Python bit.ly 链接词表生成器

Question

我一直在开发一个 Python 工具来生成 bit.ly wordlist。以下是 bit.ly 链接的特殊性:

• 包含 7 个实体
• 以数字开头(通常为 3 或 2)
• 以字母结尾
• 不能并排放置相同的实体

我已经完成了前 3 个条件，但我找不到完成最后一个的方法。

from itertools import product

def firstN(chars, length):
    for firstNumber in product(chars, repeat=length):
        yield ''.join(firstNumber)

def combiwords(chars, length):
    for letters in product(chars, repeat=length):
        yield ''.join(letters)

def lastL(chars, length):
    for lastLetter in product(chars, repeat=length):
        yield ''.join(lastLetter)

def main():
    firstNumber = "32"
    letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
    lastLetter = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"

    for wordlen1 in range(1, 2):
        for first in firstN(firstNumber, wordlen1):
            for wordlen2 in range(6, 7):
                for combo in combiwords(letters, wordlen2):
                    for wordlen3 in range(1, 2):
                        for word in lastL(lastLetter, wordlen3):
                            print('https://bit.ly/' + first + combo + word)

if __name__=="__main__":
    main()

Answer 1

解决方案

我写了几个函数可以解决你的问题:

import random
import string

def generate_alphanumeric_string_without_sequential_repetitions():
    # string.ascii_lowercase means 'abcdefghijklmnopqrstuvwxyz'. If you need it to be case unsensitive, change this to string.ascii_lowercase
    allowed_letters = string.ascii_lowercase
    allowed_chars = allowed_letters + '0123456789'
    
    result = random.choice(allowed_chars) # init first char of the sequence
    
    # generate characters (which must be alphanumeric) in positions [1,5]
    for i in range(1, 5):
        random_char = random.choice(allowed_chars)
        
        # avoid sequential repetitions by re-generating a char if is the same of the previous one
        while random_char == result[i-1]:
            random_char = random.choice(allowed_chars)
        
        result = result + random_char
        
    # generate last char (which must be a letter)
    last_char = random.choice(allowed_letters)
    
    # avoid sequential repetitions by re-generating last char if is the same of the previous one
    while last_char == result[-1]: # result[-1] is a trick in python for getting the last char of a string
        last_char = random.choice(allowed_letters)

    return result + last_char


def generate_string():
    return str(random.randint(0, 9)) + generate_alphanumeric_string_without_sequential_repetitions()

结果

执行以下代码

random.seed(10)

result = list()

for i in range(0, 10):
    result.append(generate_string())
    
print(result)

我得到了以下结果:

['9c14ano', '7rkc75k', '1pxc0it', '5y0sq3f', '4xi3p2t', '6capimj', '8xpu92n', '7eu6kon', '3c5te8c', '2yxjhgo']

如有任何疑问，请在评论中提问。希望对您有所帮助!