Yes, you can use a regular expression for this:

import re
output = re.sub(r'\d+', '', '123hello 456world')
print output  # 'hello world'

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)

Would this work for your situation?

>>> s = '12abcd405'
>>> result = ''.join([i for i in s if not i.isdigit()])
>>> result
'abcd'

This makes use of a list comprehension, and what is happening here is similar to this structure:

no_digits = []
# Iterate through the string, adding non-numbers to the no_digits list
for i in s:
    if not i.isdigit():
        no_digits.append(i)

# Now join all elements of the list with '', 
# which puts all of the characters together.
result = ''.join(no_digits)

As @AshwiniChaudhary and @KirkStrauser point out, you actually do not need to use the brackets in the one-liner, making the piece inside the parentheses a generator expression (more efficient than a list comprehension). Even if this doesn't fit the requirements for your assignment, it is something you should read about eventually :) :

>>> s = '12abcd405'
>>> result = ''.join(i for i in s if not i.isdigit())
>>> result
'abcd'