因此，對于我正在編寫的這段代碼，我試圖為三個不同的玩家生成一組三張隨機卡，其中一個是人類，兩個是模擬的。為此，我正在使用這段代碼。def shuffle_p1(): p_1_human_card_1=random.randint(1,3) p_1_human_card_2=random.randint(1,3) p_1_human_card_3=random.randint(1,3) p_1_human_cards=[p_1_human_card_1,p_1_human_card_2,p_1_human_card_3] return (p_1_human_cards)def shuffle_p2(): p_2_ai_card_1=random.randint(1,3) p_2_ai_card_2=random.randint(1,3) p_2_ai_card_3=random.randint(1,3) p_2_ai_cards=[p_2_ai_card_1,p_2_ai_card_2,p_2_ai_card_3] return (p_1_human_cards)def shuffle_p3(): p_3_ai_card_1=random.randint(1,3) p_3_ai_card_2=random.randint(1,3) p_3_ai_card_3=random.randint(1,3) p_3_ai_cards=[p_3_ai_card_1,p_3_ai_card_2,p_3_ai_card_3] return (p_1_human_cards)p_1_human_cards=shuffle_p1()p_2_ai_cards=shuffle_p2()p_3_ai_cards=shuffle_p3()這會生成三組，但玩家 1 和玩家 2 每次都擁有完全相同的牌。我什至放置了一組不同的代碼def card_auth_p1(): while p_1_human_cards[0]==p_1_human_cards[1] or p_1_human_cards[0]==p_1_human_cards[2]: p_1_human_cards[0]=random.randint(1,3) while p_1_human_cards[1]==p_1_human_cards[0] or p_1_human_cards[1]==p_1_human_cards[2]: p_1_human_cards[1]=random.randint(1,3)def card_auth_p2(): while p_2_ai_cards[0]==p_2_ai_cards[1] or p_2_ai_cards[0]==p_2_ai_cards[2]: p_2_ai_cards[0]=random.randint(1,3) while p_2_ai_cards[1]==p_2_ai_cards[0] or p_2_ai_cards[1]==p_2_ai_cards[2]: p_2_ai_cards[1]=random.randint(1,3)def card_auth_p3(): while p_3_ai_cards[0]==p_3_ai_cards[1] or p_3_ai_cards[0]==p_3_ai_cards[2]: p_3_ai_cards[0]=random.randint(1,3) while p_3_ai_cards[1]==p_3_ai_cards[0] or p_3_ai_cards[1]==p_3_ai_cards[2]: p_3_ai_cards[1]=random.randint(1,3)并且if p_1_human_cards == p_2_ai_cards or p_1_human_cards == p_3_ai_cards: p_1_human_cards=shuffle_p1()if p_2_ai_cards == p_1_human_cards or p_2_ai_cards == p_3_ai_cards: p_2_ai_cards=shuffle_p2()以確保它們并不完全相同，但打印三個列表表明玩家 1 和 2 仍然相同。此外，即使在使用此代碼在三組之間進行隨機交易的第四個代碼塊之后
2 回答
aluckdog
TA貢獻1847條經驗 獲得超7個贊
如果你有Python 3.6+，你也可以嘗試secrets
try:
import secrets as random
except ModuleNotFoundError:
import random
# Use system resources for randomization
rnd = random.SystemRandom()
# Set available card options
AVAILABLE_CARDS = [1,2,3,]
# Set number of cards per hand.
NUMBER_OF_CARDS = 3
# Create a simple key-value collection.
players = dict(
p_1_human_cards = None,
p_2_ai_cards = None,
p_3_ai_cards= None,
)
# Define a shuffler() function. Parameters are number of cards and available cards.
def shuffler(n_cards=NUMBER_OF_CARDS , available_cards = AVAILABLE_CARDS ):
return tuple([rnd.choice(available_cards) for _ in range(n_cards)])
# Iterate through players to set each hand
for hand in players:
players[hand] = shuffler()
# Print result
for player, hand in players.items():
print(f"{player}: {hand}")
輸出：
p_1_human_cards: (2, 1, 3)
p_2_ai_cards: (3, 2, 1)
p_3_ai_cards: (2, 3, 3)
翻翻過去那場雪
TA貢獻2065條經驗 獲得超14個贊
看來您復制了以前的代碼塊，并且該return部分保持不變。:) 嘗試盡可能地編寫 DRY 代碼。你也可以這樣做：
import random
def shuffle_p1():
? ? p_1_human_cards=[random.randint(1,3) for _ in range(3)]
? ? return p_1_human_cards
def shuffle_p2():
? ? p_2_ai_cards=[random.randint(1,3) for _ in range(3)]
? ? return p_2_ai_cards
def shuffle_p3():
? ? p_3_ai_cards=[random.randint(1,3) for _ in range(3)]
? ? return p_3_ai_cards
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