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

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

print(letter_count)
# {'a': 0, 'b': 0, 'c': 0, ... 'x': 0, 'y': 0, 'z': 0}

or maybe:

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

print(letter_count)
# {'a': 0, 'b': 0, 'c': 0, ... 'x': 0, 'y': 0, 'z': 0}

or even:

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

print(letter_count)
# {'a': 0, 'b': 0, 'c': 0, ... 'x': 0, 'y': 0, 'z': 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)

print(letter_counts)
# Counter({' ': 8, 'o': 4, 'e': 3, ... 'n': 1, 'x': 1, 'k': 1, 'b': 1})

I find this solution more elegant:

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

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

import collections
d = collections.defaultdict(int)

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}