行首的可选匹配

Question

I am trying to create a regular expression in Python that matches #hashtags. My definition on a hashtag is:

• It is a work that starts with a #
• It can contain all characters except [ ,\\.]
• It can be anywhere in the text

So in this text

#This string cont#ains #four, and #only four #hashtags.

The hashes here are This , four , only and hashtags .

The problem I have is the optional check for the beginning of line.

• [ \\.,]+ won't do it since it won't match the optional beginning.
• [ \\.,]? won't do it since it matches too much.

Example with +

In []: re.findall('[ \.,]+#([^ \.,]+)', '#This string cont#ains #four, and #only four #hashtags.')
Out[]: ['four', 'only', 'hashtags']

Example with ?

In []: re.findall('[ \.,]?#([^ \.,]+)', '#This string cont#ains #four, and #only four #hashtags.')
Out[]: ['This', 'ains', 'four', 'only', 'hashtags']

How can optional match the beginning of the line?

Answer 1

This seems to work:

>>> re.findall(r'\B#([^,\W]+)', '#This string cont#ains #four, and #only four #hashtags.')
['This', 'four', 'only', 'hashtags']

• \\B : Matches the empty string, but only when it is not at the beginning or end of a word. This means that r'py\\B' matches 'python' , 'py3' , 'py2' , but not 'py' , 'py.' , or 'py!' . \\B is just the opposite of \\b , so is also subject to the settings of LOCALE and UNICODE .
• \\W : When the LOCALE and UNICODE flags are not specified, matches any non-alphanumeric character; this is equivalent to the set [^a-zA-Z0-9_] . With LOCALE, it will match any character not in the set [0-9_] , and not defined as alphanumeric for the current locale. If UNICODE is set, this will match anything other than [0-9_] plus characters classied as not alphanumeric in the Unicode character properties database.

Answer 2

Before your regex you can just tell what you don't want.

(?<!\w)(#[^ \.,]+)

With negative lookbehind you can do that