python regex split string while keeping delimiter with value

Question

I'm trying to parse a text file with name:value elements in it into lists with "name:value"... Here's a twist: The values will sometimes be multiple words or even multiple lines and the delimiters are not a fixed set of words. Here's an example of what I'm trying to work with...

listing="price:44.55 name:John Doe title:Super Widget description:This widget slices, dices, and drives your kids to soccer practice\r\nIt even comes with Super Widget Mini!

What I want to return is...

["price:44.55", "name:John Doe", "title:Super Widget", "description:This widget slices, dices, and drives your kids to soccer practice\r\nIt even comes with Super Widget Mini!"]

Here's what I've tried so far...

details = re.findall(r'[\w]+:.*', post, re.DOTALL)
["price:", "44.55 name:John Doe title:Super Widget description:This widget slices, dices, and drives your kids to soccer practice\r\nIt even comes with Super Widget Mini!"]

Not what I want. Or...

details = re.findall(r'[\w]+:.*?', post, re.DOTALL)
["price:", "name:", "title:", "description:"]

Not what I want. Or...

details = re.split(r'([\w]+:)', post)
["", "price:", "44.55", "name:", "John Doe", "title:", "Super Widget", "description:", "This widget slices, dices, and drives your kids to soccer practice\r\nIt even comes with Super Widget Mini!"]

which is closer, but still no dice. Also, I can deal with an empty list item. So, basically, my question is how do you keep the delimiter with the values on a re.split() or how do you keep re.findall() from either being too greedy or too stingy?

Thanks ahead of time for reading!

Answer 1

Use a look-ahead assertion:

>>> re.split(r'\s(?=\w+:)', post)
['price:44.55',
 'name:John Doe',
 'title:Super Widget',
 'description:This widget slices, dices, and drives your kids to soccer practice\r\nIt even comes with Super Widget Mini!']

Of course, it would still fail if there are some words followed immediately by a colon in your values.