Replace first two whitespace occurrences with a comma using sed - regex

Question

I have a whitespace delimited file with a variable number of entries on each line. I want to replace the first two whitespaces with commas to create a comma delimited file with three columns.
Here's my input:
a b 1 2 3 3 2 1
c d 44 55 66 2355
line http://google.com 100 200 300
ef jh 77 88 99
z y 2 3 33
And here's my desired output:
a,b,1 2 3 3 2 1
c,d,44 55 66 2355
line,http://google.com,100 200 300
ef,jh,77 88 99
z,y,2 3 33
I'm trying to use perl regular expressions in a sed command but I can't quite get it to work. First I try capturing a word, followed by a space, then another word, but that only works for lines 1, 2, and 5:
$ cat test | sed -r 's/(\w)\s+(\w)\s+/\1,\2,/'
a,b,1 2 3 3 2 1
c,d,44 55 66 2355
line http://google.com 100 200 300
ef jh 77 88 99
z,y,2 3 33
I also try capturing whitespace, a word, and then more whitespace, but that gives me the same result:
$ cat test | sed -r 's/\s+(\w)\s+/,\1,/'
a,b,1 2 3 3 2 1
c,d,44 55 66 2355
line http://google.com 100 200 300
ef jh 77 88 99
z,y,2 3 33
I also try doing this with the .? wildcard, but that does something funny to line 4.
$ cat test | sed -r 's/\s+(.?)\s+/,\1,/'
a,b,1 2 3 3 2 1
c,d,44 55 66 2355
line http://google.com 100 200 300
ef jh,,77 88 99
z,y,2 3 33
Any help is much appreciated!

Answer 1

How about this:
sed -e 's/\s\+/,/' | sed -e 's/\s\+/,/'
It's probably possible with a single sed command, but this is sure an easy way :)
My output:
a,b,1 2 3 3 2 1
c,d,44 55 66 2355
line,http://google.com,100 200 300
ef,jh,77 88 99
z,y,2 3 33

Answer 2

Try this:
sed -r 's/\s+(\S+)\s+/,\1,/'
Just replaced \w (one "word" char) with \S+ (one or more non-space chars) in one of your attempts.

Answer 3

You can provide multiple commands to a single instance of sed by just providing multiple -e arguments.
To do the first two, just use:
sed -e 's/\s\+/,/' -e 's/\s\+/,/'
This basically runs both commands on the line in sequence, the first doing the first block of whitespace, the second doing the next.
The following transcript shows this in action:
pax$ echo 'a b 1 2 3 3 2 1
c d 44 55 66 2355
line http://google.com 100 200 300
ef jh 77 88 99
z y 2 3 33
' | sed -e 's/\s\+/,/' -e 's/\s\+/,/'
a,b,1 2 3 3 2 1
c,d,44 55 66 2355
line,http://google.com,100 200 300
ef,jh,77 88 99
z,y,2 3 33

Answer 4

Sed s/// supports a way to say which occurrence of a pattern to replace: just add the n to the end of the command to replace only the nth occurrence. So, to replace the first and second occurrences of whitespace, just use it this way:
$ sed 's/ */,/1;s/ */,/2' input
a,b ,1 2 3 3 2 1
c,d ,44 55 66 2355
line,http://google.com 100,200 300
ef,jh ,77 88 99
z,y 2,3 33
EDIT: reading another proposed solutions, I noted that the 1 and 2 after s/ */,/ is not only unnecessary but plainly wrong. By default, s/// just replaces the first occurrence of the pattern. So, if we have two identical s/// in sequence, they will replace the first and the second occurrence. What you need is just
$ sed 's/ */,/;s/ */,/' input
(Note that you can put two sed commands in one expression if you separate them by a semicolon. Some sed implementations do not accept the semicolon after the s/// command; use a newline to separate the commands, in this case.)

Answer 5

A Perl solution is:
perl -pe '$_=join ",", split /\s+/, $_, 3' some.file

Answer 6

Not sure about sed/perl, but here's an (ugly) awk solution. It just prints fields 1-2, separated by commas, then the remaining fields separated by space:
awk '{
printf("%s,", $1)
printf("%s,", $2)
for (i=3; i<=NF; i++)
printf("%s ", $i)
printf("\n")
}' myfile.txt