grep/awk：有条件地排除单词

Question

I have data formatted like this:

a cat
a dog
brown cat
brown dog
brown cow
brown sheep
brown fish

I want to filter out all of the lines starting with "brown", except brown dog . Is there an easy way to do this with grep or awk? I tried to use the carat negation like so:

grep -v "brown ^\(dog\)" corpus.txt

... but that didn't work. Any ideas would be greatly appreciated.

Eventually I want the output to be like this:

a cat
a dog
brown dog

Answer 1

Using awk :

awk '/^brown dog/ || !/^brown/' file

a cat
a dog
brown dog

---

Just as an academic exercise here is a grep command without experimental PCRE option:

grep -vE '^brown($|[^ ]| ([^d]|$)| d([^o]|$)| do([^g]|$))' file

Answer 2

Yes sir:

grep -vP '^brown (?!dog)' file

a cat
a dog
brown dog

-P for pcre engine use.
Check explanations

Answer 3

awk '/^brown/ && !/dog$/{next} 1' file

Answer 4

Ok, it's past midnight over here. I'm going to post this awk:

 $ awk '!(/brown/ && !/dog/)' file

... and think it thru in the morning. :D Good night.

Nope, couldn't sleep, had to solve it:

$ awk '!/^brown/ || /dog/' file

Output:

a cat
a dog
brown dog

Answer 5

It's not clear if you specifically want to accept "brown dog" only, but perhaps you just want something like:

sed -e '/^brown/{/dog/!d;}'

This will delete all lines that start with "brown" unless they match the string "dog". Or maybe you want to be stricter and do:

awk '!/^brown/ || $2 == "dog"'

Answer 6

Another awk :

$ awk '!(/^brown/ && $2!="dog")' file
a cat
a dog
brown dog