生成范围内的连续字符串

Question

I want to generate all strings from "aaa" down to "zzz" . Currently, I'm doing this using 3 for loops, is there a more pythonic way of doing this?

key_options = []
for n1 in range(ord('a'), ord('z')+1):
    for n2 in range(ord('a'), ord('z')+1):
        for n3 in range(ord('a'), ord('z')+1):
             key_options.append(chr(n1) + chr(n2) + chr(n3))

Answer 1

Use itertools.product and a list comprehension:

>>> from itertools import product
>>> from string import ascii_lowercase
>>> [''.join(p) for p in product(ascii_lowercase, repeat=3)]
['aaa', 'aab', 'aac', 'aad', 'aae', ..., 'zzv', 'zzw', 'zzx', 'zzy', 'zzz']

Answer 2

The itertools module is a much better way to do this sort of loop. The product function is what you'd use:

itertools.product(*iterables[, repeat])

Cartesian product of input iterables.

Equivalent to nested for-loops in a generator expression. For example, product(A, B) returns the same as ((x,y) for x in A for y in B) .

The string can provide the ASCII lowercase letters without using a range :

string.ascii_lowercase

The lowercase letters 'abcdefghijklmnopqrstuvwxyz' . This value is not locale-dependent and will not change.

Thus, you have

from itertools import product
from string import string

key_options = [''.join(n) for n in product(ascii_lowercase, repeat=3)]

Answer 3

>>> letters = [chr(i) for i in range(ord('a'), ord('z')+1)]
>>> letters
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']

import itertools
["".join(i) for i in itertools.product(letters,letters,letters)]

Output

['aaa', 'aab', 'aac', ... 'zzy', 'zzz']