正则表达式完整单词模式

Question

I want to get patterns involving complete words, not pieces of words. Eg 12345 [some word] 1234567 [some word] 123 1679 . Random text and the pattern appears again 1111 123 [word] 555 .

This should return

[[12345, 1234567, 123, 1679],[1111, 123, 555]]

I am only tolerating one word between the numbers otherwise the whole string would match. Also note that it is important to capture that 2 matches were found and so a two-element list was returned.

I am running this in python3. I have tried:

\b(\d+)\b\s\b(\w+)?\b\s\b(\d+)\b

but I am not sure how to scale this to an unrestricted number of matches.

re.findall('\b(\d+)\b\s\b(\w+)?\b\s\b(\d+)\b', string)

This matches [number] [word] [number] but not any number that might follow with or without a word in between.

Answer 1

You can't do this in one operation with the Python re engine.
But you could match the sequence with one match, then extract the
digits with another.

This matches the sequence

r"(?<!\\w)\\d+(?:(?:[^\\S\\r\\n]+[a-zA-Z](?:\\w*[a-zA-Z])*)?[^\\S\\r\\n]+\\d+)*(?!\\w)"

https://regex101.com/r/73AYLU/1

Explained

 (?<! \w )                     # Not a word behind
 \d+                           # Many digits
 (?:                           # Optional word block
      (?:                           # Optional words
           [^\S\r\n]+                    # Horizontal whitespace
           [a-zA-Z]                      # Starts with a letter
           (?: \w* [a-zA-Z] )*           # Can be digits in middle, ends with a letter
      )?                            # End words, do once
      [^\S\r\n]+                    # Horizontal whitespace
      \d+                           # Many digits
 )*                            # End word block, do many times
 (?! \w )                      # Not a word ahead

---

This gets the array of digits from the sequence matched above (use findall)

r"(?<!\\S)(\\d+)(?!\\S)"

https://regex101.com/r/BHov38/1

Explained

 (?<! \S )              # Whitespace boundary
 ( \d+ )                # (1)
 (?! \S )               # Whitespace boundary

Answer 2

Are you expecting re.findall() to return a list of lists? It will only return a list - no matter what regex you use.

One approach is to split your input string into sentences and then loop through them

import re
inputArray = re.split('<pattern>',inputText)
outputArray = []
for item in inputArray:
    outputArray.append(re.findall('\b(\d+)\b\s\b(\w+)?\b\s\b(\d+)\b', item))

the trick is to find a <pattern> to split your input.

Answer 3

This is a bit complicated, maybe this expression would be just something to look into:

(((\d+)\s*)*(?:\s*\[.*?\]\s*)((\d+)\s*)*)|([A-za-z\s]+)

and script the rest of the problem for a valid solution.

Demo