I was wondering if there was a way to delete everything after a certain line of a text file in bash. So say there's a text file with 10 lines, and I want to delete every line after line number 4, so only the first 4 lines remained, how would I go about doing that?
You can use GNU sed:
sed -i '5,$d' file.txt
That is, 5,$ means the range line 5 until the end, and d means to delete.
Only the first 4 lines will remain.
The -i flag tells sed to edit the file in-place.
If you have only BSD sed, then the -i flag requires a backup file suffix:
sed -i.bak '5,$d' file.txt
As #ephemient pointed out, while this solution is simple,
it's inefficient because sed will still read the input until the end of the file, which is unnecessary.
As #agc pointed out, the inverse logic of my first proposal might be actually more intuitive. That is, do not print by default (-n flag),
and explicitly print range 1,4:
sed -ni.bak 1,4p file.txt
Another simple alternative, assuming that the first 4 lines are not excessively long and so they easily fit in memory, and also assuming that the 4th line ends with a newline character,
you can read the first 4 lines into memory and then overwrite the file:
lines=$(head -n 4 file.txt)
echo "$lines" > file.txt
Minor refinements on Janos' answer, ephemient's answer, and cdark's comment:
Simpler (and faster) sed code:
sed -i 4q file
When a filter util can't directly edit a file, there's
sponge:
head -4 file | sponge file
Most efficient for Linux might be truncate -- coreutils sibling util to fallocate, which offers the same minimal I/O of ephemient's more portable, (but more complex), dd-based answer:
truncate -s `head -4 file | wc -c` file
The sed method that #janos is simple but inefficient. It will read every line from the original file, even ones it could ignore (although that can be fixed using 4q), and -i actually creates a new file (which it renames to replace the original file). And there's the annoying bit where you need to use sed -i '5,$d' file.txt with GNU sed but sed -i '' '5,$d' file.txt with BSD sed in order to remove the existing file instead of leaving a backup.
Another method that performs less I/O:
dd bs=1 count=0 if=/dev/null of=file.txt \
seek=$(grep -b ^ file.txt | tail -n+5 | head -n1 | cut -d: -f1)
grep -b ^ file.txt prints out byte offsets on each line, e.g.
$ yes | grep -b ^
0:y
2:y
4:y
...
tail -n+5 skips the first 4 lines, outputting the 5th and subsequent lines
head -n1 takes only the next line (e.g. only the 5th line)
After head reads the one line, it will exit. This causes tail to exit because it has nowhere to output to anymore. This causes grep to exit for the same reason. Thus, the rest of file.txt does not need to be examined.
cut -d: -f1 takes only the first part before the : (the byte offset)
dd bs=1 count=0 if=/dev/null of=file.txt seek=N
using a block size of 1 byte, seek to block N of file.txt
copy 0 blocks of size 1 byte from /dev/null to file.txt
truncate file.txt here (because conv=notrunc was not given)
In short, this removes all data on the 5th and subsequent lines from file.txt.
On Linux there is a command named fallocate which can similarly extend or truncate a file, but that's not portable.
UNIX filesystems support efficiently truncating files in-place, and these commands are portable. The downside is that it's more work to write out.
(Also, dd will print some unnecessary stats to stderr, and will exit with an error if the file has fewer than 5 lines, although in that case it will leave the existing file contents in place, so the behavior is still correct. Those can be addressed also, if needed.)
If I don't know the line number, merely the line content (I need to know that there is nothing below the line containing 'knowntext' that I want to preserve.), then I use.
sed -i '/knowntext/,$d' inputfilename
to directly alter the file, or to be cautious
sed '/knowntext/,$d' inputfilename > outputfilename
where inputfilename is unaltered, and outputfilename contains the truncated version of the input.
I am not competent to comment on the efficiency of this, but I know that files of 20kB or so are dealt with faster than I can blink.
Using GNU awk (v. 4.1.0+, see here). First we create a test file (NOTICE THE DISCLAIMER):
$ seq 1 10 > file # THIS WILL OVERWRITE FILE NAMED file WITH TEST DATA
Then the code and validation (WILL MODIFY THE ORIGINAL FILE NAMED file):
$ awk -i inplace 'NR<=4' file
$ cat file
1
2
3
4
Explained:
$ awk -i inplace ' # edit is targetted to the original file (try without -i ...)
NR<=4 # output first 4 records
' file # file
You could also exit on line NR==5 which would be quicker if you redirected the output of the program to a new file (remove # for action) which would be the same as head -4 file > new_file:
$ awk 'NR==5{exit}1' file # > new_file
When testing, don't forget the seq part first.
Related
I have a small script which basically generates a menu of all the scripts in my ~/scripts folder and next to each of them displays a sentence describing it, that sentence being the third line within the script commented out. I then plan to pipe this into fzf or dmenu to select it and start editing it or whatever.
1 #!/bin/bash
2
3 # a script to do
So it would look something like this
foo.sh a script to do X
bar.sh a script to do Y
Currently I have it run a for loop over all the files in the scripts folder and then run sed -n 3p on all of them.
for i in $(ls -1 ~/scripts); do
echo -n "$i"
sed -n 3p "~/scripts/$i"
echo
done | column -t -s '#' | ...
I was wondering if there is a more efficient way of doing this that did not involve a for loop and only used sed. Any help will be appreciated. Thanks!
Instead of a loop that is parsing ls output + sed, you may try this awk command:
awk 'FNR == 3 {
f = FILENAME; sub(/^.*\//, "", f); print f, $0; nextfile
}' ~/scripts/* | column -t -s '#' | ...
Yes there is a more efficient way, but no, it doesn't only use sed. This is probably a silly optimization for your use case though, but it may be worthwhile nonetheless.
The inefficiency is that you're using ls to read the directory and then parse its output. For large directories, that causes lots of overhead for keeping that list in memory even though you only traverse it once. Also, it's not done correctly, consider filenames with special characters that the shell interprets.
The more efficient way is to use find in combination with its -exec option, which starts a second program with each found file in turn.
BTW: If you didn't rely on line numbers but maybe a tag to mark the description, you could also use grep -r, which avoids an additional process per file altogether.
This might work for you (GNU sed):
sed -sn '1h;3{H;g;s/\n/ /p}' ~/scripts/*
Use the -s option to reset the line number addresses for each file.
Copy line 1 to the hold space.
Append line 3 to the hold space.
Swap the hold space for the pattern space.
Replace the newline with a space and print the result.
All files in the directory ~/scripts will be processed.
N.B. You may wish to replace the space delimiter by a tab or pipe the results to the column command.
I would like to extract the first line from a file, read into a variable and delete right afterwards, with a single command. I know sed can read the first line as follows:
sed '1q' file.txt
or delete it as follows:
sed '1q;d' file.txt
but can I somehow do both with a single command?
The reason for this is that multiple processes will be reading the first line of the file, and I want to minimize the chances of them getting the same line.
It's impossible.
Except you read the manpage, and have Gnu-sed:
echo -e {1..3}"\n" > input
cat input
1
2
3
sed -n '1p;2,$ Woutput' input
1
cat output
2
3
Explanation:
sed -n '1p;2,$ Woutput' input
-n no output by default
1p; print line 1
2,$ from line 2 until $ last line
W (non posix) Write buffer to file
From the man page gnu sed:
w filename
Write the current pattern space to filename.
W filename
Write the first line of the current pattern space to filename. This is a GNU extension.
However, reading and experimenting takes longer, than opening the file in a full blown office suite and deleting the line by hand, or invoking a text-to-speech framework and training it, to do the job.
It doesn't work if invoked in posix style:
sed -n --posix '1p;2,$ Woutput' input
And you still have the hard hanwork of renaming output to input again.
I didn't try to write to input in place, because that could damage my carefully crafted input file - try it on own risk:
sed -n '1p;2,$ Winput' input
However, you might set up a filesystem notify job, which always rename freshly created output files to input again. But I fear you can't do it from within the sed command. Except ... (to be continued)
I am working with plotting extremely large files with N number of relevant data entries. (N varies between files).
In each of these files, comments are automatically generated at the start and end of the file and would like to filter these out before recombining them into one grand data set.
Unfortunately, I am using MacOSx, where I encounter some issues when trying to remove the last line of the file. I have read that the most efficient way was to use head/tail bash commands to cut off sections of data. Since head -n -1 does not work for MacOSx I had to install coreutils through homebrew where the ghead command works wonderfully. However the command,
tail -n+9 $COUNTER/test.csv | ghead -n -1 $COUNTER/test.csv >> gfinal.csv
does not work. A less than pleasing workaround was I had to separate the commands, use ghead > newfile, then use tail on newfile > gfinal. Unfortunately, this will take while as I have to write a new file with the first ghead.
Is there a workaround to incorporating both GNU Utils with the standard Mac Utils?
Thanks,
Keven
The problem with your command is that you specify the file operand again for the ghead command, instead of letting it take its input from stdin, via the pipe; this causes ghead to ignore stdin input, so the first pipe segment is effectively ignored; simply omit the file operand for the ghead command:
tail -n+9 "$COUNTER/test.csv" | ghead -n -1 >> gfinal.csv
That said, if you only want to drop the last line, there's no need for GNU head - OS X's own BSD sed will do:
tail -n +9 "$COUNTER/test.csv" | sed '$d' >> gfinal.csv
$ matches the last line, and d deletes it (meaning it won't be output).
Finally, as #ghoti points out in a comment, you could do it all using sed:
sed -n '9,$ {$!p;}' file
Option -n tells sed to only produce output when explicitly requested; 9,$ matches everything from line 9 through (,) the end of the file (the last line, $), and {$!p;} prints (p) every line in that range, except (!) the last ($).
I realize that your question is about using head and tail, but I'll answer as if you're interested in solving the original problem rather than figuring out how to use those particular tools to solve the problem. :)
One method using sed:
sed -e '1,8d;$d' inputfile
At this level of simplicity, GNU sed and BSD sed both work the same way. Our sed script says:
1,8d - delete lines 1 through 8,
$d - delete the last line.
If you decide to generate a sed script like this on-the-fly, beware of your quoting; you will have to escape the dollar sign if you put it in double quotes.
Another method using awk:
awk 'NR>9{print last} NR>1{last=$0}' inputfile
This works a bit differently in order to "recognize" the last line, capturing the previous line and printing after line 8, and then NOT printing the final line.
This awk solution is a bit of a hack, and like the sed solution, relies on the fact that you only want to strip ONE final line of the file.
If you want to strip more lines than one off the bottom of the file, you'd probably want to maintain an array that would function sort of as a buffered FIFO or sliding window.
awk -v striptop=8 -v stripbottom=3 '
{ last[NR]=$0; }
NR > striptop*2 { print last[NR-striptop]; }
{ delete last[NR-striptop]; }
END { for(r in last){if(r<NR-stripbottom+1) print last[r];} }
' inputfile
You specify how much to strip in variables. The last array keeps a number of lines in memory, prints from the far end of the stack, and deletes them as they are printed. The END section steps through whatever remains in the array, and prints everything not prohibited by stripbottom.
How to get the first few lines from a gziped file ?
I tried zcat, but its throwing an error
zcat CONN.20111109.0057.gz|head
CONN.20111109.0057.gz.Z: A file or directory in the path name does not exist.
zcat(1) can be supplied by either compress(1) or by gzip(1). On your system, it appears to be compress(1) -- it is looking for a file with a .Z extension.
Switch to gzip -cd in place of zcat and your command should work fine:
gzip -cd CONN.20111109.0057.gz | head
Explanation
-c --stdout --to-stdout
Write output on standard output; keep original files unchanged. If there are several input files, the output consists of a sequence of independently compressed members. To obtain better compression, concatenate all input files before compressing
them.
-d --decompress --uncompress
Decompress.
On some systems (e.g., Mac), you need to use gzcat.
On a mac you need to use the < with zcat:
zcat < CONN.20111109.0057.gz|head
If a continuous range of lines needs be, one option might be:
gunzip -c file.gz | sed -n '5,10p;11q' > subFile
where the lines between 5th and 10th lines (both inclusive) of file.gz are extracted into a new subFile. For sed options, refer to the manual.
If every, say, 5th line is required:
gunzip -c file.gz | sed -n '1~5p;6q' > subFile
which extracts the 1st line and jumps over 4 lines and picks the 5th line and so on.
If you want to use zcat, this will show the first 10 rows
zcat your_filename.gz | head
Let's say you want the 16 first row
zcat your_filename.gz | head -n 16
This awk snippet will let you show not only the first few lines - but a range you can specify. It will also add line numbers which i needed for debugging an error message pointing to a certain line way down in a gzipped file.
gunzip -c file.gz | awk -v from=10 -v to=20 'NR>=from { print NR,$0; if (NR>=to) exit 1}'
Here is the awk snippet used in the one liner above. In awk NR is a built-in variable (Number of records found so far) which usually is equivalent to a line number. the from and to variable are picked up from the command line via the -v options.
NR>=from {
print NR,$0;
if (NR>=to)
exit 1
}
I have a file with 1 line of text, called output. I have write access to the file. I can change it from an editor with no problems.
$ cat output
1
$ ls -l o*
-rw-rw-r-- 1 jbk jbk 2 Jan 27 18:44 output
What I want to do is replace the first (and only) line in this file with a new value, either a 1 or a 0. It seems to me that sed should be perfect for this:
$ sed '1 c\ 0' output
0
$ cat output
1
But it never changes the file. I've tried it spread over 2 lines at the backslash, and with double quotes, but I cannot get it to put a 0 (or anything else) in the first line.
Sed operates on streams and prints its output to standard out.
It does not modify the input file.
It's typically used like this when you want to capture its output in a file:
#
# replace every occurrence of foo with bar in input-file
#
sed 's/foo/bar/g' input-file > output-file
The above command invokes sed on input-file and redirects the output to a new file named output-file.
Depending on your platform, you might be able to use sed's -i option to modify files in place:
sed -i.bak 's/foo/bar/g' input-file
NOTE:
Not all versions of sed support -i.
Also, different versions of sed implement -i differently.
On some platforms you MUST specify a backup extension (on others you don't have to).
Since this is an incredibly simple file, sed may actually be overkill. It sounds like you want the file to have exactly one character: a '0' or a '1'.
It may make better sense in this case to just overwrite the file rather than to edit it, e.g.:
echo "1" > output
or
echo "0" > output