Delete all rows which does not start with at least 3 digit.
I use below sed command but it removes title as well.
How can I start below command from line 2, so that title remains as it is.
sed -n '/^[0-9]\{3\}/p' my_file
I have used below as well but does not work.
sed -n '2,${/^[0-9]\{3\}/p}' my_file
You're almost there. All you need to do is:
sed -e '2,${/^[0-9]\{3\}/d}' my_file
This will begin executing the script beginning from the second line (excludes the first one, which I assume it's the title you're talking about).
Using awk:
$ cat foo
Title
11 test
222 test
test
$ awk '/^[0-9]{3}/ || NR==1' foo
Title
222 test
Edit: Title is first row. So this command should start from row number 2.
$ awk '/^[0-9]{3}/ || NR>1' foo
222 test
This might work for you (GNU sed):
sed -i '1b;/^[0-9]\{3\}/!d' file
Do not process the first line. If any line thereafter does not begin with at least 3 digits, delete it.
N.B. -i option replaces the contents of the old file with the amended file. If multiple files are placed following the sed commands, they will be treated separately i.e. each files header will be remain. To achieve the same effect without over writing the files use the -s option.
Another way:
sed '/^[0-9]\{3\}/b;1!d' file
For portability, try:
sed -n '1p;2,${/^[0-9][0-9][0-9]/p}' my_file
Or sed -in '1p;2,${/^[0-9][0-9][0-9]/p}' my_file to edit my_file directly.
Related
how to remove comment lines (as # bal bla ) and empty lines (lines without charecters) from file with one sed command?
THX
lidia
If you're worried about starting two sed processes in a pipeline for performance reasons, you probably shouldn't be, it's still very efficient. But based on your comment that you want to do in-place editing, you can still do that with distinct commands (sed commands rather than invocations of sed itself).
You can either use multiple -e arguments or separate commands with a semicolon, something like (just one of these, not both):
sed -i 's/#.*$//' -e '/^$/d' fileName
sed -i 's/#.*$//;/^$/d' fileName
The following transcript shows this in action:
pax> printf 'Line # with a comment\n\n# Line with only a comment\n' >file
pax> cat file
Line # with a comment
# Line with only a comment
pax> cp file filex ; sed -i 's/#.*$//;/^$/d' filex ; cat filex
Line
pax> cp file filex ; sed -i -e 's/#.*$//' -e '/^$/d' filex ; cat filex
Line
Note how the file is modified in-place even with two -e options. You can see that both commands are executed on each line. The line with a comment first has the comment removed then all is removed because it's empty.
In addition, the original empty line is also removed.
#paxdiablo has a good answer but it can be improved.
(1) The '/^$/d' clause only matches 100% blank lines.
If you want to also match lines that are entirely whitespace (spaces, tabs etc.) use this instead:
'/^\s*$/d'
(2) The 's/#.*$//' clause only matches lines that start with the # character in column 0.
If you want to also match lines that have only whitespace before the first # use this instead:
'/^\s*#.*$/d'
The above criteria may not be universal (e.g. within a HEREDOC block, or in a Python multi-line string the different approaches could be significant), but in many cases the conventional definition of "blank" lines include whitespace-only, and "comment" lines include whitespace-then-#.
(3) Lastly, on OSX at least, the #paxdiablo solution in which the first clause turns comment lines into blank lines, and the second clause strips blank lines (including what were originally comments) doesn't work. It seems to be more portable to make both clauses /d delete actions as I've done.
The revised command incorporating the above is:
sed -e '/^\s*#.*$/d' -e '/^\s*$/d' inputFile
This tiny jewel removes all # comments, no matter where they begin in a line (see caution below):
sed -e 's/\s*#.*$//'
Example:
text="
this is a # test
#this is a test
#this is a #test
this is # another #test
"
$echo "$text" | sed -e 's/\s*#.*$//'
this is a
this is
Next this removes any resulting blank lines:
$echo "$text" | sed -e 's/\s*#.*$//' | sed -e '/^\s*$/d'
Caution: Depending on the syntax and/or interpretation of the lines your processing, this might not be an appropriate solution, as it just stupidly removes end of lines, even if the '#' is part of your data or code. However, for use cases where you'll never use a hash except for as an end of line comment then it works fine. So just as with all coding, context must be taken into consideration.
Alternative variant, using grep:
cat file.txt | grep -Ev '(#.*$)|(^$)'
you can use awk
awk 'NF{gsub(/^[ \t]*#/,"");print}' file
First example(paxdiablo) is very good except its not change file, just output result. If you want to change it inline:
sudo sed -i 's/#.*$//;/^$/d' inputFile
On (one of) my linux boxes, sed understands extended regular expressions with the -r option, so:
sed -r '/(^\s*#)|(^\s*$)/d' squid.conf.installed
is very useful for showing all non-blank, non comment lines.
The regex matches either start of line followed by zero or more spaces or tabs followed by either a hash or end of line, and deletes those matching lines from the input.
I have a file, which has multiple lines.
For example:
a
ab#
ad.
a12fs
b
c
...
I want to use sed or awk delete the line, if the line include symbols or numbers. (For example, I want to delete: ab#, ad., a12fs.... lines)
or in another words, I just want to keep the line which include [a-z][A-Z] .
I know how to delete number line,
sed '/[0-9]/d' file.txt
but I do not know how to delete symbols lines.
Or there has any easy way to do that?
To keep blank lines:
grep '^[[:alpha:]]*$' file
sed '/[^[:alpha:]]/d' file
awk '/^[[:alpha:]]*$/' file
To remove blank lines:
grep '^[[:alpha:]]+$' file
sed -E -n '/^[[:alpha:]]+$/p' file
awk '/^[[:alpha:]]+$/' file
grep works well too and is even simpler: just do the reverse: keep the lines that interest you, which are way easier to define
grep -i '^[a-z]*$' file.txt
(match lines containing only letters and empty lines, and -i option makes grep case-insensitive)
to remove empty lines as well:
grep -i '^[a-z]+$' file.txt
caution when using Windows text files, as there's a carriage return at the end of the line, so nothing would match depending on grep versions (tested on windows here and it works)
but just in case:
grep -iP '^[a-z]*\r?$'
(note the P option to enable perl expressions or \r is not recognized)
You can use this sed:
sed '/^[A-Za-z0-9]\+$/!d' file
(OR)
sed '/[^A-Za-z0-9]/d' file
$ awk '!/[^[:alpha:]]/' file.txt
a
b
c
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.
as title said, Im trying to change only the first occurrence of word.By using
sed 's/this/that/' file.txt
though i'm not using g option it replace entire file. How to fix this.?
UPDATE:
$ cat file.txt
first line
this
this
this
this
$ sed -e '1s/this/that/;t' file.txt
first line
this // ------> I want to change only this "this" to "that" :)
this
this
this
http://www.faqs.org/faqs/editor-faq/sed/
4.3. How do I change only the first occurrence of a pattern?
sed -e '1s/LHS/RHS/;t' -e '1,/LHS/s//RHS/'
Where LHS=this and RHS=that for your example.
If you know the pattern won't occur on the first line, omit the first -e and the statement following it.
sed by itself applies the edit thru out the file and combined with "g" flag the edit is applied to all the occurrences on the same line.
e.g.
$ cat file.txt
first line
this this
this
this
this
$ sed 's/this/that/' file.txt
first line
that this
that
that
that
$ sed 's/this/that/g' file.txt
first line
that that <-- Both occurrences of "this" have changed
that
that
that
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.