“过滤掉已经使用过的密码”实际上是检查product函数的输出是否与一些使用过的密码集相匹配。（即使您编写了自己的特殊“product_with_exclusions”函数，它也需要这样工作。
您的代码已经执行了此检查，但可以使其更快。在对product的结果进行循环的每次迭代时，问题中的代码将密码列表转换为字符串，并检查输出。反复转换list -> str并检查密码是否是那个大字符串的一部分是很慢的。最好是（1）不转换为str，（2）使用set而不是list（与相同大小的列表相比，检查值是否在大集合中要快得多）。
这样，优化后的代码看起来就像这样。（这包括一些其他的整理-删除不必要的全局变量，在打开文件时与-block一起使用，并在追加模式下写入新密码。

from string import *
from itertools import product

# The password generator function.
# a little detail about the parameters:
# From = password starts from how many characters
# To = password ends with how many characters
# Fname = stands for 'file name' which is actually the database name
# cn, Base = these are just to organize different databases, not too much important in this case
def Generator(From, To, Fname, cn, Base):
    # Defining the database
    db = Fname+cn+'.txt'
    
    # Listing all the already used passwords in the 'lines_set' set for fast lookup
    with open(Fname+str(Base)+'.txt') as mainDB:
        lines_set = {l.strip() for l in mainDB.readlines()}

    # Defining the characters which (probably) can't be used as passwords(due to injection attacks and so on...)
    Blocked = ",:?/\\|<>*"+"'"+'"'
    
    # The ones that i'm not sure if they have to be blocked... but i'm blocking them anyway :|
    proBlocked = ";[]{}()+=^`"
    Blocked += proBlocked
    chars = ascii_letters + digits + punctuation
    
    # Deleting all the blocked characters from the main 'chars' string
    chars = [c for c in chars if c not in Blocked]

    # Going for generating the password trying every single password we can make and take a unique one out for use
    for i in range(From,To):
        # Starting the `product` module which is the main element of the program
        for j in product(chars, repeat=i):
            # Putting the generated word in the container(variable)
            word = "".join(j)

            # Checking if the password has already been used and added to the main database
            if word not in lines_set:
                # Adding to the database and returning the unique password
                with open(db, 'a') as p: # Open in append mode so we just write one new line
                    p.write(word + '\n')
                return word

这是输出的样子：

Type 1 to start generating, or type 0 to close the program:
1
agGM
Here's your unique password: agGM

只需要0.001ms！！！非常快，非常酷！！！a BIIIIGGGG thx to @slothrop！
我稍微清理了一下主代码：

from Generator import Generator
import os
import sys

while True:
    os.system('cls')
    inp = input("Type 1 to start generating, or type 0 to close the program:\n")
    if inp.isdigit():
        if inp == '1':
            break
        elif inp == '0':
            sys.exit()
gen = Generator(4, 10, "MainDB", '1', 1)
db = open("MainDB1.txt")
lines = set(db.readlines())
pwd = ''
for x in gen:
    print(x)
    if x not in lines:
        pwd = x
        break
print("Here's your unique password: "+pwd)

函数文件：

from string import *
from itertools import product

def Generator(From, To, Fname, cn, Base):
    db = 'MainDB1.txt'
    p = open(db, '+r')
    lines = []
    mainDB = open('MainDB1.txt', '+r')
    for x in mainDB.readlines():
        lines.append(x.replace('\n', ''))
    mainDB.close()
    lines = set(lines)
    Blocked = ",:?/\\|<>*"+"'"+'"'
    proBlocked = ";[]{}()+=^`"
    Blocked += proBlocked
    chars = ascii_letters + digits + punctuation
    for c in Blocked:
        chars = chars.replace(c,'')

    for i in range(From,To):
        for j in product(chars, repeat=i):
            word = "".join(j)
            if word not in lines:
                p.write(p.read() + word + '\n')
                yield word

我得说这真的是我收到的最好的问题！结果比我期待的要好得多！再次感谢@slothrop！