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

Answer 1

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)))

or even:

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

The preferred solution might be a different one, depending on the actual values you want in the dict.

---

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

Use Counter from the collections module:

>>> import collections
>>> the_text = 'the quick brown fox jumps over the lazy dog'
>>> letter_counts = collections.Counter(the_text)
>>> letter_counts
Counter({' ': 8, 'o': 4, 'e': 3, ... 'n': 1, 'x': 1, 'k': 1, 'b': 1})