How to create a dict with letters as keys in a concise way?

Question

I created an dictionary of the 26 alphabet letters like this:

aDict={
    "a": 1,
    "b": 2,
    "c": 3,
    "d": 4,
    etc...
}

I'm trying make my code better and my question is, is there any shorter way to do this without typing all these numbers out?

Answer 1

You can use string.ascii_lowercase and dict comprehension here.

In [4]: from string import ascii_lowercase as al

For Python 2.7+:

In [5]: dic = {x:i for i, x in enumerate(al, 1)}

For Python 2.6 or earlier:

In [7]: dic = dict((y, x) for x, y in enumerate(al, 1))

Answer 2

aDict = dict(zip('abcdefghijklmnopqrstuvwxyz', range(1, 27)))

Or instead of hard coding the alphabet:

import string
aDict = dict(zip(string.ascii_lowercase, range(1, 27)))