Replacing numbers with a placeholder in a string inclding decimals and percentages using re in Python
def remove_numbers(text):
remove = re.sub(r"\W\d\S*", " [DD]", text,)
return remove
The function works fine on this sample string. sample = "I can give you 10% of 100,000 to you. The thing went up by 10% so it costs 12.25 euros now. But if a string starts with a number, the first numer does not get replaced by the placeholder.
So looping through the replace method seems to be the easiest way to do this.
def remove_numbers(text):
nums = '123456787980'
for i in nums:
text = text.replace(i, '[DD]')
return text
\W will not match at the start of string. It appears you are using \W to make sure that the number you are replacing is not a part of a word. This makes sense. But, \W doesn't match at start-of-string. You can use \A for that. But, you probably don't want to add a space when you are replacing at start-of-string. This can be done in a single regex, but I think it results in easier-to-read code if you do it in two steps.
import re
def remove_numbers(text):
# replace internal numbers that are not a part of a word (adds a space)
remove = re.sub(r"\W\d\S*", " [DD]", text,)
# replace number at start of string (if any) (does not add a space)
remove = re.sub(r"\A\d\S*", "[DD]", remove,)
return remove
a = "3 foxes jumped over 3 fences"
b = remove_numbers(a)
print("before <{}>".format(a))
print("after <{}>".format(b))
\W requires a character to be there, so when you try it with a number at the beginning it'll look like just \d\S*.
Use '\b' instead of '\w' to match word boundaries:
def remove_numbers(text):
remove = re.sub(r"\b\d\S*", "[DD]", text,)
return remove
Or, keeping more in the spirit of your original code:
def remove_numbers(text):
remove = re.sub(r"(\s|^)\d\S*", r"\1[DD]", text,)
return remove
And use \d+ instead of \d if you want to also match multiple digits in a row.
Do this:
import re
def remove_numbers(text):
remove = re.sub(r"\W?\d\S*", " [DD]", text,)
return remove.strip()
print(remove_numbers())
The ? means 0 or more of the previous pattern
Change your regex to :
remove = re.sub("^\d+\s|\s\d+\s|\s\d+$", " [DD] ", text)
All code :
import re
def remove_numbers(text):
s = re.sub("^\d+\s|\s\d+\s|\s\d+$", " [DD] ", text)
return s
t1 = "3 foxes jumped over 3 fences"
print (remove_numbers(t1))
Output :
[DD] foxes jumped over [DD] fences
I can give 50% of 100,000 to you in cash. it went up by 2.3% and its costly?