How to remove all words which doesn't match the pattern?

Question

I've the following content:

a1a foo bar buzz
foo b2b bar buzz
foo bar c3c buzz
foo bar buzz d4d

where I can match all words consisting only alpha characters by the following pattern:

/\<[a-zA-Z]*\>

Now, how can I remove all words which doesn't match the pattern?

So the final content is:

foo bar buzz
foo bar buzz
foo bar buzz
foo bar buzz

---

I've checked Search for lines not containing pattern page, but without luck.

To remove some words patching the pattern, global command can be used:

:g/\<[a-zA-Z]*\>/;&

as per this post.

However when doing opposite:

:v/\<[a-zA-Z]*\>/;&

there is this message:

Pattern found in every line

Answer 1

You can do this with \@<! (which is sometimes called "negative lookbehind"):

/\<\w\+\>\(\<\a\+\>\)\@<!

    \@<! Matches with zero width if the preceding atom does NOT match just
         before what follows.

It is a bit tricky to understand at first:

1. Initially we try searching for all words (including ones with numbers) /\<\w\+\>.
2. Since we want some words to match and not others, we use a "zero-width" in this position to force certain text to match. In this case, the text we want is words without a digit: \<\a\+\>.
3. We wrap the second pattern in \(...\)\@<!, which denotes the zero-width mentioned above. We use \(...\) to create a single atom.

Now, to remove those words, we can use :substitute, but add \s\? to take care of the spaces.

:%s/\<\w\+\>\(\<\a\+\>\)\@<!\s\?//g

We can make this slightly shorter using \v very magic:

:%s/\v<\w+>(<\a+>)@<!\s?//g