Let's break down exactly what's happening here to understand that confusing empty string:

1. How re.split behaves with capture groups #

When you use a capture group in your regex pattern for re.split, two key rules kick in:

• The string gets split at every position where the pattern matches
• Every matched separator (from the capture group) is added as a separate element in the resulting list

2. Step-by-step breakdown of your example #

Let's walk through splitting "Hello! How are you?" with r'([!\s])' step by step:

• First match: the ! right after "Hello".
  • The substring before this match is "Hello" → added to the list.
  • The matched separator ! → added to the list.
• Next, we start looking for the next match immediately after the !. The very next character is a space ( ), which matches our pattern.
  • The substring between the ! and this space is... nothing. There are zero characters between them, so this becomes the empty string '' → added to the list.
  • The matched space → added to the list.
• From there, the rest of the splits work as expected: the space before "are" splits out "How", then the space itself, then "are", and so on.

That's exactly where that third empty string comes from—it's the zero-length gap between two consecutive separators (! followed immediately by a space).

3. Why filtering empty strings still reconstructs the original string #

You noticed that both ''.join(t) and ''.join([ch for ch in t if ch]) give the original string. That's because empty strings contribute nothing to the joined result. But re.split doesn't omit them automatically because it follows a strict rule: every segment between split points (even empty ones) gets included, along with the matched separators.

This is different from the non-capture group scenario you mentioned (like splitting '/segment/segment/' with '/'), where empty strings at the start/end are necessary to fully reconstruct the original string. In your capture group case, the empty string is just a byproduct of two separators being adjacent, but since it doesn't affect the join result, you can safely filter it out without losing any data.

内容的提问来源于stack exchange，提问作者robertspierre