Split on delimiter except when followed by space

Question

I'm trying to figure out how to split a string, keeping the delimiters, except when the delimiter is followed by a space. I seem to be most of the way there, except that the character immediately following the delimiter is retained with the delimiter.

What I have so far is the following:

>>> s='\nm222 some stuff \n more stuff'
>>> re.split('(\n[^ ])',s)
['', '\nm', '222 some stuff \n more stuff']

The result i need is

['', '\n', 'm222 some stuff \n more stuff']

What am I missing here? Thanks for the help.

Answer 1

Use a negative lookahead:

>>> s='\nm222 some stuff \n more stuff'
>>> re.split(r'(\n(?! ))', s)
['', '\n', 'm222 some stuff \n more stuff']

Your code,

re.split('(\n[^ ])',s)

Doesn't work because (\\n[^ ]) puts the "not a space" character in the same capturing group as \\n , giving you \\nm . (\\n(?! )) avoids consuming the "not a space" character, placing it in the next capturing group but still using it to split.

You can read more about lookaheads on the python regex documentation page .

Answer 2

Use \\n(?! ) . This is a negative lookahead

This will ensure the \\n is not followed by a space

---

If you wanted, you could even use \\n(?!\\s) . \\s includes a variety of whitespace characters like

• ' ' (a single space)
• \\t (tab)
• \\n (newline)
• \\r (carriage return)

Answer 3

你需要一个先行断言。

re.split('(\n(?=[^ ]))', s)