Match all punctuation not surrounded by alphanumeric characters?

Question

I am trying to write a regular expression that removes all non alphanumeric characters from a string, except for those that are surrounded by alphanumeric characters.

For example, consider the following three examples.

1.

it's -> it's

2.

its. -> its

3.

It's a: beautiful day? I'm =sure it is. The coca-cola (is frozen right?

It's a beautiful day I'm sure it is The coca-cola is frozen right

I am using Python's re module, and can match the opposite of what I am looking for with the following expression.

(?<=[a-zA-Z])[^a-zA-Z ](?=[a-zA-Z])

Any ideas?

Answer 1

Use

[^a-zA-Z\s](?!(?<=[a-zA-Z].)[a-zA-Z])

Regex proof

EXPLANATION

PATTERN	DETAILS
[^a-zA-Z\s]	non-letter and non-whitespace
(?!(?<=[a-zA-Z].)[a-zA-Z])	unmatch if followed and preceded with letter

Answer 2

If alphanumeric characters can also be word characters like \w (including an underscore) you can use word boundaries:

[^a-zA-Z\s](?<!\b.\b)

Explanation

• [^a-zA-Z\s]
• (?<!\b.\b) Negative lookbehind, assert not a char to the left surrounded by word boundaries

Regex demo

Or another alternative using a case insensitive match excluding chars A-Z and digits on the left and right:

[^a-zA-Z\s](?<![A-Z\d].(?=[A-Z\d]))

Regex demo