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

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6

1

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“.)

regular-expression

share|improve this question

asked May 14 at 7:26

myrddmyrdd

1636

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Check out our Code of Conduct.

add a comment | 

6

1

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“.)

regular-expression

share|improve this question

asked May 14 at 7:26

myrddmyrdd

1636

New contributor

myrdd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

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“.)

regular-expression

share|improve this question

asked May 14 at 7:26

myrddmyrdd

1636

New contributor

myrdd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

1 x 3 x
x 3 x 1

share|improve this question

asked May 14 at 7:26

myrddmyrdd

1636

New contributor

myrdd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked May 14 at 7:26

myrddmyrdd

1636

New contributor

myrdd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 14 at 7:26

myrddmyrdd

1636

New contributor

myrdd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 14 at 7:26

myrddmyrdd

1636

3 Answers
3

active

oldest

votes

7

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...).*$

share|improve this answer

edited May 14 at 8:14

answered May 14 at 8:05

dedowsdidedowsdi

1,182411

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  I've never really understood what /& is for. Now I do. Which is weird, because this is essentially identical to the "Peter and Bob" example given in the :help. Still, reading your answer, it finally clicked in my brain. Thanks! +1
  
  – Rich
  May 14 at 9:52
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This is way easier to type in general, and I (like Rich) learned something. +1; this should be the accepted answer.
  
  – D. Ben Knoble
  May 14 at 17:11
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  it works a bit like && in most coding language: is brilliant!
  
  – Maxim Kim
  May 15 at 8:20
  

  
  
  
  

add a comment | 

4

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

share|improve this answer

edited May 14 at 9:47

answered May 14 at 9:41

RichRich

15.9k12066

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  nicely explained, therefore +1
  
  – myrdd
  May 15 at 8:12
  

  
  
  
  

add a comment | 

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)@!
 ‾ ‾ ‾

share|improve this answer

edited yesterday

answered May 14 at 7:26

myrddmyrdd

1636

New contributor

myrdd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I'd consider removing the <, >, <, and > end-of-word atoms from this answer. They're not required for the simple example given in your question, and your regular expressions are easier to understand without them!
  
  – Rich
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you @Rich for your edit and your suggestion!
  
  – myrdd
  21 hours ago
  

  
  
  
  

add a comment | 

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7

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...).*$

share|improve this answer

edited May 14 at 8:14

answered May 14 at 8:05

dedowsdidedowsdi

1,182411

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  I've never really understood what /& is for. Now I do. Which is weird, because this is essentially identical to the "Peter and Bob" example given in the :help. Still, reading your answer, it finally clicked in my brain. Thanks! +1
  
  – Rich
  May 14 at 9:52
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This is way easier to type in general, and I (like Rich) learned something. +1; this should be the accepted answer.
  
  – D. Ben Knoble
  May 14 at 17:11
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  it works a bit like && in most coding language: is brilliant!
  
  – Maxim Kim
  May 15 at 8:20
  

  
  
  
  

add a comment | 

7

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...).*$

share|improve this answer

edited May 14 at 8:14

answered May 14 at 8:05

dedowsdidedowsdi

1,182411

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  I've never really understood what /& is for. Now I do. Which is weird, because this is essentially identical to the "Peter and Bob" example given in the :help. Still, reading your answer, it finally clicked in my brain. Thanks! +1
  
  – Rich
  May 14 at 9:52
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This is way easier to type in general, and I (like Rich) learned something. +1; this should be the accepted answer.
  
  – D. Ben Knoble
  May 14 at 17:11
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  it works a bit like && in most coding language: is brilliant!
  
  – Maxim Kim
  May 15 at 8:20
  

  
  
  
  

add a comment | 

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...).*$

share|improve this answer

edited May 14 at 8:14

answered May 14 at 8:05

dedowsdidedowsdi

1,182411

pattern0&pattern1

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

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

share|improve this answer

edited May 14 at 8:14

answered May 14 at 8:05

dedowsdidedowsdi

1,182411

answered May 14 at 8:05

dedowsdidedowsdi

1,182411

answered May 14 at 8:05

dedowsdidedowsdi

1,182411

4

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

share|improve this answer

edited May 14 at 9:47

answered May 14 at 9:41

RichRich

15.9k12066

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  nicely explained, therefore +1
  
  – myrdd
  May 15 at 8:12
  

  
  
  
  

add a comment | 

4

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

share|improve this answer

edited May 14 at 9:47

answered May 14 at 9:41

RichRich

15.9k12066

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  nicely explained, therefore +1
  
  – myrdd
  May 15 at 8:12
  

  
  
  
  

add a comment | 

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

share|improve this answer

edited May 14 at 9:47

answered May 14 at 9:41

RichRich

15.9k12066

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

share|improve this answer

edited May 14 at 9:47

answered May 14 at 9:41

RichRich

15.9k12066

answered May 14 at 9:41

RichRich

15.9k12066

answered May 14 at 9:41

RichRich

15.9k12066

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)@!
 ‾ ‾ ‾

share|improve this answer

edited yesterday

answered May 14 at 7:26

myrddmyrdd

1636

New contributor

myrdd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I'd consider removing the <, >, <, and > end-of-word atoms from this answer. They're not required for the simple example given in your question, and your regular expressions are easier to understand without them!
  
  – Rich
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you @Rich for your edit and your suggestion!
  
  – myrdd
  21 hours ago
  

  
  
  
  

add a comment | 

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)@!
 ‾ ‾ ‾

share|improve this answer

edited yesterday

answered May 14 at 7:26

myrddmyrdd

1636

New contributor

myrdd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I'd consider removing the <, >, <, and > end-of-word atoms from this answer. They're not required for the simple example given in your question, and your regular expressions are easier to understand without them!
  
  – Rich
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you @Rich for your edit and your suggestion!
  
  – myrdd
  21 hours ago
  

  
  
  
  

add a comment | 

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)@!
 ‾ ‾ ‾

share|improve this answer

edited yesterday

answered May 14 at 7:26

myrddmyrdd

1636

New contributor

myrdd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

magic: ^(.*1)@=(.*3)@=
 ‾ ‾

very magic: v^(.*1)@=(.*3)@=
 ‾ ‾

magic: ^(.*1)@=(.*3)@=(.*x)@!
 ‾ ‾ ‾

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

share|improve this answer

edited yesterday

answered May 14 at 7:26

myrddmyrdd

1636

New contributor

myrdd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered May 14 at 7:26

myrddmyrdd

1636

New contributor

myrdd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered May 14 at 7:26

myrddmyrdd

1636