Need Regex that matches all patterns with format as `{word}{.,#}{word}` with strict matching

Question

So I have been trying to construct a regex that can detect the pattern {word}{.,#}{word} and seperate it into [word,',' (or '.','#'), word].

But i am not able to create one that does strict matching for this pattern and ignores everything else.

I used the following regex

r"[\w]+|[.]"

this one is doing well , but it doesnt do strict matching, as in if (,, # or .) characters dont occur in text, it will still give me words, which i dont want.

I would like to have a regex which strictly matches the above pattern and gives me the splits(using re.findall) and if not returns the whole word as it is.

Please Note: word on either side of the {,.#} , both words are not strictly to be present but atleast one should be present

Some example text for reference:

no.16         would give me ['no','.','16']
#400          would give me ['#,'400']
word1.word2   would give me ['word1','.','word2']

Looking forward to some help and assistance from all regex gurus out there

EDIT:

I forgot to add this. @viktor's version works as needed with only one problem, It ignores ALL other words during re.findall

eg. ONE TWO THREE #400 with the viktor's regex gives me ['','#','400']

but what was expected was ['ONE','TWO','THREE','#',400]

this can be done with NLTK or spacy, but use of those is a limitation.

Answer 1

I suggest using

(\w+)?([.,#])((?(1)\w*|\w+))

See the regex demo.

Details

• (\w+)? - An optional group #1: one or more word chars
• ([.,#]) - Group #2: ., , or #
• ((?(1)\w*|\w+)) - Group #3: if Group 1 matched, match zero or more word chars (the word is optional on the right side then), else, match one or more word chars (there must be a word on the right side of the punctuation chars since there is no word before them).

See the Python demo:

import re
pattern = re.compile(r'(\w+)?([.,#])((?(1)\w*|\w+))')
strings = ['no.16', '#400', 'word1.word2', 'word', '123']
for s in strings:
    print(s, ' -> ', pattern.findall(s))

Output:

no.16  ->  [('no', '.', '16')]
#400  ->  [('', '#', '400')]
word1.word2  ->  [('word1', '.', 'word2')]
word  ->  []
123  ->  []

The answer to your edit is

if re.search(r'\w[.,#]|[.,#]\w', text): 
    print( re.findall(r'[.,#]|[^\s.,#]+', text) )

If there is a word char, then any of the three punctuation symbols, and then a word char again in the input string, you can find and extract all occurrences of the [.,#]|[^\s.,#]+ pattern, namely a ., , or #, or one or more occurrences of any one or more chars other than whitespace, ., , and #.

Answer 2

I hope this code will solve your problem if you want to split the string by any of the mentioned special characters:

a='no.16'
b='#400'
c='word1.word2'

lst=[a, b, c]

for elem in lst:
    result= re.split('(\.|#|,)',elem)
    while('' in result):
        result.remove('')
    print(result)

Answer 3

You could do something like this:

import re

str = "no.16"

pattern = re.compile(r"(\w+)([.|#])(\w+)")

result = list(filter(None, pattern.split(str)))

The list(filter(...)) part is needed to remove the empty strings that split returns (see Python - re.split: extra empty strings that the beginning and end list).

However, this will only work if your string only contains these two words separated by one of the delimiters specified by you. If there is additional content before or after the pattern, this will also be returned by split.