges for the output to be saved under the name of the original file.
should show the lines except the first line :
cat textfile.txt | tail -n +2
Could use vim to do this:
vim -u NONE +'1d' +'wq!' /tmp/test.txt
This should be faster, since vim won't read whole file when process.
How about using csplit?
man csplit
csplit -k file 1 '{1}'
This one liner will do:
echo "$(tail -n +2 "$FILE")" > "$FILE"
It works, since tail is executed prior to echo and then the file is unlocked, hence no need for a temp file.
Since it sounds like I can't speed up the deletion, I think a good approach might be to process the file in batches like this:
While file1 not empty
file2 = head -n1000 file1
process file2
sed -i -e "1000d" file1
end
The drawback of this is that if the program gets killed in the middle (or if there's some bad sql in there - causing the "process" part to die or lock-up), there will be lines that are either skipped, or processed twice.
(file1 contains lines of sql code)
tail +2 path/to/your/file
works for me, no need to specify the -n flag. For reasons, see Aaron's answer.
You can use the sed command to delete arbitrary lines by line number
# create multi line txt file
echo """1. first
2. second
3. third""" > file.txt
deleting lines and printing to stdout
$ sed '1d' file.txt
2. second
3. third
$ sed '2d' file.txt
1. first
3. third
$ sed '3d' file.txt
1. first
2. second
# delete multi lines
$ sed '1,2d' file.txt
3. third
# delete the last line
sed '$d' file.txt
1. first
2. second
use the -i option to edit the file in-place
$ cat file.txt
1. first
2. second
3. third
$ sed -i '1d' file.txt
$cat file.txt
2. second
3. third
If what you are looking to do is recover after failure, you could just build up a file that has what you've done so far.
if [[ -f $tmpf ]] ; then
rm -f $tmpf
fi
cat $srcf |
while read line ; do
# process line
echo "$line" >> $tmpf
done
Based on 3 other answers, I came up with this syntax that works perfectly in my Mac OSx bash shell:
line=$(head -n1 list.txt && echo "$(tail -n +2 list.txt)" > list.txt)
Test case:
~> printf "Line #%2d\n" {1..3} > list.txt
~> cat list.txt
Line # 1
Line # 2
Line # 3
~> line=$(head -n1 list.txt && echo "$(tail -n +2 list.txt)" > list.txt)
~> echo $line
Line # 1
~> cat list.txt
Line # 2
Line # 3
Would using tail on N-1 lines and directing that into a file, followed by removing the old file, and renaming the new file to the old name do the job?
If i were doing this programatically, i would read through the file, and remember the file offset, after reading each line, so i could seek back to that position to read the file with one less line in it.