创建一个填充自定义格式文本的文本文件

Egco 发表于 Dev

爱可

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

例子： aaa00aaa bbb11bbb

我创建了一个代码，它使用较小的字符数组执行此操作，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？

奥马尔·阿弗拉克

您可以使用 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))

虽然这将花费相当多的时间......

此外，您不需要保留所有生成的密码的列表。您可以在生成密码时将密码写入 for 循环内的文件中。

编辑：

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

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