我正试图在我的Tic-Tac-Toe程序中成为一个"聪明"的对手。为了做到这一点，我创建了一个"可能获胜"函数，它将决定下一轮是否有可能获胜。运行这段代码时，我的问题是，在for循环的每一次迭代中，变量板似乎都会发生变化。
我想在每次迭代开始时将潜在板重置为原始板，这就是为什么我在循环开始时包含potential_board=board[：]。然后我编辑potential_board，但每次循环重复时，这个变量都不会重置，事实上board也发生了变化。为什么会这样？
非常感谢！
import random,copy
board = [['o','o',' '],[' ',' ',' '],[' ',' ',' ']]
cols = [['o',' ',' '],['o','',''],['o','','']]
def possible_win(board,player):
""" This function should predict whether a winning move is possible in
the next turn. This could be done by simulating every possible next move
and running check_win() on those positions.
:param board,player: checks a win for the specified player
:return:
"""
spaces = empty_spaces(board)
print('Spaces',spaces)
winning_moves = []
for space in spaces:
potential_board = board[:]
print('PBoard',potential_board)
print(space[0],space[1])
potential_board[space[0]][space[1]] = 'o'
if check_win(potential_board,'o'):
winning_moves.append(space)
return winning_moves
def choose_space(board):
a = True
while a:
col = int(input('Choose your column of 1,2,3: ')) - 1
row = int(input('Choose your row of 1,2,3: ')) - 1
if board[row][col] == ' ':
board[row][col] = 'o'
a = False
else: print('Sorry, try again')
return board
def empty_spaces(board):
empty_spaces = []
ind = 0
for row in board:
ind1 = 0
for space in row:
if space == ' ':
empty_spaces.append((ind, ind1))
ind1 += 1
ind += 1
return empty_spaces
def comp_choose_space(board):
choice = random.choice(empty_spaces(board))
board[choice[0]][choice[1]] = 'x'
return board
def check_win(board,player):
rows = board
columns = construct_cols(board)
for row in board:
# if player fills row win = True
a = ind = 0
for space in row:
if rows[board.index(row)][ind] != player: break
else: a += 1
ind += 1
if a == 3:
return True
for col in columns:
a = ind = 0
for space in col:
if rows[columns.index(col)][ind] != player:
break
else:
a += 1
ind += 1
if a == 3:
return True
if rows[0][0] == player and rows[1][1] == player and rows[2][2] == player
or rows[0][2] == player and rows[1][1] == player and rows[2][0] == player:
return True
return False
def construct_cols(board):
cols = [['','',''],['','',''],['','','']]
for row in range(len(board)):
for col in range(row):
cols[col][row] = board[row][col] # sounds like this should work
return cols
def print_board(board):
for row in board:
print('| {} {} {} |'.format(row[0],row[1],row[2]))
def main():
turns = 0
board = [[' ',' ',' '],[' ',' ',' '],[' ',' ',' ']]
print_board(board)
win = False
while win == False and turns < 9:
turns += 1
board = choose_space(board)
if check_win(board,'o'): win,winner = True,'won'
board = comp_choose_space(board)
if check_win(board,'x'): win,winner = True,'lost'
print_board(board)
if turns == 9: print('You drew!')
else:
print('{}, you {}'.format('Congratulations' if winner == 'won' else 'Sorry',winner))
print(possible_win(board,'o'))
# print(empty_spaces(board))
# print(check_win(board,'o'))
# print_board(board)
# print(comp_choose_space(board))
# main()
# Future project - make the computer smarter than just randomly choosing a space
# ie seeing how close i am to winning
编辑：通过使用copy.deepcopy((，我设法解决了这个问题，但我不明白为什么它有效，而copy.copy((和board[：]不起作用？有人能解释一下吗？
这就是copy.deepcopy的作用。它将遍历该结构，在其中创建每个可变对象的副本。使用切片[:]或浅层copy仅复制顶层，使每行的列表共享。
基本上，如果我们从一个列表开始：
l = [a, b, c]
shallow = l[:]
shallow2 = copy(l)
deep = deepcopy(l)
两个shallow副本仅在l上运行，而未在a、b或c上运行。它们都具有值[a, b, c]，但是是不同的列表。它们都引用了相同的a、b和c对象(从它们的角度来看，唯一的变化是有更多的引用(。
deep拷贝更深入，拷贝了每个元素；它是一个形状为[deepcopy(a), deepcopy(b), deepcopy(c)]的新列表，无论这些值变成什么。