Split string based on regex

What is the best way to split a string like "HELLO there HOW are YOU" by upper case words (in Python)?

So I'd end up with an array like such: results = ['HELLO there', 'HOW are', 'YOU']

EDIT:

I have tried:

p = re.compile("\b[A-Z]{2,}\b") print p.split(page_text) 

It doesn't seem to work, though.

3 Answers

I suggest

l = re.compile("(?<!^)\s+(?=[A-Z])(?!.\s)").split(s) 

Check this demo.

You could use a lookahead:

re.split(r'[ ](?=[A-Z]+\b)', input) 

This will split at every space that is followed by a string of upper-case letters which end in a word-boundary.

Note that the square brackets are only for readability and could as well be omitted.

If it is enough that the first letter of a word is upper case (so if you would want to split in front of Hello as well) it gets even easier:

re.split(r'[ ](?=[A-Z])', input) 

Now this splits at every space followed by any upper-case letter.

Your question contains the string literal "\b[A-Z]{2,}\b", but that \b will mean backspace, because there is no r-modifier.

Try: r"\b[A-Z]{2,}\b".