Does lookbehind work in sed?

Question

I created a test using grep but it does not work in sed.

grep -P '(?<=foo)bar' file.txt

This works correctly by returning bar.

sed 's/(?<=foo)bar/test/g' file.txt

I was expecting footest as output, but it did not work.

Imprint

sed does not support lookaround assertions. – Segregate

For what it's worth, grep -P is also a nonstandard extension, though typically available on Linux (but not other platforms). – Metagenesis

Answer 1

GNU sed does not have support for lookaround assertions. You could use a more powerful language such as Perl or possibly experiment with ssed which supports Perl-style regular expressions.

perl -pe 's/(?<=foo)bar/test/g' file.txt

Segregate

The text accompanying your solution doesn't quite make sense since Perl doesn't support PCRE either (at least not natively). – Askance

Answer 2

Note that most of the time you can avoid a lookbehind (or a lookahead) using a capture group and a backreference in the replacement string:

sed 's/\(foo\)bar/\1test/g' file.txt

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Simulating a negative lookbehind is more subtile and needs several substitutions to protect the substring you want to avoid. Example for (?<!foo)bar:

sed 's/#/##/g;s/foobar/foob#ar/g;s/bar/test/g;s/foob#ar/foobar/g;s/##/#/g' file.txt

• choose an escape character and repeat it (for example # => ##).
• include this character in the substring you want to protect (foobar here, => foob#ar or ba => b#a).
• make your replacement.
• replace foob#ar with foobar (or b#a with ba).
• replace ## with #.

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Obviously, you can also describe all that isn't foo before bar in a capture group:

sed -E 's/(^.{0,2}|[^f]..|[^o].?)bar/\1test/g' file.txt

But it will quickly become tedious with more characters.

Ganglion

But this does not work for "negative" lookbehind, e.g. you want "bar" NOT preceded by "foo" to be replaced with "test", what would be done (if it worked) with /(?<!foo)bar/test/. Has anyone a solution to this? (I want to use uniq on the 5th field of an SQL file but preceding fields may contain spaces so I have no better idea than to replace all spaces NOT between ...', '... by "_"...) – Ionian

@Max: 1) choose an escape character and repeat it (for example # => ##). 2) include this character in the substring you want to protect (foobar here, => foob#ar). 3) make your replacement. 4) replace foob#ar with foobar. 5) replace ## with #. Example with sed: sed 's/#/##/g;s/foobar/foob#ar/g;s/bar/test/g;s/foob#ar/foobar/g;s/##/#/g' <<<'abc foobar # foob#ar foo bar' – Ganglion

OK, yes that works, essentially you remove what you want to protect (maybe foobar=>-#- (instead foob#ar) would be clearer), then you find & replace all others, then you put the "protected" ones back – Ionian

@Max: foobar => -#-: if you want (and only if you have replaced all # with ## before). – Ganglion

@MichaelChirico: Thanks Chirico (and the sorceress) for your edit. – Ganglion

Answer 3

sed doesn't support lookarounds but choose (I'm the author) does. It uses PCRE2 syntax.

For example:

$ echo "hello bar foobar" | choose -r --sed '(?<=foo)bar' --replace test
hello bar footest

It's speed is comparable to sed.

Deliberate