CODEWITHSUNDEEP
regexpython-3.xnlpnumeric-ranges
kawai
kawai

Reputation: 7

Numeric Ranges with a Regular Expression python

So I am working on a text analytics problem and I am trying to remove all the numbers between 0 and 999 with regular expression in Python. I have tried Regex Numeric Range Generator to get the regular expression but I didn't succed. I can only remove all the numbers.

I have tried several Regex but it didn't work. here's what I tried

# Remove numbers starting from 0 ==> 999
data_to_clean = re.sub('[^[0-9]{1,3}$]', ' ', data_to_clean)

I have tried this also:

# Remove numbers starting from 0 ==> 999
data_to_clean = re.sub('\b([0-9]|[1-8][0-9]|9[0-9]|[1-8][0-9]{2}|9[0-8][0-9]|99[0-9])\b', ' ', data_to_clean)

this one:

^([0-9]|[1-8][0-9]|9[0-9]|[1-8][0-9]{2}|9[0-8][0-9]|99[0-9])$

and this:

def clean_data(data_to_clean):
    # Remove numbers starting from 0 ==> 999
    data_to_clean = re.sub('[^[0-9]{1,3}$]', ' ', data_to_clean)  
    return data_to_clean

I have a lot of numbers but I need to delete just the ones under 3 decimals and keep the other.

Thank You for your help

Upvotes: 0

Views: 404

Answers (3)

krisz
krisz

Reputation: 2695

You need precede the pattern string with an r to prevent escaping so the interpeter won't swap \b with a backspace. Plus you can simplify the pattern like this:

data_to_clean = re.sub(r'\b([0-9]|[1-9][0-9]{1,2})\b', ' ', data_to_clean)

Upvotes: 1

JGNI
JGNI

Reputation: 4013

Numbers from 0 to 999 are

  1. A single character [0-9]
  2. Two characters [1-9][0-9]
  3. Three characters [1-9][0-9][0-9]

This gives a naive regex of /\b(?:[0-9]|[1-9][0-9]|[1-9][0-9][0-9])\b/ However we have duplicated characters classes in the options so we can factor them out

/(?!\b0[0-9])\b[0-9]{1,3}\b/

This works by using a negative lookahead (?!\b0[0-9]) to check for the start of a word followed by a 0 followed by a digit to disregard 01 etc. and then looks for 1 to three 0 - 9 characters. Because the negative lookahead needs at least 2 characters a single 0 still passes as valid.

Upvotes: 0

gaw
gaw

Reputation: 1960

I think you can use a combination of your try with word boundaries (\b) and your last try ([0-9]{1,3}).

So the resulting regex should look like: \b[0-9]{1,3}\b

If you check the demo: regex101.com/r/qDrobh/6 It should replace all 1-digit, 2-digit and 3-digit numbers and ignore higher numbers and other words.

Upvotes: 0

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