简化 Python If 语句

Question

我是 python 的初学者，我正在终端中创建一个两人井字游戏。这段代码总共占用了 139 行，（下面是我遇到问题的相关代码部分），但是，这个 CheckWin 函数占用了大约 40 行代码，与数量相比，我认为这是相当多的这段代码中的行数，并考虑到它执行了一些基本功能。基本上，在游戏中，此函数会检查一行、一列或对角线是否有三个 X 或三个 O，如果有，它会将 X 分配给获胜者，将 O 分配给获胜者。无论如何，这是代码。

X = "X"
O = "O"
empty = " " 
S = [" ", " ", " ", " ", " ", " ", " ", " ", " "]

def CheckWin(S):
    global winner
    winner = ""
    if S[0] == S[1] == S[2] != empty:
        if S[0] == X:
            winner = X
        if S[0] == O:
            winner = O
    if S[3] == S[4] == S[5] != empty:
        if S[3] == X:
            winner = X
        if S[3] == O:
            winner = O
    if S[6] == S[7] == S[8] != empty:
        if S[6] == X:
            winner = X
        if S[6] == O:
            winner = O
    if S[0] == S[3] == S[6] != empty:
        if S[0] == X:
            winner = X
        if S[0] == O:
            winner = O
    if S[1] == S[4] == S[7] != empty:
        if S[1] == X:
            winner = X
        if S[1] == O:
            winner = O
    if S[2] == S[5] == S[8] != empty:
        if S[2] == X:
            winner = X
        if S[2] == O:
            winner = O
    if S[0] == S[4] == S[8] != empty:
        if S[0] == X:
            winner = X
        if S[0] == O:
            winner = O
    if S[2] == S[4] == S[6] != empty:
        if S[2] == X:
            winner = X
        if S[2] == O:
            winner = O

基本上，我需要帮助使功能更简单。但是，我不想消除X、O和winner变量，也不想消除列表S的列表索引方法。尽管如此，有没有办法简化所有这些If语句，保留这些东西？如果是这样，如何？

Answer 1

您的代码查找 "trios" 个位置；您还不如拥有一个包含此信息的对象：

trios = ((0,1,2), (3,4,5), (6,7,8), (0,3,6), (1,4,7), (2,5,8), (0,4,8), (2,4,6))

然后 CheckWin 会循环遍历每个三重奏，检查您正在做的事情，如果三重奏匹配，return 就会获胜。这样，CheckWin 将少于 10 行代码。

我不想放弃一切，因为我相信你能做到:)

此外，您不需要在 CheckWin 中使用名为 "winner" 的全局变量；只需让 CheckWin return 获胜者（或“”），并将结果存储在函数本身之外的变量中。
即

winner = CheckWin(S)

Answer 2

您尝试过使用循环代替吗？

X, O = 'X', 'O'
S    = [X,O,X,O,X,O,O,X,O] # Test values
def CheckWin(S):
    index      = 0
    has_winner = False
    while index < len(S):
        print index
        if index <= 6: # after this point you won't find a winner (or you run this after every turn?)
            if (index % 3 == 0 and S[index] == S[index + 1] and S[index] == S[index + 2]): # horizontals
                has_winner = True
            elif index < 3: # verticals and diagonals (you only need the first row to know this)
                if (S[index] == S[(index + 3)] and S[index] == S[index + 6]) or \
                   (index == 0 and S[index] == S[4] and S[index] == S[8]) or \
                   (index == 2 and S[index] == S[4] and S[index] == S[6]):
                    has_winner = True
        if has_winner: # I'm using this to prevent duplicating code above (one if less)
            if S[index] in [X,O]:
                return S[index]
        index += 1
    return has_winner # If we have a winner but the X or O criterion isn't met, it returns False
winner = CheckWin(S)

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