Regex to group words separated by space

Question

I have a data frame, in which one column is a Series of strings, in which distinct phrases are either single words or multiple words separated by spaces; and the first letter of each individual word is upper case (e.g. "Strawberry" or "Strawberry Jam", respectively). In contrast, if not part of the same phrase, the words are not spaced out (e.g. "JamApple").

df = pd.DataFrame({
    'foo': ['Strawberry JamApple', 'BananaPear CrumblePotato', 'Almond Cake'],
    'bar': ['A', 'B', 'C'],
    'baz': [1, 2, 3],
    'zoo': ['x', 'y', 'z'],
})


                        foo bar  baz zoo
0       Strawberry JamApple   A    1   x
1  BananaPear CrumblePotato   B    2   y
2               Almond Cake   C    3   z

How could I use regex to separate phrases in a string based on the rule above (into "Strawberry Jam", "Apple", "Banana", "Pear Crumble", "Potato", "Almond Cake"). and extract them? I.e., get the following data frame:

   foo
0  Strawberry Jam
0  Apple
1  Banana
1  Pear Crumble
1  Potato
2  Almond Cake

I started with the following code:

df.loc[:, 'foo'].str.extractall('([A-Z]{1}[a-z]+)').copy()

However, this separates all words and doesn't use space to "connect" them. How would I include the latter?

Thanks.

Answer 1

Series.str.split + explode

df['foo'].str.split(r'(?<=[a-z])(?=[A-Z])').explode()

---

0    Strawberry Jam
0             Apple
1            Banana
1      Pear Crumble
1            Potato
2       Almond Cake
Name: foo, dtype: object

Regex details:

• (?<=[a-z]) : Positive Lookbehind matches the single character in the range a to z

• (?=[A-Z]) : Positive Lookahead matches the single character in the range A to Z

See the regex demo