Is there a fast way to generate a dict of the alphabet in Python?

Question

I want to generate a dict with the letters of the alphabet as the keys, something like

letter_count = {'a': 0, 'b': 0, 'c': 0}

what would be a fast way of generating that dict, rather than me having to type it in?

Thanks for your help.

EDIT
Thanks everyone for your solutions :)

nosklo's solution is probably the shortest

Also, thanks for reminding me about the Python string module.

Answer 1

There's this too:

import string
letter_count = dict((letter, 0) for letter in string.ascii_lowercase)

Answer 2

Here's a compact version, using a list comprehension:

>>> import string
>>> letter_count = dict( (key, 0) for key in string.ascii_lowercase )
>>> letter_count
{'a': 0, 'c': 0, 'b': 0, 'e': 0, 'd': 0, 'g': 0, 'f': 0, 'i': 0, 'h': 0, 'k': 0,
 'j': 0, 'm': 0, 'l': 0, 'o': 0, 'n': 0, 'q': 0, 'p': 0, 's': 0, 'r': 0, 'u': 0, 
't': 0, 'w': 0, 'v': 0, 'y': 0, 'x': 0, 'z': 0}

Answer 3

import string
letter_count = dict(zip(string.ascii_lowercase, [0]*26))

Or maybe:

import string
import itertools
letter_count = dict(zip(string.lowercase, itertools.repeat(0)))

Can't decide which one I like more at the moment.

EDIT: I've decided that I like nosklo's the best :-)

import string
letter_count = dict.fromkeys(string.ascii_lowercase, 0)

---

I'll take a guess here: do you want to count occurences of letters in a text (or something similar)? There are better ways to do this than starting with an initialized dictionary.

For one, there is defaultdict in the collections module:

>>> import collections
>>> letter_count = collections.defaultdict(int)
>>> the_text = 'the quick brown fox jumps over the lazy dog'
>>> for letter in the_text: letter_count[letter] += 1
... 
>>> letter_counts
defaultdict(<type 'int'>, {' ': 8, 'a': 1, 'c': 1, 'b': 1, ... 'z': 1})

Then there is this way:

>>> dict((c, s.count(c)) for c in string.ascii_lowercase)
{'a': 1, 'b': 1, 'c': 1, ... 'z': 1}

Horrible performance, but short and easy to read. Might be worth it, if performance doesn't matter.

Answer 4

import string
letters = string.ascii_lowercase
d = dict(zip(letters, [0]*len(letters))

Answer 5

Yet another 1-liner Python hack:

letter_count = dict([(chr(i),0) for i in range(97,123)])

Answer 6

If you plan to use it for counting, I suggest the following:

import collections
d = collections.defaultdict(int)

Answer 7

I find this solution more elegant:

import string
d = dict.fromkeys(string.ascii_lowercase, 0)