Delete digits in Python (Regex)

Question

I'm trying to delete all digits from a string. However the next code deletes as well digits contained in any word, and obviously I don't want that. I've been trying many regular expressions with no success.

Thanks!

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s = "This must not b3 delet3d, but the number at the end yes 134411"
s = re.sub("\d+", "", s)
print s

Result:

This must not b deletd, but the number at the end yes

Answer 1

Add a space before the \d+.

>>> s = "This must not b3 delet3d, but the number at the end yes 134411"
>>> s = re.sub(" \d+", " ", s)
>>> s
'This must not b3 delet3d, but the number at the end yes '

Edit: After looking at the comments, I decided to form a more complete answer. I think this accounts for all the cases.

s = re.sub("^\d+\s|\s\d+\s|\s\d+$", " ", s)

Answer 2

Try this:

"\b\d+\b"

That'll match only those digits that are not part of another word.

Answer 3

Using \s isn't very good, since it doesn't handle tabs, et al. A first cut at a better solution is:

re.sub(r"\b\d+\b", "", s)

Note that the pattern is a raw string because \b is normally the backspace escape for strings, and we want the special word boundary regex escape instead. A slightly fancier version is:

re.sub(r"$\d+\W+|\b\d+\b|\W+\d+$", "", s)

That tries to remove leading/trailing whitespace when there are digits at the beginning/end of the string. I say "tries" because if there are multiple numbers at the end then you still have some spaces.

Answer 4

To handle digit strings at the beginning of a line as well:

s = re.sub(r"(^|\W)\d+", "", s)

Answer 5

If your number is allways at the end of your strings try : re.sub("\d+$", "", s)

otherwise, you may try re.sub("(\s)\d+(\s)", "\1\2", s)

You can adjust the back-references to keep only one or two of the spaces (\s match any white separator)

Answer 6

Non-regex solution:

>>> s = "This must not b3 delet3d, but the number at the end yes 134411"
>>> " ".join([x for x in s.split(" ") if not x.isdigit()])
'This must not b3 delet3d, but the number at the end yes'

Splits by " ", and checks if the chunk is a number by doing str().isdigit(), then joins them back together. More verbosely (not using a list comprehension):

words = s.split(" ")
non_digits = []
for word in words:
    if not word.isdigit():
        non_digits.append(word)

" ".join(non_digits)

Answer 7

I don't know what your real situation looks like, but most of the answers look like they won't handle negative numbers or decimals,

re.sub(r"(\b|\s+\-?|^\-?)(\d+|\d*\.\d+)\b","")

The above should also handle things like,

"This must not b3 delet3d, but the number at the end yes -134.411"

But this is still incomplete - you probably need a more complete definition of what you can expect to find in the files you need to parse.

Edit: it's also worth noting that '\b' changes depending on the locale/character set you are using so you need to be a little careful with that.

Answer 8

You could try this

s = "This must not b3 delet3d, but the number at the end yes 134411"
re.sub("(\s\d+)","",s) 

result:

'This must not b3 delet3d, but the number at the end yes'

the same rule also applies to

s = "This must not b3 delet3d, 4566 but the number at the end yes 134411" 
re.sub("(\s\d+)","",s) 

result:

'This must not b3 delet3d, but the number at the end yes'

Answer 9

>>>s = "This must not b3 delet3d, but the number at the end yes 134411"
>>>s = re.sub(r"\d*$", "", s)
>>>s

"This must not b3 delet3d, but the number at the end yes "

This will remove the numericals at the end of the string.