Question

我试图实现一个简单的TicTacToe游戏，但我无法制作某个列表的浅层副本。这是我脚本的最小摘录，它提出了我的问题：

SIZE = 3 # size of game board
PLAYERS = (PLAYER, COMPUTER) = "X", "O"

class TicTacToe(list):
    """ Class handling game board """
    def __init__(self, size=SIZE, players=PLAYERS):
        """ Board size defaults to 3 """
        super().__init__([[" " for idx in range(size)] for jdx in range(size)])
        self._size = size

    def winner_move(self, letter, range_):
        """ Determine winner move - range_ members are tuples """
        for tile in range_:
            sim = self[:] # shallow copy of game board for simulation
            sim[tile[0]][tile[1]] = letter
            print("Current board:")
            print(self)
            # code makes some decisions here and continues

    def get_empty(self):
        """ Return empty places on board as a set of (row, col) """
        return {(row, col) for row in range(self._size)\
                for col in range(self._size) if self[row][col] == " "}

board = TicTacToe()
board.winner_move(PLAYER, board.get_empty())

虽然我只为sim分配新值，但更改反映在＆＃34; main＆＃34;板。我也试过

sim = list(self)

和

sim = [item for item in self]

但两者都给出了相同的结果。

有什么建议我做错了吗？

Answer 1

Arpegius' comment是对的。你有一个嵌套列表。使用sim = self[:]，sim = list(self)或sim = [item for item in self]，您将获得一个新的外部列表，但它将是内部列表的相同引用列表。

sim = [lst[:] for lst in self]

或

sim = [list(lst) for lst in self]

会有效，但

from copy import deepcopy
# ...
sim = deepcopy(self)

更优雅，其目的更容易理解。请参阅its documentation。

无法在Python3中制作列表的浅表副本

1 个答案: