I have a file that looks
a
b
c
d
Suppose I want to add N lines (in the example 3, but I actually need 20 or 100 depending on the file)
a
b
c
d
I can add one blank line between all of them with sed
sed -i '0~1 a\\' file
But sed -i '0~3 a\\' file inserts one line every 3 rows.
You may use with GNU sed:
sed -i 'G;G;G' file
The G;G;G will append three empty lines below each non-final line.
Or, awk:
awk 'BEGIN{ORS="\n\n\n"};1'
See an online sed and awk demo.
If you need to set the number of newlines dynamically use
nl="
"
awk -v nl="$nl" 'BEGIN{for(c=0;c<3;c++) v=v""nl;ORS=v};1' file > newfile
With GNU awk:
awk -i inplace -v lines=3 '{print; for(i=0;i<lines;i++) print ""}' file
Update with Ed's hints (see comments):
awk -i inplace -v lines=3 '{print; for(i=1;i<=lines;i++) print ""}' file
Update (without trailing empty lines):
awk -i inplace -v lines=3 'NR==1; NR>1{for(i=1;i<=lines;i++) print ""; print}' file
Output to file:
a
b
c
d
With sed and corutils:
N=4
sed "\$b;$(yes G\; | head -n$N)" infile
Similar trick with awk:
N=4
awk 1 RS="$(yes \\n | head -n$N | tr -d '\n')" infile
This might work for you (GNU sed):
sed ':a;G;s/\n/&/2;Ta' file
This will add 2 blank lines following each line.
Change 2 to what ever number you desire between each line.
An alternative (more efficient?):
sed '1{x;:a;/^.\{2\}/!s/^/\n/;ta;s/.//;x};G' file
I have csv files that need to be changed f -> 0 and t -> 1 only between commas for every single csv if it matches. From:
,t,t,f,f,a,t,f,t,f,f,t,f,
tftf
to:
,1,1,0,0,a,1,0,1,0,0,1,0,
tftf
Works this way, but want to know better way that could reduce the replacing time consume
for i in 1 2 3 4 5 6
do
echo "converting tables for mariaDB"
find ./ -type f -name "*.csv" -print0 | xargs -0 sed -i 's/\,t\,/\,1\,/g'
find ./ -type f -name "*.csv" -print0 | xargs -0 sed -i 's/\,f\,/\,0\,/g'
echo "$i time(s) changed "
done
I except , one single command will change the line
Could you please try following. Though it is not perfect solution but would be simplest use it in case you don't have gawk's latest version where -inplace edit option is present.
for file in *.csv
awk '{gsub(/,t,/,",1,");gsub(/,f,/,",0,");gsub(/,t,/,",1,");gsub(/,f,/,",0,")} 1' "$file" > temp && mv temp"$file"
done
OR
for file in *.csv
awk -v t_val="1" -v f_val="0" 'BEGIN{FS=OFS=","}{for(i=2;i<NF;i++){$i=($i=="t"?t_val:$i=="f"?f_val:$i)}} 1' "$file" > temp && mv temp "$file"
done
2nd solution: Using gawk's latest version where we could save edit into Input_file itself.
gawk -i inplace '{gsub(/,t,/,",1,");gsub(/,f,/,",0,");gsub(/,t,/,",1,");gsub(/,f,/,",0,")} 1' *.csv
OR
gawk -i inplace -v t_val="1" -v f_val="0" 'BEGIN{FS=OFS=","}{for(i=2;i<NF;i++){$i=($i=="t"?t_val:$i=="f"?f_val:$i)}} 1' Input_file
The main problem, in this case, is that a regular expression does not allow overlap when parsing it with sed 's/ere/str/g' or awk '{gsub(ere,str,$0)}'. This comment nicely explains how you can circumvent this in sed using the t<label> command, which means: if a change happened to the pattern space, move to <label>. The comment shows a generic way of doing it. The awk alternative to this rule would be:
$ awk '{while(match($0,ere)) gsub(ere,str)}'
An alternative sed solution in the case of the OP's example could use the following idea:
duplicate all commas. Since we are searching for strings of the form ",t,", this duplication avoid overlap using s.
since no overlap is possible, replace all ",f," with ",0," and all ",t," with ",1,".
We can now revert all duplicated commas again. As no overlap is allowed, sequences like ,,,, will be nicely converted to ,, and not ,
In POSIX sed this looks like:
$ sed -e 's/,/,,/g' -e 's/,f,/,0,/g' \
-e 's/,t,/,1,/g' -e 's/,,/,/g' file > file.tmp
$ mv file.tmp file
With GNU sed we can do it in one go:
$ sed -i 's/,/,,/g;s/,f,/,0,/g;s/,t,/,1,/g;s/,,/,/g' file
With awk, this would look like:
$ awk 'BEGIN{FS=",";OFS=FS FS}
{$1=$1;gsub(/,f,/,",0,");gsub(/,t,/,",1,");gsub(OFS,FS)}1' file > file.tmp
$ mv file.tmp file
I have a text file with the following structure:
bla1
bla2
bla3
bla4
bla5
So you can see that some lines of text are preceeded by an empty line.
I understand that sed has the concept of two buffers, a pattern space buffer and a hold space buffer, so I'm guessing these need to come in to play here, but I'm unclear how to specify them to accomplish what I need.
In my contrived example above, I'd expect to see the following lines outputted:
bla3
bla5
sed is for doing s/old/new on individual lines, that is all. Any time you start talking about buffers or doing anything related to multi-lines comparisons you're using the wrong tool.
You could do this with awk:
$ awk -v RS= -F'\n' 'NR>1{print $1}' file
bla3
bla5
but it would fail to print the first non-empty line if the first line(s) in the file were empty so this may be what you want if you want lines of all space chars considered to be empty lines:
$ awk 'NF && !p{print} {p=NF}' file
bla3
bla5
and this otherwise:
$ awk '($0!="") && (p==""){print} {p=$0}' file
bla3
bla5
All of the above will work even if there are multiple empty lines preceding any given non-empty line.
To see the difference between the 3 approaches (which you won't see given the sample input in the question):
PS1> printf '\nfoo\n \nbar\n\netc\n' | cat -E
$
foo$
$
bar$
$
etc$
PS1> printf '\nfoo\n \nbar\n\netc\n' | awk -v RS= -F'\n' 'NR>1{print $1}'
etc
PS1> printf '\nfoo\n \nbar\n\netc\n' | awk 'NF && !p{print} {p=NF}'
foo
bar
etc
PS1> printf '\nfoo\n \nbar\n\netc\n' | awk '($0!="") && (p==""){print} {p=$0}'
foo
etc
You can use the hold buffer easily to print the line before the blank like this:
sed -n -e '/^$/{x; p;}' -e h input
But I don't see an easy way to use it for your use case. For your case, instead of using the hold buffer, you could do:
sed -n -e '/^$/ba' -e d -e :a -e n -e p input
But I would do this with awk.
awk 'NR!=1{print $1}' RS= FS=\\n input-file
awk 'p;{p=/^$/}' file
above command does these for each line:
if p is 1, print line;
if line is empty, set p to 1.
if lines consisting of one or more spaces are also considered empty:
awk 'p;{p=!NF}' file
to print non-empty lines each coming right after an empty line, you can use this:
awk 'p*!(p=/^$/)' file
if p is 1 and this line is not empty (1*!(0) = 1*1 = 1), print this line;
otherwise (1*!(1) = 1*0 = 0, 0*anything = 0), don't print anything.
note that this one may not work with all awks, a portable version of this would look like:
awk 'p*(/./);{p=/^$/}' file
if lines consisting of one or more spaces are also considered empty:
awk 'p*NF;{p=!NF}' file
see them online here, and here.
If sed/awk is not mandatory, you can do it with grep:
grep -A 1 '^$' input.txt | grep -v -E '^$|--'
You can use sed to match a range of lines and do sub-matches inside the matches, like so:
# - use the "-n" option to omit printing of lines
# - match lines between a blank line (/^$/) and a non-blank one (/^./),
# then print only the line that contains at least a character,
# i.e, the non-blank line.
sed -ne '
/^$/,/^./ {
/^./{ p; }
}' input.txt
tested by gnu sed, your data in 'a':
$ sed -nE '/^$/{N;s/\n(.+)/\1/p}' a
bla3
bla5
add -i option precedes -n to real editing
I have a csv file in which every other line is blank. I have tried everything, nothing removes the lines. What should make it easier is that the the digits 44 appear in each valid line. Things I have tried:
grep -ir 44 file.csv
sed '/^$/d' <file.csv
cat -A file.csv
sed 's/^ *//; s/ *$//; /^$/d' <file.csv
egrep -v "^$" file.csv
awk 'NF' file.csv
grep '\S' file.csv
sed 's/^ *//; s/ *$//; /^$/d; /^\s*$/d' <file.csv
cat file.csv | tr -s \n
Decided I was imagining the blank lines, but import into Google Sheets and there they are still! Starting to question my sanity! Can anyone help?
sed -n -i '/44/p' file
-n means skip printing
-i inplace (overwrite same file)
- /44/p print lines where '44' exists
without '44' present
sed -i '/^\s*$/d' file
\s is matching whitespace, ^startofline, $endofline, d delete line
Use the -i option to replace the original file with the edited one.
sed -i '/^[ \t]*$/d' file.csv
Alternatively output to another file and rename it, which is doing the exactly what -i does.
sed '/^[[:blank:]]*$/d' file.csv > file.csv.out && mv file.csv.out file.csv
Given:
$ cat bl.txt
Line 1 (next line has a tab)
Line 2 (next has several space)
Line 3
You can remove blank lines with Perl:
$ perl -lne 'print unless /^\s*$/' bl.txt
Line 1 (next line has a tab)
Line 2 (next has several space)
Line 3
awk:
$ awk 'NF>0' bl.txt
Line 1 (next line has a tab)
Line 2 (next has several space)
Line 3
sed + tr:
$ cat bl.txt | tr '\t' ' ' | sed '/^ *$/d'
Line 1 (next line has a tab)
Line 2 (next has several space)
Line 3
Just sed:
$ sed '/^[[:space:]]*$/d' bl.txt
Line 1 (next line has a tab)
Line 2 (next has several space)
Line 3
Aside from the fact that your commands do not show that you capture their output in a new file to be used in place of the original, there's nothing wrong with them, EXCEPT that:
cat file.csv | tr -s \n
should be:
cat file.csv | tr -s '\n' # more efficient alternative: tr -s '\n' < file.csv
Otherwise, the shell eats the \ and all that tr sees is n.
Note, however, that the above only eliminates only truly empty lines, whereas some of your other commands also eliminate blank lines (empty or all-whitespace).
Also, the -i (for case-insensitive matching) in grep -ir 44 file.csv is pointless, and while using -r (for recursive searches) will not change the fact that only file.csv is searched, it will prepend the filename followed by : to each matching line.
If you have indeed captured the output in a new file and that file truly still has blank lines, the cat -A (cat -et on BSD-like platforms) you already mention in your question should show you if any unusual characters are present in the file, in the form of ^<char> sequences, such as ^M for \r chars.
If you like awk, this should do:
awk '/44/' file
It will only print lines that contains 44
here's my scanerio:
my input file like:
/tmp/abc.txt
/tmp/cde.txt
/tmp/xyz/123.txt
and i'd like to obtain the following output in 2 files:
first file
/tmp/
/tmp/
/tmp/xyz/
second file
abc.txt
cde.txt
123.txt
thanks a lot
Here is all in one single awk
awk -F\/ -vOFS=\/ '{print $NF > "file2";$NF="";print > "file1"}' input
cat file1
/tmp/
/tmp/
/tmp/xyz/
cat file2
abc.txt
cde.txt
123.txt
Here we set input and output separator to /
Then print last field $NF to file2
Set the last field to nothing, then print the rest to file1
I realize you already have an answer, but you might be interested in the following two commands:
basename
dirname
If they're available on your system, you'll be able to get what you want just piping through these:
cat input | xargs -l dirname > file1
cat input | xargs -l basename > file2
Enjoy!
Edit: Fixed per quantdev's comment. Good catch!
Through grep,
grep -o '.*/' file > file1.txt
grep -o '[^/]*$' file > file2.txt
.*/ Matches all the characters from the start upto the last / symbol.
[^/]*$ Matches any character but not of / zero or more times. $ asserts that we are at the end of a line.
The awk solution is probably the best, but here is a pure sed solution :
#n sed script to get base and file paths
h
s/.*\/\(.*.txt\)/\1/
w file1
g
s/\(.*\)\/.*.txt/\1/
w file2
Note how we hold the buffer with h, and how we use the write (w) command to produce the output files. There are many other ways to do it with sed, but I like this one for using multiple different commands.
To use it :
> sed -f sed_script testfile
Here is another oneliner that uses tee:cat f1.txt | tee >(xargs -n 1 dirname >> f2.txt) >(xargs -n 1 basename >> f3.txt) &>/dev/random