创建一个填充自定义格式文本的文本文件答案

【问题标题】：Creating a text file filled with custom formatted text创建一个填充自定义格式文本的文本文件
【发布时间】：2021-05-10 22:48:34
【问题描述】：

我正在尝试创建一个密码列表，其中包含以下格式的所有组合：

3 个字符 2 个数字 3 个字符

例如：aaa00aaabbb11bbb

我创建了一个代码，它使用较小的字符数组，但是当我尝试使用 ASCII 系统中的所有字符时，代码会出现内存错误。

import string

numbers = ["0","1","2","3","4","5","6","7","8","9"]
characters = ["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"]

characterCombs = []
numbCombs = []
allCombs = []

for el1 in characters:
    for el2 in characters:
        for el3 in characters:
            characterCombs.append(el1+el2+el3)

for n1 in numbers:
    for n2 in numbers:
        numbCombs.append(n1 + n2)

for ch1 in characterCombs:
    for no in numbCombs:
        for ch2 in characterCombs:
            allCombs.append(ch1+no+ch2)
for i in allCombs:
    f.write(i+"\n")
f.close()

有没有办法优化这段代码，或者我应该改变我的方法并找到不同的方法来组合所有这些字符？

【问题讨论】：

标签： python python-3.x passwords combinations

【解决方案1】：

可以使用pythonitertools模块生成组合。

from itertools import permutations

numbers = ["0","1","2","3","4","5","6","7","8","9"]
characters = ["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"]

for a in permutations(characters, 3):
    for b in permutations(numbers, 2):
        for c in permutations(characters, 3):
            print("".join(a + b + c))

不过这需要相当长的时间......

此外，您不需要保留所有生成密码的列表。您可以在生成密码时将密码写入文件中的 for 循环中。

编辑：

正如下面提到的@Andreas，它确实应该是所有排列（而不是组合）。我更新了代码。

【讨论】：

我认为你需要排列而不是组合，它应该是 30.891.577.600 种可能的组合。
是的，我更新了答案，感谢您的关注。
感谢您的回答。我不知道 itertools 并试图自己创建一切。这让我的工作变得更轻松了。

【解决方案2】：

你的问题是你对for的作用缺乏了解。
出于某种原因，您认为它随机选择一个值，但这不是它的作用。 for 循环遍历给定迭代中的每个元素。
根据第一部分的结果有多大，您会得到一个内存错误是有道理的。

for el1 in characters:
    for el2 in characters:
        for el3 in characters:
            characterCombs.append(el1+el2+el3)

这里发生的事情是这样的：

el1 = "a"
    el2 = "a"
        el3 = "a"
            characterCombs[aaa]
        el3 = "b"
            characterCombs[aaa, aab]
        el3 = "c"
            characterCombs[aaa, aab, aac]
        ...
    el2 = "b"
        el3 = "a"
            characterCombs[aaa, aab, aac, ..., aba]

这很容易使用随机化来解决。这是使用random.choice

import random

def main():
    numbers = ["0","1","2","3","4","5","6","7","8","9"]
    characters = ["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"]

pass_format = ((3, characters), (2, numbers), (3, characters))

pass_choices = tuple(random.choice(lst[1]) for lst in pass_format for _ in range(lst[0]))
print(pass_choices)


if __name__ == '__main__':
    main()

它打印： ('r', 'w', 'c', '3', '9', 'n', 'j', 'x')

另一个选项，使用random.choices：

def main():
    numbers = ["0","1","2","3","4","5","6","7","8","9"]
    characters = ["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"]

    pass_format = ((3, characters), (2, numbers), (3, characters))

    pass_choices = []
    for section in pass_format:
        pass_choices.extend(random.choices(section[1], k=section[0]))
    print(pass_choices)


if __name__ == '__main__':
    main()

【讨论】：