Removing numbers from string
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'
And, just to throw it in the mix, is the oft-forgotten str.translate
which will work a lot faster than looping/regular expressions:
For Python 2:
from string import digits
s = 'abc123def456ghi789zero0'
res = s.translate(None, digits)
# 'abcdefghizero'
For Python 3:
from string import digits
s = 'abc123def456ghi789zero0'
remove_digits = str.maketrans('', '', digits)
res = s.translate(remove_digits)
# 'abcdefghizero'
Not sure if your teacher allows you to use filters but...
filter(lambda x: x.isalpha(), "a1a2a3s3d4f5fg6h")
returns-
'aaasdffgh'
Much more efficient than looping...
Example:
for i in range(10):
a.replace(str(i),'')