regular expression to find lines containing multiple specific words or patterns in any arbitrary order

Question

Suppose we have the following very simple file:

1 x 3 x
x 3 x 1

Now I'd like to have a pattern that matches all lines containing both 1 and 3. The order and position should not matter.

(Note that I used a very simple file on purpose. The file could contain many more lines, and the ”words“ could be words like ”hello“ and ”world“.)

Answer 1

Another way is to use :h /\& , it works a bit like && in most coding language:

pattern0\&pattern1

Above expression will match pattern1 if pattern0 also match at the same position.

To solve your example:

\v^(.*1&.*3).*$

• \v very magic
• ^ start of line
• (.*1&.*3) match .*3 if .*1 matches.
• .*$ match anything until line end

If you have multiple patterns, you can write it as:

\v^(.*pattern0&.*pattern1&.*pattern2...).*$

Answer 2

Here's another alternative: use :h /bar to match either …1…3… or …3…1…:

\v.*(1.*3|3.*1).*

• \v very magic
• .* anything (or nothing)
• (...) a group, containing either:
  • 1.*3 1 followed by anything followed by 3
  • | or
  • 3.*1 3 followed by anything followed by 1
• .* anything

Answer 3

One option is to use a pattern with multiple lookaheads:

magic:        ^\(.*1\)\@=\(.*3\)\@=
                   ‾         ‾

very magic:   \v^(.*1)@=(.*3)@=
                    ‾      ‾

Since lookaheads (and lookarounds in general) do not consume characters, the same line can be searched for multiple different patterns/words.

---

additional hint:

This even allows one to add another condition to the regular expression: matching lines that contain 1 and 3, but do not contain x:

magic:        ^\(.*1\)\@=\(.*3\)\@=\(.*x\)\@!
                   ‾         ‾         ‾

very magic:   \v^(.*1)@=(.*3)@=(.*x)@!
                    ‾      ‾      ‾