如何在不删除这些字符的情况下拆分带有特殊字符的字符串？

Question

I'm writing this function which needs to return an abbreviated version of a str . The return str must contain the first letter, number of characters removed and the, last letter;it must be abbreviated per word and not by sentence, then after that I need to join every word again with the same format including the special-characters. I tried using the re.findall() method but it automatically removes the special-characters so I can't use " ".join() because it will leave out the special-characters.

Here's my code:

import re
def abbreviate(wrd):
    return " ".join([i if len(i) < 4 else i[0] + str(len(i[1:-1])) + i[-1] for i in re.findall(r"[\w']+", wrd)]) 

print(abbreviate("elephant-rides are really fun!"))

The output would be:

e6t r3s are r4y fun

But the output should be:

e6t-r3s are r4y fun!

Answer 1

No need for str.join . Might as well take full advantage of what the re module has to offer.

re.sub accepts a string or a callable object (like a function or lambda), which takes the current match as an input and must return a string with which to replace the current match.

import re

pattern = "\\b[a-z]([a-z]{2,})[a-z]\\b"
string = "elephant-rides are really fun!"

def replace(match):
    return f"{match.group(0)[0]}{len(match.group(1))}{match.group(0)[-1]}"

abbreviated = re.sub(pattern, replace, string)

print(abbreviated)

Output:

e6t-r3s are r4y fun!
>>> 

Maybe someone else can improve upon this answer with a cuter pattern, or any other suggestions. The way the pattern is written now, it assumes that you're only dealing with lowercase letters, so that's something to keep in mind - but it should be pretty straightforward to modify it to suit your needs. I'm not really a fan of the repetition of [az] , but that's just the quickest way I could think of for capturing the "inner" characters of a word in a separate capturing group. You may also want to consider what should happen with words/contractions like "don't" or "shouldn't" .

Answer 2

Thank you for viewing my question. After a few more searches, trial, and error I finally found a way to execute my code properly without changing it too much. I simply substituted re.findall(r"[\w']+", wrd) with re.split(r'([\W\d\_])', wrd) and also removed the whitespace in "".join() for they were simply not needed anymore.

    import re
    def abbreviate(wrd):
        return "".join([i if len(i) < 4 else i[0] + str(len(i[1:-1])) + i[-1] for i in re.split(r'([\W\d\_])', wrd)])

     print(abbreviate("elephant-rides are not fun!"))

Output:

     e6t-r3s are not fun!