Remove All Punctuation From String, Except If It's Between Digits

I have a text that contains words and numbers. I'll give a representative example of the text: string = 'This is a 1example of the text. But, it only is 2.5 percent of all data' I

Solution 1:

You can use regex lookarounds like this:

(?<!\d)[.,;:](?!\d)

Working demo

The idea is to have a character class gathering the punctuation you want to replace and use lookarounds to match punctuation that does not have digits around

regex = r"(?<!\d)[.,;:](?!\d)"

test_str = "This is a 1example of the text. But, it only is 2.5 percent of all data"

result = re.sub(regex, "", test_str, 0)

Result is:

This is a 1example of the text But it only is2.5 percent of all data

Solution 2:

Okay folks, here is an answer (the best ? I don't know but it seems to work) :

item = "This is a 1example 2Ex of the text.But, it only is 2.5 percent of all data?"#if there is two strings contatenated with the second starting with capital letter
item = ' '.join(re.sub( r"([A-Z])", r" \1", item).split())
#if a word starts with a digit like "1example"
item = ' '.join(re.split(r'(\d+)([A-Za-z]+)', item) )
#Magical line that removes punctuation apart from floats
item = re.sub('\S+', lambda m: re.match(r'^\W*(.*\w)\W*$', m.group()).group(1), item)
item = item.replace("  "," ")
print item

Solution 3:

I am out of touch with Python, but have some insight into the regexps. My I suggest the usage of or? I would use this regexp: "(\d+)([a-zA-Z])|([a-zA-Z])(\d+)", and then as the replacement string use: "\1 \2"If some corner cases plague you, you can pass the back-reference to a procedure, and then deal 1-by-1, probably by checking if your "\1\2" can translate to float. TCL has such built-in functionality, Python should too.

Solution 4:

I tried this and it worked very well.

a = "This is a 1example of the text. But, it only is 2.5 percent of all data" a.replace(". ", " ").replace(", "," ")

Notice that, in replace function there is space after punctuation. I just replaced punctuation and space with only space.

Solution 5:

Code:

from itertools import groupby

s1 = "This is a 1example of the text. But, it only is 2.5 percent of all data"
s2 = [''.join(g) for _, g in groupby(s1, str.isalpha)]
s3 = ' '.join(s2).replace("   ", "  ").replace("  ", " ")

#you can keep adding a replace for each ponctuation
s4 = s3.replace(". ", " ").replace(", "," ").replace("; "," ").replace(", "," ").replace("- "," ").replace("? "," ").replace("! "," ").replace(" ("," ").replace(") "," ").replace('" '," ").replace(' "'," ").replace('... '," ").replace('/ '," ").replace(' “'," ").replace('” '," ").replace('] '," ").replace(' ['," ")

s5 = s4.replace("  ", " ")
print(s5)

Output:

'Thisis a 1 example of the text But it only is2.5 percent of all data'

P.s.: You can take a look at Punctuation Marks and keep adding them inside the .replace() function.