简单Blackjack游戏

Question

这是我用Python制作的简单Blackjack游戏。我希望有一些反馈来编写更好的代码。

from art import logo    
import random

input("Do you want to play black-jack ? 'y' or 'no': ")
print()
print(logo)

cards = [11, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 10, 10]
user_hand = []
com_hand = []


def print_score(score_to_print):

    if score_to_print == "user":
        print(f"user_hand {user_hand} your score = {get_score(user_hand)}")
    elif score_to_print == "com":
        print(f"com_hand{com_hand} com_score = {get_score(com_hand)}")


def get_score(hand):
    score = 0
    for i in range(len(hand)):
        score += hand[i]

    return score


def restart_game(result):

    print(result)
    print_score("user")
    print_score("com")

    user_hand.clear()
    com_hand.clear()
    user_input = input("Do you want to play black-jack ? 'y' or 'no': ")
    if user_input == 'y':
        deal_cards(cards)
    else:
        print("ok bye bye have fun in the afterlife")


def check_for_blackjack(hand):
    if get_score(com_hand) == 21:
        restart_game("You Lose")
    elif get_score(user_hand) == 21:
        restart_game("you Win")


def check_for_score_over_21():
    if get_score(user_hand) > 21:
        restart_game("You Lose")
    elif get_score(com_hand) > 21:
        restart_game("You Win")


def check_for_ace(card, hand):
    if get_score(hand) + card > 21:
        card = 1
    return card


def check_for_winner():
    user_score = get_score(user_hand)
    com_score = get_score(com_hand)
    if user_score == com_score:
        restart_game("Draw")
    elif user_score < com_score:
        restart_game("You Lose")
    else:
        restart_game("You Win")


def deal_computer_cards():
    while get_score(com_hand) < 17:
        new_card = random.choice(cards)
        com_hand.append(check_for_ace(new_card, com_hand))


def ask_for_new_card():
    print_score("user")
    user_choice = input("Do you want another card ?: ")

    if user_choice == 'y':
        new_card = random.choice(cards)
        user_hand.append(check_for_ace(hand=user_hand, card=new_card))
        if get_score(user_hand) < 21:
            ask_for_new_card()
        elif get_score(user_hand) == 21:
            restart_game("You Win")
        elif get_score(user_hand) > 21:
            restart_game("You Lose")
        else:
            print_score("user")
    elif user_choice == 'n':
        deal_computer_cards()
        check_for_winner()


def add_cards_to_hand(hand):
    card1 = random.choice(cards)
    card2 = random.choice(cards)
    hand.append(check_for_ace(card1, hand))
    hand.append(check_for_ace(card2, hand))


def deal_cards(cards):
    add_cards_to_hand(user_hand)
    print_score("user")
    
    add_cards_to_hand(com_hand)
    check_for_blackjack(com_hand)
    check_for_blackjack(user_hand)
    
    print(f"com_hand {com_hand} com score = {get_score(com_hand)}")
    print(f"computer first card is a {com_hand[0]}")
    
    ask_for_new_card()

deal_cards(cards)

Answer 1

input("Do you want to play black-jack ? 'y' or 'no': ")

这一行不执行任何操作，因为您从未将input函数的结果分配给名称或将其用作参数。

def print_score(score_to_print):
    if score_to_print == "user":
        print(f"user_hand {user_hand} your score = {get_score(user_hand)}")
    elif score_to_print == "com":
        print(f"com_hand{com_hand} com_score = {get_score(com_hand)}")

为了避免不必要的分支，可以将其分成两部分：

def print_user_info():
    print(f"Your hand: {*user_hand}")
    print(f"Your score: {get_score(user_hand)}")


def print_bot_info():
    print(f"Bot's hand: {*bot_hand}")
    print(f"Bot's score: {get_score(bot_hand)}")

或为一般情况重写：

def print_info(player_name: str, player_hand: List[int]):
    print(f"{player_name}'s hand: {*player_hand}")
    print(f"{player_name}'s score: {get_score(player_hand)}")

def get_score(hand):
    score = 0
    for i in range(len(hand)):
        score += hand[i]

    return score

这可以重写为

def get_score(hand: List[int]):
    return sum(hand)

实际上，您实际上并不需要这样的函数，在代码中使用sum(hand)很好地传达了这个意思。

ask_for_new_card()要求用户输入并检查它是y还是n。它从来没有告诉球员，这些是唯一的选择。例如，如果键入no，程序就会退出。

您需要说明其他可能的输入。例如，再次询问用户是否没有输入其中一个选项。您还可以尝试将输入更改为标准表单。例如，对于以y开头的任何单词，这一行将返回"Y"：

response = input()[0].upper()

def check_for_ace(card, hand):
    if get_score(hand) + card > 21:
        card = 1
    return card

这并不能真正检查这张卡是否是王牌。因为你每次向别人的手上加一张卡片，你都不可能输掉：

user_hand [7, 7] your score = 14
Do you want another card ?: y
user_hand [7, 7, 1] your score = 15
Do you want another card ?: y
user_hand [7, 7, 1, 1] your score = 16
Do you want another card ?: y
user_hand [7, 7, 1, 1, 1] your score = 17
Do you want another card ?: y
user_hand [7, 7, 1, 1, 1, 1] your score = 18
Do you want another card ?: y
user_hand [7, 7, 1, 1, 1, 1, 1] your score = 19
Do you want another card ?: y
user_hand [7, 7, 1, 1, 1, 1, 1, 1] your score = 20
Do you want another card ?: y
You Win
user_hand [7, 7, 1, 1, 1, 1, 1, 1, 1] your score = 21

请测试您的代码，然后再张贴在这里。