Python 如何从列表中删除报价？

我正在为我的一个朋友制作一个

discord.py

bot。我想这样做，你可以添加愚蠢的报价列表，然后有他们随机选择。我已经完成了这么多

然而，我也希望能够删除不相关的引用、错误的引用等。我想知道如何才能做到这一点

以下是到目前为止我得到的信息：

import discord
from discord.ext.commands import Bot
from discord.ext import commands
import random
import pickle
import os

Client = discord.Client()
bot_prefix = "."
bot = commands.Bot(command_prefix=bot_prefix)

@bot.event
async def on_ready():
    print("Bot Online!")
    print("Name: {}".format(bot.user.name))
    print("ID: {}".format(bot.user.id))

class Main_Commands():
    def __init__(self, bot):
     self.bot = bot

# .addquote
@bot.command(pass_context = True)
async def addquote(ctx, *answer):
    answer=" ".join(answer)
    if not os.path.isfile("quote_file.pk1"):
        quote_list = []
    else:
        with open("quote_file.pk1", "rb") as quote_file:
            quote_list = pickle.load(quote_file)
    quote_list.append(answer)
    with open("quote_file.pk1", "wb") as quote_file:
        pickle.dump(quote_list, quote_file)
    await bot.say('Quote: "{}" added!'.format(answer))

# .quote
@bot.command(pass_context = True)
async def quote(ctx):
    with open("quote_file.pk1", "rb") as quote_file:
        quote_list = pickle.load(quote_file)
    await bot.say(random.choice(quote_list))

bot.run("TOKEN")

---

基本上，我想添加另一个函数，

.removequote（）

，该函数可以删除特定的引号，同时保持其他引号的完整性。

remove quotes与add quote非常相似，只需替换一行。使用列表comp而不是.append来过滤掉不需要的报价。其他一切都可以保持不变

# .removequote
@bot.command(pass_context = True)
async def removequote(ctx, *answer):
    answer=" ".join(answer)
    if not os.path.isfile("quote_file.pk1"):
        quote_list = []
    else:
        with open("quote_file.pk1", "rb") as quote_file:
            quote_list = pickle.load(quote_file)

    # This line:
    quote_list = [quote for quote in quote_list if quote != answer]

    with open("quote_file.pk1", "wb") as quote_file:
        pickle.dump(quote_list, quote_file)
    await bot.say('Quote: "{}" deleted!'.format(answer))

对于每台服务器的pickle，您可以为每台服务器使用不同的文件名，或者为pickle使用dict（就个人而言，我认为存储在单独的文件中更有效，因为您不必加载其他服务器的引号）：

---

谢谢有没有办法为每台服务器创建一个单独的列表？@根据消息来自哪个服务器创建单独的pickle文件，或者将信息存储在某种数据库中（并将原始服务器位置信息与报价本身一起存储）。

# .removequote
@bot.command(pass_context = True)
async def removequote(ctx, *answer):
    answer=" ".join(answer)
    filename = "quote_file_{}.pkl".format(ctx.message.server.id)
    if not os.path.isfile(filename):
        quote_list = []
    else:
        with open(filename, "rb") as quote_file:
            quote_list = pickle.load(quote_file)

    quote_list = [quote for quote in quote_list if quote != answer]

    with open(filename, "wb") as quote_file:
        pickle.dump(quote_list, quote_file)
    await bot.say('Quote: "{}" deleted!'.format(answer))