How can I replace multiple empty lines with a single empty line in bash?

Question

I have a file that contains:

something



something else

something else again

I need a bash command, sed/grep w.e that will produce the fo

Answer 1

I just solved this problem by sed. Even if this is a 7 years old question, someone may can here for help, so I am writing my solution by sed here:

sed 'N;/^\n$/D;P;D;'

Answer 2

grep -A1 . <yourfile> | grep -v "^--$"

This grep solution works assuming you want the following:

Input

line1

line2
line3


line4



line5

Output

line1

line2
line3

line4

line5

Answer 3

Use python:

s = file("filename.txt").read()
while "\n\n\n" in s: s = s.replace("\n\n\n", "\n\n")
import sys
sys.stdout.write(s)

Answer 4

Actually, if you replace multiple newlines with a single newline, the output would be:

something
something else
something else again

You can achieve this by:

sed /^$/d FILE

Answer 5

This uses marco's solution on multiple files:

for i in *; do FILE=$(cat -s "$i"); echo "$FILE" > "$i"; done

Answer 6

Python, with regular expression:

import re
import sys
sys.stdout.write(re.sub('\n{2,}','\n\n', sys.stdin.read()))