python regular expression, pulling all letters out

Question

Is there a better way to pull A and F from this: A13:F20

a="A13:F20"
import re
pattern = re.compile(r'\D+\d+\D+')
matches = re.search(pattern, a)
num = matches.group(0)
print num[0]
print num[len(num)-1]

output

A F

note: the digits are of unknown length

Answer 1

You don't have to use regular expressions, or re at all. Assuming you want just letters to remain, you could do something like this:

a = "A13:F20"
a = filter(lambda x: x.isalpha(), a)

Answer 2

Use a simple list comprehension, as a filter and get only the alphabets from the actual string.

print [char for char in input_string if char.isalpha()]
# ['A', 'F']

Answer 3

I'd do it like this:

>>> re.findall(r'[a-z]', a, re.IGNORECASE)
['A', 'F']

Answer 4

You could use re.sub:

>>> a="A13.F20"
>>> re.sub(r'[^A-Z]', '', a)     # Remove everything apart from A-Z
'AF'
>>> re.sub(r'[A-Z]', '', a)      # Remove A-Z
'13.20'
>>> 

Answer 5

If you're working with strings that all have the same format, you can just cut out substrings:

a="A13:F20"
print a[0], a[4]

More on python slicing in this answer: Is there a way to substring a string in Python?