我正在构建一个学习玩tic tac toe的小蟒蛇脚本。我的过程是存储在游戏中进行的每个动作，并根据它是否导致获胜结果进行评分。最终我会尝试在很多轮比赛中训练这个。
我的问题在于我的update_weights()方法。我希望它采取存储的移动（从board对象访问并表示为列表[row，col]）并遍历该移动列表。然后，该方法应引用板的存储权重（（3,3）numpy数组的字典）并更新相应的权重以进行适当的移动。
e.g。假设获胜。在获胜序列中，移动＃2位于棋盘位置[0,1]。该方法应该访问权重字典（键是移动＃）并将数组[0,1]的位置乘以因子1.05。
问题是我的方法正在改变我的权重字典中的所有数组，而不仅仅是与正确的移动＃键相关联的数组。我无法弄清楚这是怎么回事。
import numpy as np
import random
class ttt_board():
def __init__(self):
self.board_state = np.array([[0,0,0],[0,0,0],[0,0,0]])
self.board_weight = self.reset_board_weights()
self.moves = []
def reset_board_weights(self):
board_weight_instance = np.zeros((3,3))
board_weight_instance[board_weight_instance >= 0] = 0.5
board_weight = {0: board_weight_instance,
1: board_weight_instance,
2: board_weight_instance,
3: board_weight_instance,
4: board_weight_instance}
return board_weight
def reset_board(self):
self.board_state = np.array([[0,0,0],[0,0,0],[0,0,0]])
def reset_moves(self):
self.moves = []
def is_win(self):
board = self.board_state
if board.trace() == 3 or np.flipud(board).trace() == 3:
return True
for i in range(3):
if board.sum(axis=0)[i] == 3 or board.sum(axis=1)[i] == 3:
return True
else:
return False
def is_loss(self):
board = self.board_state
if board.trace() == 12 or np.flipud(board).trace() == 12:
return True
for i in range(3):
if board.sum(axis=0)[i] == 12 or board.sum(axis=1)[i] == 12:
return True
else:
return False
def is_tie(self):
board = self.board_state
board_full = True
for i in range(len(board)):
for k in range(len(board)):
if board[i][k] == 0:
board_full = False
if board_full and not self.is_win() and not self.is_loss():
return True
else:
return False
def update_board(self,player,space):
#takes player as 1 or 4
#takes space as list [0,0]
self.board_state[space[0],space[1]] = player
if player == 1:
self.store_move(space)
return
def get_avail_spots(self):
avail_spots = []
board = self.board_state
for i in range(len(board)):
for k in range(len(board)):
if board[i][k] == 0:
avail_spots.append([i,k])
return avail_spots
def gen_next_move(self):
avail_spots = self.get_avail_spots()
move = random.randrange(len(avail_spots))
return avail_spots[move]
def update_weights(self,win):
moves = self.moves
if win:
factor = 1.05
else:
factor= 0.95
for i in range(len(moves)):
row = moves[i][0]
col = moves[i][1]
old_weight = self.board_weight[i][row,col]
new_weight = old_weight*factor
self.board_weight[i][row,col] = new_weight
return
def store_move(self,move):
self.moves.append(move)
return
if __name__ == '__main__':
board = ttt_board()
while not board.is_win() and not board.is_loss() and not board.is_tie():
try:
board.update_board(1,board.gen_next_move())
board.update_board(4,board.gen_next_move())
except ValueError:
break
if board.is_win():
board.update_weights(1)
print('Player 1 wins: {w}'.format(w=board.is_win()))
elif board.is_loss():
board.update_weights(0)
print('Player 2 wins: {l}'.format(l=board.is_loss()))
elif board.is_tie():
print('Game ends in tie: {t}'.format(t=board.is_tie()))
print('Here is the final board')
print(board.board_state)
print(board.board_weight)
print(board.moves)
正如您通过运行脚本所看到的，单个游戏后打印的权重字典对于每个键具有相同的数组值。 我希望每个数组只能在一个位置进行更改，因为它只能用于与其关联的键对应的移动＃。
答案 0 :(得分：2)
问题是您在字典
中的board_weight_instance数组上共享相同的引用
board_weight_instance = np.zeros((3,3))
board_weight_instance[board_weight_instance >= 0] = 0.5
board_weight = {0: board_weight_instance,
1: board_weight_instance,
2: board_weight_instance,
3: board_weight_instance,
4: board_weight_instance}
我会在字典理解中这样做，使用辅助方法为每个元素创建一个新的引用：
@staticmethod
def create_element():
board_weight_instance = np.zeros((3,3))
board_weight_instance[:] = 0.5 # simpler than your method
return board_weight_instance
board_weight = {i:self.create_element() for i in range(0,5)}
在你的情况下，为什么甚至在你可以使用list时使用字典：没有散列，更快的处理：
board_weight = [self.create_element() for _ in range(0,5)]
你可以用同样的方式访问它