How do I iterate through a string in Python?

Question

As an example, lets say I wanted to list the frequency of each letter of the alphabet in a string. What would be the easiest way to do it?

This is an example of what I'm thinking of... the question is how to make allTheLetters equal to said letters without something like allTheLetters = "abcdefg...xyz". In many other languages I could just do letter++ and increment my way through the alphabet, but thus far I haven't come across a way to do that in python.

def alphCount(text):   lowerText = text.lower()   for letter in allTheLetters:       print letter + ":", lowertext.count(letter) 

Answer 1

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!") 

Answer 2

the question is how to make allTheLetters equal to said letters without something like allTheLetters = "abcdefg...xyz"

That's actually provided by the string module, it's not like you have to manually type it yourself ;)

import string  allTheLetters = string.ascii_lowercase  def alphCount(text):   lowerText = text.lower()   for letter in allTheLetters:       print letter + ":", lowertext.count(letter)