根据胜利者编写扑克函数的Pythonic方法来开经销商和百叶窗

Question

我正在编写一个扑克项目，并刚刚编写了一个函数来重新描述经销商，小百叶窗和大百叶窗的基础上(赛前)。代码工作正常，但看上去不像pythonic：

players = {1:'Jose',2:'Sam',3:'John',4:'Victor'}
## Each player receives a card, John gets an Ace and wins the dealer spot
winner = 'John'

for num in players:
    if players[num] == winner:
        dealer = num
        
        if len(players) == 2:
            if num+1 <= len(players):
                small_blind = num+1
            else:
                small_blind = 1
                
        elif len(players) >= 3:
            if num+1 <=len(players):
                small_blind = num+1
                if num+2 <= len(players):
                    big_blind = num+2
                else:
                    big_blind = 1
            else:
                small_blind = 1
                big_blind = 2

print(f'{players[dealer]} will be dealing cards')
print(f'{players[small_blind]} will be small blind')
print(f'{players[big_blind]} will be big blind')

从指定索引开始循环整个列表的最有效方法是什么？

Answer 1

您可以直接使用索引(列表的mod长度)：

players = ['Jose', 'Sam', 'John', 'Victor']
winner = 'John'

dealer = players.index(winner)
small_blind = (dealer+1)%len(players)
big_blind   = (dealer+2)%len(players)