I have a file like this:


I want to replace all occurences of \ with #, except for \N, such that the output looks like this:


I am trying to write a sed command which looks like-

sed -e 's@\\([^N])@#\1/g' filename

but this will fail for all values wherever there is a \N at the start for ex. \Nato in above example.

Could somebody please help me figure out a sed command(regex) for my requirement.


I'm not that well-versed in sed. But this can be done easily in perl using regex lookaround:

perl -pe 's/\\(?!N\b)/#/g' your_file_here

This will print the modified file to stdout. To make the substitutions in place, you can do:

perl -pi -e 's/\\(?!N\b)/#/g' your_file_here

Basically, this substitutes any backslash not followed by N\b (the character N at a word boundary) with #.


If you're absolutely sure that your entries are comma-delimited, you can do this to cover the case when \N is at the end of a word (as in SOLEM\N in the example you supplied):

perl -pe ' s/\\/#/g; s/(\A|,)\s*#N\s*(\Z|,)/$1\\N$2/g;' your_file_here

It simply substitutes any backslash with a hash mark then looks for #N found between two commas, between the beginning of the string and a comma or between a comma and the end of the string.

  • @user2118349 Answer updated to take the edge case you supplied into account. – Joseph R. Jul 29 '13 at 19:13
$ echo '1,2,subjects,\mat\hs,unix\,\Nato,\N,123,\N' \
  | sed -r -e 's@\\([^N]|N[^,])@#\1@g'

Edit for the \N as part of a word case: (see comments)

$ echo '1,2,subjects,\mat\hs,unix\,\SOLEM\N,\N,123,\N' \
  | sed -r -e 's@,\\N$@,:SINGLE_N_PLACEHOLDER:@g' \
           -e 's@^\\N,@:SINGLE_N_PLACEHOLDER:,@g' \
           -e 's@,\\N,@,:SINGLE_N_PLACEHOLDER:,@g' \
           -e 's@\\@#@g' \
           -e 's@:SINGLE_N_PLACEHOLDER:@\\N@g'

This is ugly and unreliable (placeholder string needs to be unique, will stop working if that pattern appears in the text), but I didn't find a working way to make sed use assertions the way PCRE/Perl support them.

You could make it shorter like so:

$ echo '1,2,subjects,\mat\hs,unix\,\SOLEM\N,\N,123,\N' \
  | sed -r -e 's@\\@#@g' \
           -e 's@(,|^)#N(,|$)@\1\\N\2@g'

But in this case, a single #N is the placeholder that must not already appear in the string.

  • Note that this will break if there's a space between \N and the , following it. – Joseph R. Jul 28 '13 at 20:13
  • I was thinking of using a negation of end-of-word \> after N but cannot seem to figure it out. Any ideas? – unxnut Jul 28 '13 at 20:18
  • @unxnut I don't think this is possible because \> is not really a character that can be negated: it matches a place in the regex (the place between a word character and a non-word one). I know perl offers a negated word boundary anchor (\B) but I'm not sure if sed's BREs support it. – Joseph R. Jul 28 '13 at 20:21
  • I am sure there should be a way to negate an entire pattern or search string; in this case, the search string is \\N\>. I am just hoping that some guru type person can provide a way of doing it or say why it cannot be done. – unxnut Jul 28 '13 at 20:25
  • @frostschutz Appreciate the quick help that offered, i am a new to regular expression. Wouldn't sed -r -e 's@\([^N]|N[^,])@#\1@g' fail whenvever \N would appear at the end of string for example- 1,2,subjects,\mat\hs,unix\,\SOLEM\N,\N,123,\N – user2118349 Jul 28 '13 at 20:54
sed -e 's@\\\([^N]\|\(N[a-zA-Z]\|$\)\)@#\1@g' your_file_here

Explaining: Replace all \ followed by

  • anything not N,
  • or N followed by any letter in [a-zA-Z] (extend this if needed)
  • end of line

with # and the rest of the matching pattern.

$ echo '1,2,subjects,\mat\hs,unix\,\Nato,\N,123,\N'   | 
         sed -r -e 's/\\N/XELI/g' | 
         sed -e 's/\\/#/g' | 
         sed -e 's/XELI/\\N/g'

I find it more readably to split it up:

  • replace \N by some non occuring character,
  • then replace / by #
  • and replace the /N back
  • 2
    Welcome to SX :). You haven't worked around the \Nato case mentioned in the OP. – Joseph R. Jul 28 '13 at 20:39

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