convert vim regex to python for re.sub

Question

I have a working regex under vim: /^ \\{-}\\a.*$\\n

I implement a global search and replace as :%s/^ \\{-}\\a.*$\\n// This works great -- removes all lines that start with any number of spaces (matched non-greedily), followed by a letter and anything else to the end of the line including the newline.

I cannot (to save my soul) figure out the analogous regex in Python. Here's what make sense to me:

x = re.sub("^ *?\a.$\n","",y)

But this doesn't do anything.

Many thanks for your sagacious replies.

Answer 1

\\a means the bell character (0x07) in Python, and $\\n is a redundant bad idea, so:

x = re.sub(r"^ *[A-Za-z].*\n","",y)

Also, there's no reason to write ' *?' instead of ' *' here, as it's always going to be followed by a non-space if it's matching.

Answer 2

If you want to match any number of whitespace, you can also use the \\s sequence.

Any letter will be matched by the [a-zA-Z] character class. You also don't need to use the $ and the \\n , either will do.

Suggest the following:

x = re.sub(r"^\s*[a-zA-Z].*(\r|\n)","",y)

If you want at least one whitespace, use \\s+ instead of \\s*