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'

You can use:

>>> line = "1234567 7852853427.111 https://en.wikipedia.org/wiki/Dictionary_(disambiguation)" 
>>> print re.sub(r'\b\d+(?:\.\d+)?\s+', '', line)

https://en.wikipedia.org/wiki/Dictionary_(disambiguation)

Regex \b\d+(?:\.\d+)?\s+ will match an integer or decimal number followed by 1 or more spaces. \b is for word boundary.

Yes, you can use a regular expression for this:

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

Why not just use word boundaries?

\b\d+\b

Here is an example:

>>> import re
>>> words = ['python 3', 'python3', '1something', '2', '434', 'python 35', '1 ', ' 232']
>>> for word in words:
...     print("'{}' => '{}'".format(word, re.sub(r'\b\d+\b', '', word)))
...
'python 3' => 'python '
'python3' => 'python3'
'1something' => '1something'
'2' => ''
'434' => ''
'python 35' => 'python '
'1 ' => ' '
' 232' => ' '

Note that this will not remove spaces before and after. I would advise using strip(), but if not you can probably do \b\d+\b\s* (for space after) or something similar.

The pattern '1-9' will match the string 1-9. If you want to specify a set of characters, you should enclose it with []:

print(re.sub('[1-9]', '', 'Name 1with 12numbers'))
# >> Name with numbers

your regex is wrong:

re.sub(r'[0-9]',"",str)

should work:

>>> str="aaaa22222111111kkkkk"
>>> re.sub(r'[0-9]',"",str)
'aaaakkkkk'