Replace characters using regex grouping with sed

Question

I have a text file that is like this:

FOO BAR PIPPO PLUTO 31337 1010
FOOZ BAZ 130
VERY LONG LINE LIKE THIS THEN A NUMBER LIKE 42

I need to turn it into:

FOO-BAR-PIPPO-PLUTO 31337 1010
FOOZ-BAZ 130
VERY-LONG-LINE-LIKE-THIS-THEN-A-NUMBER-LIKE 42

The best I could do is:

sed -re 's/([A-Z]+)( )([A-Z]+)/\1-\3/g'

but the output is

FOO-BAR PIPPO-PLUTO 31337 1010
FOOZ-BAZ 130
VERY-LONG LINE-LIKE THIS-THEN A-NUMBER LIKE 42

Close, but no cigar. Any idea on why my regex doesn't work?

Answer 1

You can't have overlapping matches. "BAR PIPPO" isn't detected because "BAR" was already consumed when matching "FOO BAR".

FOO BAR PIPPO PLUTO 31337 1010
------- ===========
   1         2

Try this instead:

$ sed -re 's/ ([A-Z])/-\1/g'

Note that this doesn't have overlapping matches:

FOO BAR PIPPO PLUTO 31337 1010
   --  ==    --
   1   2     3

Answer 2

sed 's/ \([^0-9]\)/-\1/g'

Just look for space followed by not a number and replace that space with a -. The advantage of this is that it will work for lines that have non-alphanumeric characters.

Proof of Concept

$ cat ./infile
FOO BAR PIPPO PLUTO 31337 1010
FOOZ BAZ 130
VERY LONG LINE LIKE THIS THEN A NUMBER LIKE 42
THIS LINE HAS $ODD$ #CHARS# IN %IT% 42

$ sed 's/ \([^0-9]\)/-\1/g' ./infile
FOO-BAR-PIPPO-PLUTO 31337 1010
FOOZ-BAZ 130
VERY-LONG-LINE-LIKE-THIS-THEN-A-NUMBER-LIKE 42
THIS-LINE-HAS-$ODD$-#CHARS#-IN-%IT% 42

Answer 3

Very close. You don't need to match more than one letter though - you just want letter space letter:

sed -Ee 's/([A-Z])( )([A-Z])/\1-\3/g' foo.txt 
FOO-BAR-PIPPO-PLUTO 31337 1010
FOOZ-BAZ 130
VERY-LONG-LINE-LIKE-THIS-THEN-A NUMBER-LIKE 42

(sed params adjusted for BSD sed)