创建按字母顺序升序的列表

你可以利用 itertools.product()

from itertools import product
from string import ascii_lowercase

for i in range(1, 4):
    for x in product(ascii_lowercase, repeat=i):
        print(''.join(x))

首先,您需要所有字母,然后是所有对,然后是所有三元组,等等。这就是为什么我们首先需要迭代所有您想要的字符串长度( for i in range(...)

然后,我们需要所有可能的关联 i product(ascii_lowercase) for 循环重复 我 时代。

我 需要,最后才是 join()

对于 while :

def generate():
    i = 0
    while True:
        i += 1
        for x in product(ascii_lowercase, repeat=i):
            yield ''.join(x)

generator = generate()
next(generator)  # 'a'
next(generator)  # 'b'
...

对于一般解决方案,我们可以使用生成器和 islice 从…起 itertools

import string
from itertools import islice
def generate():
    base = ['']
    while True:
        next_base = []
        for b in base:
            for i in range(26):
                next_base.append(b + string.ascii_lowercase[i])
                yield next_base[-1]
        base = next_base

print('\n'.join(islice(generate(), 1000)))

以及输出:

a
b
c
...
z
aa
ab
...
zz
aaa
aab
...

你可以用 小岛 需要多少线就拿多少线。

尝试:

>>import string
>>string.ascii_lowercase
'abcdefghijklmnopqrstuvwxyz'
>>len(string.ascii_lowercase)
26

当您在下线的索引超过26时,将引发异常

div = int(i/26)

,因为ascii_小写长度:

for i in range(26*26):  # <--- 26 is string.ascii_lowercase
    mod = int(i%26)
    div = int(i/26)
    print(string.ascii_lowercase[div]+string.ascii_lowercase[mod])

编辑:

或者您可以使用:

import string

n = 4  # number of chars
small_limit = len(string.ascii_lowercase)
limit = small_limit ** n
i = 0
while i < limit:
    s = ''
    for c in range(n):
        index = int(i/(small_limit**c))%small_limit
        s += string.ascii_lowercase[index]
    print(s)
    i += 1

from string import ascii_lowercase
l = list(ascii_lowercase) + [letter1+letter2 for letter1 in ascii_lowercase for letter2 in ascii_lowercase]+ [letter1+letter2+letter3 for letter1 in ascii_lowercase for letter2 in ascii_lowercase for letter3 in ascii_lowercase]

上面提供了这个问题的答案 Code Review SE

def increment_char(c):
    return chr(ord(c) + 1) if c != 'z' else 'a' 

def increment_str(s):
    lpart = s.rstrip('z')
    num_replacements = len(s) - len(lpart)
    new_s = lpart[:-1] + increment_char(lpart[-1]) if lpart else 'a' 
    new_s += 'a' * num_replacements
    return new_s

s = ''
for _ in range(1000):
    s = increment_str(s)
    print(s)