你知道用Python缩短4个函数的方法吗？

Question

在Python中有任何方法来缩短这4个函数吗？

def right_check(board: List[List[dict]], rowid: int, colid: int) -> bool:
    """This function checks rows for 5 continuously symbols.

    Args:
        board (List[List[dict]]): The board
        rowid (int): The row ID where the win function is currently at.
        colid (int): The col ID where the win function is currently at.

    Returns:
        bool: True when function description applies
    """
    try:
        if board[rowid][colid]["val"] == "X" and board[rowid][colid+1]["val"] == "X" and board[rowid][colid+2]["val"] == "X" and board[rowid][colid+3]["val"] == "X" and board[rowid][colid+4]["val"] == "X":
            print("X won")
            return True
        if board[rowid][colid]["val"] == "O" and board[rowid][colid+1]["val"] == "O" and board[rowid][colid+2]["val"] == "O" and board[rowid][colid+3]["val"] == "O" and board[rowid][colid+4]["val"] == "O":
            print("O won")
            return True
    except IndexError:
        return False
 
def down_check(board: List[List[dict]], rowid: int, colid: int) -> bool:
    """This function checks columns for 5 continuously symbols.

    Args:
        board (List[List[dict]]): The board
        rowid (int): The row ID where the win function is currently at.
        colid (int): The col ID where the win function is currently at.

    Returns:
        bool: True when function description applies
    """
    try:
        if board[rowid][colid]["val"] == "X" and board[rowid+1][colid]["val"] == "X" and board[rowid+2][colid]["val"] == "X" and board[rowid+3][colid]["val"] == "X" and board[rowid+4][colid]["val"] == "X":
            print("X won")
            return True
        if board[rowid][colid]["val"] == "O" and board[rowid+1][colid]["val"] == "O" and board[rowid+2][colid]["val"] == "O" and board[rowid+3][colid]["val"] == "O" and board[rowid+4][colid]["val"] == "O":
            print("O won")
            return True
    except IndexError:
        return False
 
def down_right_check(board: List[List[dict]], rowid: int, colid: int) -> bool:
    """This function checks down-right direction for 5 continuously symbols.

    Args:
        board (List[List[dict]]): The board
        rowid (int): The row ID where the win function is currently at.
        colid (int): The col ID where the win function is currently at.

    Returns:
        bool: True when function description applies
    """
    try:
        if board[rowid][colid]["val"] == "X" and board[rowid+1][colid+1]["val"] == "X" and board[rowid+2][colid+2]["val"] == "X" and board[rowid+3][colid+3]["val"] == "X" and board[rowid+4][colid+4]["val"] == "X":
            print("X won")
            return True
        if board[rowid][colid]["val"] == "O" and board[rowid+1][colid+1]["val"] == "O" and board[rowid+2][colid+2]["val"] == "O" and board[rowid+3][colid+3]["val"] == "O" and board[rowid+4][colid+4]["val"] == "O":
            print("O won")
            return True
    except IndexError:
        return False
 
def down_left_check(board: List[List[dict]], rowid: int, colid: int) -> bool:
    """This function checks down-left direction for 5 continuously symbols.

    Args:
        board (List[List[dict]]): The board
        rowid (int): The row ID where the win function is currently at.
        colid (int): The col ID where the win function is currently at.

    Returns:
        bool: True when function description applies
    """
    try:
        if board[rowid][colid]["val"] == "X" and board[rowid+1][colid-1]["val"] == "X" and board[rowid+2][colid-2]["val"] == "X" and board[rowid+3][colid-3]["val"] == "X" and board[rowid+4][colid-4]["val"] == "X":
            print("X won")
            return True
        if board[rowid][colid]["val"] == "O" and board[rowid+1][colid-1]["val"] == "O" and board[rowid+2][colid-2]["val"] == "O" and board[rowid+3][colid-3]["val"] == "O" and board[rowid+4][colid-4]["val"] == "O":
            print("O won")
            return True
    except IndexError:
        return False

我最感兴趣的是缩短colid+1 +2 +3、rowid+1 +2 +3部分，而不必重复每个if值为"O“的部分。

Answer 1

首先，您只是不需要重复使用"O“。条件基本上检查所有的单元格都有相同的值--不管是X还是O，只要检查它们是否都一样。如果是的话，打印这个重复的值。

此外，您总是检查连续的索引，因此可以使用一个循环而不是许多and。

最后，实现check if multiple items are identical的一种方法是使用一个集合：

def right_check(board, rowid, colid):
    try:
        if len(set(board[rowid][colid+i]["val"] for i in range(5))) == 1:
            print(f"{board[rowid][colid]["val"]} won")
            return True
    except IndexError:
        return False

同样的也适用于其他方向的函数。