discord.py Blackjack抽牌函数返回None问题及去重算法优化求助

原代码存在两个核心问题：

• 每次调用deck()都会重新生成并洗牌整副牌，完全违背真实抽牌逻辑——你应该维护一副全局的、已洗牌的牌组，而非每次抽牌都重建牌组。
• 递归分支无返回值：当抽到重复牌时，调用Test.deck(self)但未返回该调用的结果，导致当前函数执行到return时无有效返回值，最终返回None。

正确思路是在类初始化时生成并洗牌完整牌组，后续抽牌直接从该牌组取牌，天然避免重复，无需额外去重检查。

修改后代码示例：

import random

class BlackJackGame:
    def __init__(self):
        # 初始化时生成整副牌并完成洗牌
        card_suits = ["Club ", "Diamond ", "Spade ", "Heart "]
        card_ranks = ["2", "3", "4", "5", "6", "7", "8", "9", "10", "J", "K", "Q", "A"]
        self.deck = [suit + rank for suit in card_suits for rank in card_ranks]
        random.shuffle(self.deck)
    
    def draw_card(self):
        # 牌组抽完时自动重新洗牌
        if not self.deck:
            card_suits = ["Club ", "Diamond ", "Spade ", "Heart "]
            card_ranks = ["2", "3", "4", "5", "6", "7", "8", "9", "10", "J", "K", "Q", "A"]
            self.deck = [suit + rank for suit in card_suits for rank in card_ranks]
            random.shuffle(self.deck)
        return self.deck.pop()

如果非要保留原思路（不建议，效率极低），需修复递归的返回逻辑：

def deck(self):
    card_type = ["Club ", "Diamond ", "Spade ", "Heart "]
    card_nums = ["2", "3", "4", "5", "6", "7", "8", "9", "10", "J", "K", "Q", "A"]
    cards_total = [i + j for i in card_type for j in card_nums]
    random.shuffle(cards_total)
    pop_card = cards_total.pop()

    if pop_card in self.overlap_check:
        # 必须返回递归调用的结果
        return self.deck()
    self.overlap_check.append(pop_card)
    return pop_card

此方案弊端明显：每次抽牌都要重建、洗牌整副牌，牌组快抽完时递归深度会急剧上升，容易触发Python递归深度限制。

• 完全贴合真实二十一点游戏规则：一副牌抽完后重新洗牌。
• 彻底规避重复抽牌问题，无需额外去重逻辑。
• 性能更优，无冗余的牌组生成和递归操作。

内容的提问来源于stack exchange，提问作者Luminous Lunetzsche