The question you've asked (how to iterate through the alphabet) is not the same question as the problem you're trying to solve (how to count the frequency of letters in a string).
You can use string.lowercase, as other posters have suggested:
import string
allTheLetters = string.lowercase
To do things the way you're "used to", treating letters as numbers, you can use the "ord" and "chr" functions. There's absolutely no reason to ever do exactly this, but maybe it comes closer to what you're actually trying to figure out:
def getAllTheLetters(begin='a', end='z'):
beginNum = ord(begin)
endNum = ord(end)
for number in xrange(beginNum, endNum+1):
yield chr(number)
You can tell it does the right thing because this code prints True
:
import string
print ''.join(getAllTheLetters()) == string.lowercase
But, to solve the problem you're actually trying to solve, you want to use a dictionary and collect the letters as you go:
from collections import defaultdict
def letterOccurrances(string):
frequencies = defaultdict(lambda: 0)
for character in string:
frequencies[character.lower()] += 1
return frequencies
Use like so:
occs = letterOccurrances("Hello, world!")
print occs['l']
print occs['h']
This will print '3' and '1' respectively.
Note that this works for unicode as well:
# -*- coding: utf-8 -*-
occs = letterOccurrances(u"héĺĺó, ẃóŕĺd!")
print occs[u'l']
print occs[u'ĺ']
If you were to try the other approach on unicode (incrementing through every character) you'd be waiting a long time; there are millions of unicode characters.
To implement your original function (print the counts of each letter in alphabetical order) in terms of this:
def alphCount(text):
for character, count in sorted(letterOccurrances(text).iteritems()):
print "%s: %s" % (character, count)
alphCount("hello, world!")