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Delete empty lines using sed

I am trying to delete empty lines using sed:

sed '/^$/d'

but I have no luck with it.

For example, I have these lines:

xxxxxx


yyyyyy


zzzzzz

and I want it to be like:

xxxxxx
yyyyyy
zzzzzz

What should be the code for this?

your sed command looks fine, it should work
The above command wouldn't work even if you don't have a space/tab but CR+LF line endings.

K
Kent

You may have spaces or tabs in your "empty" line. Use POSIX classes with sed to remove all lines containing only whitespace:

sed '/^[[:space:]]*$/d'

A shorter version that uses ERE, for example with gnu sed:

sed -r '/^\s*$/d'

(Note that sed does NOT support PCRE.)


@HuStmpHrrr gnu sed doesn't support PCRE at all. it is ERE with -r
OS X needed sed -i "" '/^[[:space:]]*$/d' <filename>,
@BernieReiter ^\s*$ will match all "empty" lines, empty here means, the line contains no chars, or the line contains only empty strings (E.g. spaces). All matched lines will be removed by sed, with the d command.
Perhaps sed '/\S/!d' file
R
Ramy

I am missing the awk solution:

awk 'NF' file

Which would return:

xxxxxx
yyyyyy
zzzzzz

How does this work? Since NF stands for "number of fields", those lines being empty have 0 fields, so that awk evaluates 0 to False and no line is printed; however, if there is at least one field, the evaluation is True and makes awk perform its default action: print the current line.


Whoah. Is even running with BSD's "minimized" version of awk (version 20121220 (FreeBSD). Thanks :-)
@BernieReiter you are welcome :) Yes, this is a very basic idiomatic thing all awk versions allow.
And it is so much quicker although - for a quick and dirty test - I am invoking awk twice: $ time (topic companies <data.tpx | awk 'NF' - | awk -f dialog_menu.awk -) real 0m0.006s user 0m0.000s sys 0m0.008s $ time (topic companies <data.tpx | gsed '/^\s*$/d' | awk -f dialog_menu.awk -) real 0m0.014s user 0m0.002s sys 0m0.006s Would you know of a nifty way to include this into an awk-script like, e.g., a pattern? awk '/mypattern/ {do stuff...}'
Note that this will also ignore lines with whitespace only.
that's so cool, thanks
C
Community

sed '/^$/d' should be fine, are you expecting to modify the file in place? If so you should use the -i flag.

Maybe those lines are not empty, so if that's the case, look at this question Remove empty lines from txtfiles, remove spaces from start and end of line I believe that's what you're trying to achieve.


yes. i am modifying a file. *.csv. how should the -i be placed to the sed command?
sed -i '/^$/d' is one way of doing it.
w
wisbucky

sed

'/^[[:space:]]*$/d'

'/^\s*$/d'

'/^$/d'

-n '/^\s*$/!p'

grep

.

-v '^$'

-v '^\s*$'

-v '^[[:space:]]*$'

awk

/./

'NF'

'length'

'/^[ \t]*$/ {next;} {print}'

'!/^[ \t]*$/'


These show up correctly in your online tool, but [] should not be escaped in a bracket expression, so the code here isn't correct for \[\[:space:\]\] or \[ \t\] - should be [[:space:]] and [ \t].
@BenjaminW. Thanks for catching that. Those were not from the original author, but came from Edit 3 when it was changed from regular text to "code", which then "exposed" the `\` escaping. I have fixed them now.
V
Vadim

I believe this is the easiest and fastest one:

cat file.txt | grep .

If you need to ignore all white-space lines as well then try this:

cat file.txt | grep '\S'

Example:

s="\
\
a\
 b\
\
Below is TAB:\
    \
Below is space:\
 \
c\
\
"; echo "$s" | grep . | wc -l; echo "$s" | grep '\S' | wc -l

outputs

7
5

No need for cat, grep takes files as well: grep . file.txt
Yes, I know, but the initial question did not mention whether the source is a file or something else, so the solution is what comes after "|", and before it just an example of a source. Simply to distinguish the solution from the source of lines.
grep '\S' is definitely not portable. If you have grep -P then you can use grep -P '\S' but it's not supported on all platforms, either.
The downside of grep . compared to the other solutions is that it will highlight all the text in red. The other solutions can preserve the original colors. Compare unbuffer apt search foo | grep . to unbuffer apt search foo | grep -v ^$
C
Community

With help from the accepted answer here and the accepted answer above, I have used:

$ sed 's/^ *//; s/ *$//; /^$/d; /^\s*$/d' file.txt > output.txt

`s/^ *//`  => left trim
`s/ *$//`  => right trim
`/^$/d`    => remove empty line
`/^\s*$/d` => delete lines which may contain white space

This covers all the bases and works perfectly for my needs. Kudos to the original posters @Kent and @kev


u
user319660

Another option without sed, awk, perl, etc

strings $file > $output

strings - print the strings of printable characters in files.


Do you mean strings instead of string ?
"For each file given, GNU strings prints the printable character sequences that are at least 4 characters long..." so very short lines might give you a surprise if you're unaware of this. There is a --bytes=min-len option to allow shorter lines.
f
fedorqui

You can say:

sed -n '/ / p' filename    #there is a space between '//'

.. which means print all lines except the empty one(s)and be quiet
S
Samuel Kenneth

The command you are trying is correct, just use -E flag with it.

sed -E '/^$/d'

-E flag makes sed catch extended regular expressions. More info here


There is nothing in this particular regex which requires the -E flag.
B
Benjamin W.

You are most likely seeing the unexpected behavior because your text file was created on Windows, so the end of line sequence is \r\n. You can use dos2unix to convert it to a UNIX style text file before running sed or use

sed -r "/^\r?$/d"

to remove blank lines whether or not the carriage return is there.


Hi, what is the -r flag doing and is it possible to combine it with -i to modify the file directly and avoid printing to screen. In addition, I think that this command would also work as sed -r "/^\r$/d"
C
Claes Wikner

This works in awk as well.

awk '!/^$/' file
xxxxxx
yyyyyy
zzzzzz

L
Lowbit

You can do something like that using "grep", too:

egrep -v "^$" file.txt

j
justincbagley

My bash-specific answer is to recommend using perl substitution operator with the global pattern g flag for this, as follows:

$ perl -pe s'/^\n|^[\ ]*\n//g' $file
xxxxxx
yyyyyy
zzzzzz

This answer illustrates accounting for whether or not the empty lines have spaces in them ([\ ]*), as well as using | to separate multiple search terms/fields. Tested on macOS High Sierra and CentOS 6/7.

FYI, the OP's original code sed '/^$/d' $file works just fine in bash Terminal on macOS High Sierra and CentOS 6/7 Linux at a high-performance supercomputing cluster.


K
Kpym

If you want to use modern Rust tools, you can consider:

ripgrep: cat datafile | rg '.' line with spaces is considered non empty cat datafile | rg '\S' line with spaces is considered empty rg '\S' datafile line with spaces is considered empty (-N can be added to remove line numbers for on screen display)

cat datafile | rg '.' line with spaces is considered non empty

cat datafile | rg '\S' line with spaces is considered empty

rg '\S' datafile line with spaces is considered empty (-N can be added to remove line numbers for on screen display)

sd cat datafile | sd '^\n' '' line with spaces is considered non empty cat datafile | sd '^\s*\n' '' line with spaces is considered empty sd '^\s*\n' '' datafile inplace edit

cat datafile | sd '^\n' '' line with spaces is considered non empty

cat datafile | sd '^\s*\n' '' line with spaces is considered empty

sd '^\s*\n' '' datafile inplace edit


N
Niket Srivastav

NF is the command of awk you can use to delete empty lines in a file awk NF filename and by using sed sed -r "/^\r?$/d"


Both these answers were already given years ago.
f
fedorqui

For me with FreeBSD 10.1 with sed worked only this solution:

sed -e '/^[     ]*$/d' "testfile"

inside [] there are space and tab symbols.

test file contains:

fffffff next 1 tabline ffffffffffff

ffffffff next 1 Space line ffffffffffff

ffffffff empty 1 lines ffffffffffff

============ EOF =============

N
Nilesh Shukla

Using vim editor to remove empty lines

:%s/^$\n//g