UNIX command to add comma and space

Question

I have a file that looks like

$ cat IP 10.3.1.1 10.4.1.1 10.6.3.1 10.19.4.2 10.22.3.4 

How do I make it look like:

$ cat IP 10.3.1.1, 10.4.1.1, 10.6.3.1, 10.19.4.2, 10.22.3.4 

Answer 1

With perl:

perl -pe 's/\n/, / unless eof' IP 

Add the -i option to edit the file inplace. Or -i.orig to edit it in place but keep the original as IP.orig.

With the zsh shell, you can also load those IPs into an array with:

ips=( ${(f)"$(<IP)"} ) 

And then write them back joined with commas with:

print -r -- ${(j[, ])ips} > IP 

Contrary to the perl one which processes one line at a time, that one loads the file whole in memory.

Two other differences:

• empty lines are discarded
• a newline character is always added at the end even if the original file didn't contain any like for an empty file or a file whose last line was not properly delimited.

Answer 2

With the GNU implementation of sed:

sed -z 's/\n*$//;s/\n/, /g;s/$/\n/' IP 

Answer 3

Perhaps

perl -e 'print join(", ", map { chomp; $_ } <>), "\n"' IP 

Or

awk '{ printf "%s%s", c, $0; c = ", " } END { print "" }' IP 

Or for a quick and dirty hack, subject to constraints on the number of lines in your file (potentially as few as 8000), and bearing in mind that xargs wasn't really designed for this

xargs <IP | sed 's/ /, /g' 

All these produce the same output from your example data set:

10.3.1.1, 10.4.1.1, 10.6.3.1, 10.19.4.2, 10.22.3.4 

You can then redirect that output to a temporary file and then overwrite the original. Here, replace command_from_above with your choice of any command (here in this answer or anyone else's) that writes the changes to stdout rather than to the original file:

command_from_above >"IP.$$.tmp" && mv -f -- "IP.$$.tmp" IP rm -f -- "IP.$$.tmp" 

Answer 4

With bash, zsh, or ksh:

( x=$(cat IP) && echo "${x//$'\n'/, }" > IP ) 

• Create a subshell to prevent the x variable from leaking out
• Set x to the contents of the IP file, including all newlines except the last
• Replace newlines in x with , (comma,space)
• Echo the result back to the IP file, adding (back) a final newline

It works even if there are spaces in the lines, but it sounds like you don't need that. However in zsh if the data contains backslashes they are interpreted, often producing wrong results, unless you use echo -E.

Answer 5

Using awk:

awk '{printf "%s%s", (NR==1) ? "" : ", ", $0}END{print ""}' 

This will prefix every line except for the first with a comma.

Answer 6

Adding comma with sed but excluding the last line, then replacing every new line with space:

sed '$!s/$/,/' file | paste -sd ' ' - 

Answer 7

Using AWK:

$ awk 'NR>1{printf "%s", a "," OFS} {a=$0}END {print a}' 

If only comma is needed without space and all records have only one fields, then datamash may be used.

$ datamash -W collapse 1 <file 

Answer 8

If you can't remember / don't want to learn the more powerful tools like awk and sed (and don't mind a trailing comma) here's a simple and understandable solution:

OUT=""; for i in $(cat IP); do OUT="$OUT $i,"; done; echo $OUT 

Output:

10.3.1.1, 10.4.1.1, 10.6.3.1, 10.19.4.2, 10.22.3.4, 

Answer 9

If you don’t actually care about the space after each comma, this is what paste is good at:

paste -sd , - 

You could restore the space by piping to sed if you know that no values contain a comma:

… | sed s/,/,\ /g 

Answer 10

Using Raku (formerly known as Perl_6)

~$ raku -ne '$++ && print ", "; .print;' file #OR ~$ raku -e 'slurp.chomp.subst(:global, / \n /, ", ").put;' file 

Herein are answers written in Raku, a member of the Perl-family of programming languages.

In the first example, the file is read-in using the -ne non-autoprinting linewise flags, which auto-chomps (removes) the trailing newline by default. When the anonymous $++ post-incrementing counter variable reads the first line (i.e. $++ equals zero), the ", " comma-space is skipped while the line is output without a newline terminator using .print. Note: .print is short for $_.print meaning to print the $_ topic variable. Otherwise, for every subsequent line, a comma-space is printed followed by the line (i.e. topic) itself.

In the second example, the file is slurped, i.e. read-into memory all at once, preserving internal/trailing newlines, etc. The trailing newline is chomped off, internal \n newlines are substituted globally with ", " comma-space, and the resultant file is output, which (in contrast to print) adds a newline at the end of the string (here, the end of the file).

Sample Input:

10.3.1.1 10.4.1.1 10.6.3.1 10.19.4.2 10.22.3.4 

Sample Output (both code examples):

10.3.1.1, 10.4.1.1, 10.6.3.1, 10.19.4.2, 10.22.3.4 

https://docs.raku.org
https://raku.org

Answer 11

with sed ( I don't know if it's posix ):

sed -n -e '1h;1!H;${g;s/\n/, /g;p;}' data 

Save all lines to the hold space and then, at the end, substitute all newlines with a , .

---

With awk (this should be posix portable. I don't remember if all awk let you unset the RS variable but i tested it with the --posix and --lint set to fatal in gawk and it worked ):

awk -v RS= 'gsub(/\n/, ", ")' data 

Unset the Record Separator variable to get all the file content as a single line and then replace all the new line charcter's with a ,

Answer 12

This would be a simple way to do it:

sed 's/$/,/' |xargs echo |sed 's/,$//' 

However, it would fail:

• for very long files (that exceed the maximum size of a program's arguments on your system),
• if the first line of your file looks like an option to your system's echo program.

Edit: see the comments for more failure modes. Only use this for short files with no characters that are special to xargs or echo.

Answer 13

Some more simple variations:

cat IP | sed 's/$/, /' | tr -d '\n' | sed 's/, $/\n/' 

or

cat IP | perl -pe 's/\n/, /' | perl -pe 's/, $/\n/'