Capture all consecutive all-caps words with regex in python?

Question

I am trying to match all consecutive all caps words/phrases using regex in Python. Given the following:

    text = \"The following words are ALL CAPS. The follow         


        

Answer 1

Keeping your regex, you can use strip() and filter:

string = "The following words are ALL CAPS. The following word is in CAPS."
result = filter(None, [x.strip() for x in re.findall(r"\b[A-Z\s]+\b", string)])
# ['ALL CAPS', 'CAPS']

Answer 2

Your regex is relying on explicit conditions(space after letters).

matches = re.findall(r"([A-Z]+\s?[A-Z]+[^a-z0-9\W])",text)

Capture A to Z repetitions if there are no trailing lowercase or none-alphabet character.

Answer 3

This one does the job:

import re
text = "tHE following words aRe aLL CaPS. ThE following word Is in CAPS."
matches = re.findall(r"(\b(?:[A-Z]+[a-z]?[A-Z]*|[A-Z]*[a-z]?[A-Z]+)\b(?:\s+(?:[A-Z]+[a-z]?[A-Z]*|[A-Z]*[a-z]?[A-Z]+)\b)*)",text)
print matches

Output:

['tHE', 'aLL CaPS', 'ThE', 'Is', 'CAPS']

Explanation:

(           : start group 1
  \b        : word boundary
  (?:       : start non capture group
    [A-Z]+  : 1 or more capitals
    [a-z]?  : 0 or 1 small letter
    [A-Z]*  : 0 or more capitals
   |        : OR
    [A-Z]*  : 0 or more capitals
    [a-z]?  : 0 or 1 small letter
    [A-Z]+  : 1 or more capitals
  )         : end group
  \b        : word boundary
  (?:       : non capture group
    \s+     : 1 or more spaces
    (?:[A-Z]+[a-z]?[A-Z]*|[A-Z]*[a-z]?[A-Z]+) : same as above
    \b      : word boundary
  )*        : 0 or more time the non capture group
)           : end group 1

Answer 4

Assuming you want to start and end on a letter, and only include letters and whitespace

\b([A-Z][A-Z\s]*[A-Z]|[A-Z])\b

|[A-Z] to capture just I or A