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)

Asked By: Lawrence Johnston

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Source

Answer 1

Something like this?

for letter in range(ord('a'), ord('z') + 1):
  print chr(letter) + ":", lowertext.count(chr(letter))

Answered By: Jacob Krall

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)

Answered By: Matthew Trevor

Answer 3

Do you mean using:

import string
string.ascii_lowercase

then,

counters = dict()
for letter in string.ascii_lowercase:
    counters[letter] = lowertext.count(letter)

All lowercase letters are accounted for, missing counters will have zero value.

using generators:

counters = 
    dict( (letter,lowertext.count(letter)) for letter in string.ascii_lowercase )

Answered By: gimel

Answer 4

If you just want to do a frequency count of a string, try this:

s = 'hi there'
f = {}

for c in s:
        f[c] = f.get(c, 0) + 1

print f

Answered By: Adam Pierce

Answer 5

Main question is “iterate through the alphabet”:

import string
for c in string.lowercase:
    print c

How get letter frequencies with some efficiency and without counting non-letter characters:

import string

sample = "Hello there, this is a test!"
letter_freq = dict((c,0) for c in string.lowercase)

for c in [c for c in sample.lower() if c.isalpha()]:
    letter_freq[c] += 1

print letter_freq

Answered By: mhawke

Answer 6

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!")

Answered By: Glyph

Answer 7

For counting objects, the obvious solution is the Counter

from collections import Counter
import string

c = Counter()
for letter in text.lower():
    c[letter] += 1

for letter in string.lowercase:
    print("%s: %d" % (letter, c[letter]))

Answered By: Kimvais

Answer 8

This is what I do:

import string
for x in list(string.lowercase):
    print x

Answered By: masnun

Answer 9

How about this, to use letters, figures and punctuation (all usable to form a Django key):

import random
import string

chars = string.letters + string.digits + string.punctuation
chars_len = len(chars)
n = 40

print(''.join([chars[random.randint(0, chars_len)] for i in range(n)]))

Example result: coOL:V!D+P,&S*hzbO{a0_6]2!{4|OIbVuAbq0:

Answered By: user1956583

Answer 10

Just use:

import string
string.lowercase  
string.uppercase

or

string.letters[:26]  
string.letters[26:]

Answered By: guest

How do I iterate through a string in Python?

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