正则表达式仅提取字母数字单词

Question

I am looking for a regex to extract the word that ONLY contain alphanumeic characters:

string = 'This is a $dollar sign !!'
matches = re.findall(regex, string)
matches = ['This', 'is', 'sign']

This can be done by tokenizing the string and evaluate each token individually using the following regex:

^[a-zA-Z0-9]+$

Due to performance issues, I want to able to extract the alphanumeric tokens without tokenizing the whole string. The closest I got to was

regex = \b[a-zA-Z0-9]+\b

, but it still extracts substrings containing alphanumeric characters:

string = 'This is a $dollar sign !!'
matches = re.findall(regex, string)
matches = ['This', 'is', 'dollar', 'sign']

Is there a regex able to pull this off? I've tried different things but can't come up with a solution.

Answer 1

Instead of word boundaries, lookbehind and lookahead for spaces (or the beginning/end of the string):

(?:^|(?<= ))[a-zA-Z0-9]+(?= |$)

https://regex101.com/r/TZ7q1c/1

Note that "a" is a standalone alphanumeric word, so it's included too.

['This', 'is', 'a', 'sign']

Answer 2

There is no need to use regexs for this, python has a built in isalnum string method. See below:

string = 'This is a $dollar sign !!'

matches = [word for word in string.split(' ') if word.isalnum()]

Answer 3

[Edited thanks to Khabz's comment. I misunderstood the question]

Depending on your intention, you could also "split" instead of "match".

 >>> matches = re.split(r'(?:\s*\S*[\$\!]+\S*\s*|\s+)', string)

 ['This', 'is', 'a', 'sign', '']

And in case you need to remove leading or trailing empty string:

>>> matches = [x for x in re.split(r'(?:\s*\S*[\$\!]+\S*\s*|\s+)', a) if x ]
['This', 'is', 'a', 'sign']

CertainPerformance's respond using look behind and ahead is the most compact. Using split is sometimes advantages when the exclusion is specified, ie, the regex above describes what needs to be excluded. In this case, however, it is the inclusion of alpha-numeric that is specified, so using split() is not a good idea.