Question

我有这个游戏，您需要用从列表发送的5个随机表情来做出反应。问题是，有时random.randint()会吐出两次相同的表情符号，因此它不可能对具有相同表情符号的同一消息做出两次反应。有更好的方法来处理多个random.randints吗？

async def food_loop():
    await client.wait_until_ready()
    channel = client.get_channel("523262029440483329")
    while not client.is_closed:
        foodtime = random.randint(1440, 1880)
        food = ['','','','','','','','','','','','','','','','','','','','','',
                '','','','','','','','','','','','','','','','','','',
                '','','','','','','','','','','','','','','','','','','',
                '','','','','','','','','','','','','☕','','','','','','','',
                '']
        food1 = food[random.randint(0,79)]
        food2 = food[random.randint(0,79)]
        food3 = food[random.randint(0,79)]
        food4 = food[random.randint(0,79)]
        food5 = food[random.randint(0,79)]
        foodmonies = random.randint(350,750)
        up = 'order up'
        def orderup(m):
            return m.content.lower() == up
        foodmsg = 'Customer has ordered {}, {}, {}, {}, and {}! Fulfill their order ASAP!'.format(food1, food2, food3, food4, food5)
        foodmsgsend = await client.send_message(channel, foodmsg)
        foodpay1 = await client.wait_for_reaction(emoji=food1, message=foodmsgsend, timeout=3600,
                                             check=lambda reaction, user: user != client.user)
        foodpay2 = await client.wait_for_reaction(emoji=food2, message=foodmsgsend, timeout=3600,
                                             check=lambda reaction, user: user != client.user)
        foodpay3 = await client.wait_for_reaction(emoji=food3, message=foodmsgsend, timeout=3600,
                                             check=lambda reaction, user: user != client.user)
        foodpay4 = await client.wait_for_reaction(emoji=food4, message=foodmsgsend, timeout=3600,
                                             check=lambda reaction, user: user != client.user)
        foodpay5 = await client.wait_for_reaction(emoji=food5, message=foodmsgsend, timeout=3600,
                                             check=lambda reaction, user: user != client.user)
        foodguess = await client.wait_for_message(timeout=3600, channel=channel, check=orderup)
        if foodpay1 and foodpay2 and foodpay3 and foodpay4 and foodpay5 and foodpay3.user.id in blacklist:
            pass
        else:
            if foodpay1 and foodpay2 and foodpay3 and foodpay4 and foodpay5 and foodguess:
                await client.delete_message(foodmsgsend)
                await client.send_message(channel, "{} fulfills the order and earns ${}".format(foodpay5.user.mention, foodmonies))
                add_dollars(foodpay5.user, foodmonies)
                await asyncio.sleep(int(foodtime))

Answer 1

根据定义，随机数可以重复，因为对randint的任何调用都独立于前一个。您可以替换以下内容：

food1 = food[random.randint(0,79)]
food2 = food[random.randint(0,79)]
food3 = food[random.randint(0,79)]
food4 = food[random.randint(0,79)]
food5 = food[random.randint(0,79)]

与此：

food1, food2, food3, food4, food5 = random.sample(food, 5)

来自docs（重点是我）：

random.sample(population, k)

返回从总体序列或集合中选择的唯一元素的k长度列表。

话虽如此，最好重构该部分并改用列表而不是声明5个变量（如果需要50个或500个变量，则会更麻烦）。

Random.Randint（）重复

1 个答案: