$ perl -le '$_="1 2 3 4 5 6"; s/^((?:[^\s]*\s){3}[^\s]*)\s/$1\n/ and p
+rint'
1 2 3 4
5 6

[download]

That might be more readable as:

s/^([^\s]*(?:\s[^\s]*){3})\s/$1\n/

[download]

or expanded:

s/^([^\s]*\s[^\s]*\s[^\s]*\s[^\s]*)\s/$1\n/

The idea is to match anything that is not a space followed by a space and 0 or more non-spaces 3 times and capture it. Then match another space and replace all of that with what you captures and a newline at the end.

Expanding and commenting gives us:

s/
    ^             # Start at the beginning.
     (            # Start capturing.
       [^\s]*     # 0 or more non-space.
       \s [^\s]*  # A space and 0 or more non-spaces... once.
       \s [^\s]*  # twice.
       \s [^\s]*  # three times.
     )            # Stop capturing.
    \s            # Match another space.
/$1\n/x           # Replace with what we captured and a newline.

[download]

Of course, that all looks rather ugly with the negated char classes written like that. You should probably use \S instead. So, finally:

s/^(\S*(?:\s\S*){3})\s/$1/n/ 
-sauoq
"My two cents aren't worth a dime.";

In reply to Re: Need help with regex to replace 4th \s with \n in data line by sauoq
in thread Need help with regex to replace 4th \s with \n in data line by Popcorn Dave

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