Yes, you can use a regular expression for this:

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

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 regex:

>>> import re
>>> strs = "c1309, IF1306, v1309, p1209, a1309, mo1309"
>>> re.sub(r'\d','',strs)
'c, IF, v, p, a, mo'

or a faster version:

>>> re.sub(r'\d+','',strs)
'c, IF, v, p, a, mo'

timeit comparisons:

>>> strs = "c1309, IF1306, v1309, p1209, a1309, mo1309"*10**5

>>> %timeit re.sub(r'\d','',strs)
1 loops, best of 3: 1.23 s per loop

>>> %timeit re.sub(r'\d+','',strs)
1 loops, best of 3: 480 ms per loop

>>> %timeit ''.join([c for c in strs if not c.isdigit()])
1 loops, best of 3: 1.07 s per loop

#winner
>>> %timeit from string import digits;strs.translate(None, digits)
10 loops, best of 3: 20.4 ms per loop

Instead of strip, select for the numbers with a regular expression:

import re

numbers = re.compile(r'\d+(?:\.\d+)?')
numbers.findall("It took 2.3 seconds")

Demo:

>>> import re
>>> numbers = re.compile(r'\d+(?:\.\d+)?')
>>> numbers.findall("It took 2.3 seconds")
['2.3']

This returns a list of all matches; this lets you find multiple numbers in a string too:

>>> numbers.findall("It took between 2.3 and 42.31 seconds")
['2.3', '42.31']

You can remove all digits, dots, dashes and spaces from the start using str.lstrip():

string1.lstrip('0123456789.- ')

The argument to str.strip() is treated as a set, e.g. any character at the start of the string that is a member of that set is removed until the string no longer starts with such characters.

Demo:

>>> samples = """\
... 123.123.This is a string some other numbers
... 1. This is a string some numbers
... 12-3-12.This is a string 123
... 123-12This is a string 1234
... """.splitlines()
>>> for sample in samples:
...     print 'From: {!r}\nTo:   {!r}\n'.format(
...         sample, sample.lstrip('0123456789.- '))
...
From: '123.123.This is a string some other numbers'
To:   'This is a string some other numbers'

From: '1. This is a string some numbers'
To:   'This is a string some numbers'

From: '12-3-12.This is a string 123'
To:   'This is a string 123'

From: '123-12This is a string 1234'
To:   'This is a string 1234'

>>> import re
>>> re.sub("[^0-9]", "", "sdkjh987978asd098as0980a98sd")
'987978098098098'
>>> # or
>>> re.sub(r"\D", "", "sdkjh987978asd098as0980a98sd")
'987978098098098'

No, split would not work, because split only can work with a fixed string to split on.

You could use the str.rstrip() method:

import string

cleaned = yourstring.rstrip(string.digits)

This uses the string.digits constant as a convenient definition of what needs to be removed.

or you could use a regular expression to replace digits at the end with an empty string:

import re

cleaned = re.sub(r'\d+$', '', yourstring)