Python regex to remove apostrophe in contractions

Question

I'd like to use regex to remove the apostrophes in common contractions. For example, I'd like to map

test1 test2 can't test3 test4 won't

to

test1 test2 cant test3 test4 wont

My current naive approach is just to manually sub all the contractions I want to use.

def remove_contraction_apostrophes(input):
    text = re.sub('can.t', 'cant', input)                                       
    text = re.sub('isn.t', 'isnt', text)                                       
    text = re.sub('won.t', 'wont', text)                                       
    text = re.sub('aren.t', 'arent', text)  
    return text

(I'm using can.t because in the text I am parsing, it can use multiple characters for the apostrophe, like can't and can`t).

This is pretty unwieldy as I want to add all the common contractions. Is there a better way of doing this with regex, where I could construct a regex of this type by inputting a list of contractions? Or am I better off just listing them all like this?

It also may be possible to just work with the endings, like 'll, n't etc, but I'm a afraid of catching other things besides contractions with this.

Answer 1

I would do something like this.

import re

def remove_contraction_apostraphes(input):
    text = re.sub('([A-Za-z]+)[\'`]([A-Za-z]+)', r'\1'r'\2', input)                                       
    return text

print(remove_contraction_apostraphes("can't"))

---

1. It matches one or more letters [A-Za-z]+

• things in square brackets means one of these characters, the plus means at least one or more of what comes before

2. followed by one of the following ' or `

3. followed by one or more letters

and replaces it with

1. what was found in the first set of parenthesis r'\1'

• r'\1' returns the pattern that was matched by the first ([A-Za-z]+)

2. followed by what was found in the second set of parenthesis r'\2'

---

If you have other characters, such as �, and you know what they all are you can place them with the square brackets. This line will match any of those characters, and account for the chance of white spaces by the apostrophe

text = re.sub('([A-Za-z]+)\s?[\'`�]\s?([A-Za-z]+)', r'\1'r'\2', input)       

• /s : Any white space
• ? : 0 or 1 of the previous

---

You could also use [^A-Za-z0-9]

    text = re.sub('([A-Za-z]+)[^A-Za-z0-9]([A-Za-z]+)', r'\1'r'\2', input)     

to match any any number of character's followed by any character which isn't a letter or a number, followed by any number of character's. If you want to add the \s? in there, I would recommend adding \., \?, \!, \: ... to you regex making it '([A-Za-z]+)\s?[^A-Za-z0-9\.\!\?\:]s?([A-Za-z]+)' because otherwise your regex will match things like the ends of sentences, which are not contractions

---

This will match any contraction, no matter how letters before or after the apostrophe there are. You will need to put all the different apostrophe's that you have within the ['`] block

Answer 2

Use look arounds to check for letters either side of an "apostrophe":

text = re.sub("(?<=\w)[‘`’'](?=\w)", '', input) 
                              

Look arounds assert, without consuming, preceding/following input.

—--

import re
input = "I can’t understand what's wrong"
text = re.sub("(?<=\w)[‘`’'](?=\w)", '', input)
print(text)

Produces

I cant understand whats wrong

See live demo.

Answer 3

Regular expressions let you easily list a set of alternatives.

def remove_contraction_apostrophes(input):
    text = re.sub(r'\b(are|ca|is|wo)n.t\b', r'\1nt', input)
    text = re.sub(r'\b(I|[hw]e|it|she|they|you).ll\b', r'\1ll', text)

In re.sub, the back reference \1 recalls the text which matched the first parenthesized subexpression, in the replacement too. (\2 gets the second, etc.)

Notice also the addition of word-boundary anchors \b to prevent the regex from matching in the middle of a longer word, like volcanity.

Answer 4

You can just simply do like this:

t="test1 test2 can't test3 test4 won't"
re.sub("\'","",t)