shell replace cr\lf by comma

Question

I have input.txt

1
2
3
4
5

I need to get such output.txt

1,2,3,4,5

How to do it?

Answer 1

Try this:

tr '\n' ',' < input.txt > output.txt

Answer 2

With sed, you could use:

sed -e 'H;${x;s/\n/,/g;s/^,//;p;};d'

The H appends the pattern space to the hold space (saving the current line in the hold space). The ${...} surrounds actions that apply to the last line only. Those actions are: x swap hold and pattern space; s/\n/,/g substitute embedded newlines with commas; s/^,// delete the leading comma (there's a newline at the start of the hold space); and p print. The d deletes the pattern space - no printing.

You could also use, therefore:

sed -n -e 'H;${x;s/\n/,/g;s/^,//;p;}'

The -n suppresses default printing so the final d is no longer needed.

This solution assumes that the CRLF line endings are the local native line ending (so you are working on DOS) and that sed will therefore generate the local native line ending in the print operation. If you have DOS-format input but want Unix-format (LF only) output, then you have to work a bit harder - but you also need to stipulate this explicitly in the question.

It worked OK for me on MacOS X 10.6.5 with the numbers 1..5, and 1..50, and 1..5000 (23,893 characters in the single line of output); I'm not sure that I'd want to push it any harder than that.

Answer 3

In response to @Jonathan's comment to @eumiro's answer:

tr -s '\r\n' ',' < input.txt | sed -e 's/,$/\n/' > output.txt

Answer 4

Use paste command. Here is using pipes:

echo "1\n2\n3\n4\n5" | paste -s -d, /dev/stdin

Here is using a file:

echo "1\n2\n3\n4\n5" > /tmp/input.txt
paste -s -d, /tmp/input.txt

Per man pages the s concatenates all lines and d allows to define the delimiter character.

Answer 5

tr and sed used be very good but when it comes to file parsing and regex you can't beat perl (Not sure why people think that sed and tr are closer to shell than perl... )

perl -pe 's/\n/$1,/' your_file

if you want pure shell to do it then look at string matching

${string/#substring/replacement}

Answer 6

• Awk versions:
  
  • awk '{printf("%s,",$0)}' input.txt
  • awk 'BEGIN{ORS=","} {print $0}' input.txt
  • Output - 1,2,3,4,5,

Since you asked for 1,2,3,4,5, as compared to 1,2,3,4,5, (note the comma after 5, most of the solutions above also include the trailing comma), here are two more versions with Awk (with wc and sed) to get rid of the last comma:

• i='input.txt'; awk -v c=$(wc -l $i | cut -d' ' -f1) '{printf("%s",$0);if(NR<c){printf(",")}}' $i

• awk '{printf("%s,",$0)}' input.txt | sed 's/,\s*$//'

Answer 7

printf "1\n2\n3" | tr '\n' ','

if you want to output that to a file just do

printf "1\n2\n3" | tr '\n' ',' > myFile

if you have the content in a file do

cat myInput.txt | tr '\n' ',' > myOutput.txt

Answer 8

• python version:

  python -c 'import sys; print(",".join(sys.stdin.read().splitlines()))'

Doesn't have the trailing comma problem (because join works that way), and splitlines splits data on native line endings (and removes them).

Answer 9

cat input.txt | sed -e 's|$|,|' | xargs -i echo "{}"