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

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 2

You can use dictionary and range directly, so you can create your own function and easily customize it.

def gen_alphabet(start, value):
    return {chr(ord('a') + i) : 0 for i in range(value)}

print(gen_alphabet('a', 26))

OUTPUT:

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

Answer 4

My variant :

Note : range A-Z in unicode => 65-90 (decimal)

d = dict.fromkeys([chr(j) for j in range(65, 90)], 0)
print(d)

OUTPUT :

>>> {'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}

Answer 5

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})

Answer 6

I find this solution more elegant:

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