Delete digits in Python (Regex)

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)", "12", s)

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

Try this:

"bd+b"

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

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

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

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"bd+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"$d+W+|bd+b|W+d+$", "", 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.

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)

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.

>>>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.

You could try this

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

result:

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

To match only pure integers in a string:

b(?<![0-9-])(d+)(?![0-9-])b

It does the right thing with this, matching only everything after million:

max-3 cvd-19 agent-007 8-zoo 2ab c3d ef4 55g h66i jk77 
8m9n o0p2     million     0 22 333  4444

All of the other 8 regex answers on this page fail in various ways with that input.

The dash at the end by that first 0-9 … [0-9-] … preserves -007 and the dash in the second set preserves 8-.

Or d in place of 0-9 if you prefer

at regex101

Can it be simplified?

I had a light-bulb moment, I tried and it works:

sol = re.sub(r'[~^0-9]', '', 'aas30dsa20')

output:

aasdsa