I have a very large file that has zero-width spaces scattered throughout. It takes too long to open and edit using vi so I'd like to delete all instances of the character using sed. The problem is, I can't figure out how to match the character! I've tried using \u200B, \x{200b}. Any ideas?
I'm running CentOS 5 if that helps at all.
23 Answers
This seems to work for me:
sed 's/\xe2\x80\x8b//g' inputfile Demonstration:
$ /usr/bin/printf 'X\u200bY\u200bZ' | hexdump -C 00000000 58 e2 80 8b 59 e2 80 8b 5a |X...Y...Z| $ /usr/bin/printf 'X\u200bY\u200bZ' | sed 's/\xe2\x80\x8b//g' | hexdump -C 00000000 58 59 5a |XYZ| Edit:
Based partially on Gilles' answer:
tr -d $(/usr/bin/printf "\u200b") < inputfile GNU sed's behavior with UTF-8 doesn't seem to be very well-defined. Experimentally, you can make it replace the bytes of the UTF-8 representation:
<old sed 's/\xe2\x80\e8b//g' >new Alternatively, you can type the character into your shell and use any of the standard commands in a UTF-8 locale:
<old tr -d '' >new <old sed 's///g' >new In zsh, you can also enter the character through an escape sequence:
<old tr -d $'\u200B' >new Well, unless anyone has any ideas for how to get sed to do this (which I'm still interested in, by the way) its Python to the rescue...
import sys, re pattern = re.compile(u"\u200b") f = open(sys.stdin, "rb") for line in f: a = pattern.sub("", line.decode("utf8")) print a.encode("utf8"), f.close() 
