我正在尝试画一个假的3x3 tic tac toe board。我是python的新手，我不明白为什么这不起作用。帮助将不胜感激。谢谢！
def draw():
for i in range(4):
board = (" ___ " * 3)
for i in board:
("| " * 4).join(board)
print(board)
draw()
编辑：
最终代码：
def draw():
board = ''
for i in range(-1,6):
if i%2==0:
board += '| ' * 4
board += '\n| | | |'
else:
board += ' _____ ' * 3
board += '\n'
print (board)
draw()
输出：
_____ _____ _____
| | | |
| | | |
_____ _____ _____
| | | |
| | | |
_____ _____ _____
| | | |
| | | |
_____ _____ _____
Double Edit：
另一种方式：
def drawsmall():
a = (' ___' * 3 )
b = ' '.join('||||')
print('\n'.join((a, b, a, b, a, b, a, )))
drawsmall()
输出：
___ ___ ___
| | | |
___ ___ ___
| | | |
___ ___ ___
| | | |
___ ___ ___
答案 0 :(得分：1)
我发现在一个循环中执行此操作更容易，每次迭代打印一行板。您可以使用%运算符检查当前迭代是偶数还是奇数，从而在垂直和水平条之间切换。
使用字符串，您不需要使用连接 - 使用+=运算符附加更清晰。
def draw():
# initialize an empty board
board = ""
# there are 5 rows in a standard tic-tac-toe board
for i in range(5):
# switch between printing vertical and horizontal bars
if i%2 == 0:
board += "| " * 4
else:
board += " --- " * 3
# don't forget to start a new line after each row using "\n"
board += "\n"
print(board)
draw()
输出：
| | | |
--- --- ---
| | | |
--- --- ---
| | | |
答案 1 :(得分：1)
试试这个代码：
bool mysort(int arg1,int arg2){
return arg1 % 2 == 0 && arg2 % 2 != 0;
}
输出：
def draw():
a=('\n _____ _____ _____ ')
b= ('\n| | | |')
print(a,b,b,a,b,b,a,b,b,a)
draw()
为了更好地使用视图：
_____ _____ _____
| | | |
| | | |
_____ _____ _____
| | | |
| | | |
_____ _____ _____
| | | |
| | | |
_____ _____ _____
输出：
def print_tic_tac_toe():
print("\n")
print("\t | |")
print("\t | | ")
print('\t_____|_____|_____')
print("\t | |")
print("\t | | ")
print('\t_____|_____|_____')
print("\t | |")
print("\t | | ")
print("\t | |")
print("\n")
print_tic_tac_toe()
答案 2 :(得分：0)
查看join函数的工作原理。首先，它接受给定的字符串并将其用于“glue”，即连接其他字符串的字符串。其次，返回构造的字符串;您的join操作无法保存结果。
首先尝试使用嵌套循环执行此操作：打印一行框，然后打印水平分隔符等。然后，逐位将其转换为所需的单字符串输出。
答案 3 :(得分：0)
你可以试试这个：
def draw():
return [["__" for b in range(3)] for i in range(3)]
现在您有一个包含您的电路板的列表列表。要打印出来，你可以这样做：
the_board = draw()
for i in the_board:
for b in i:
print('|'.join(i), end="")
print()
print(" | | ")
答案 4 :(得分：0)
我认为我会简化事情，以便自己理解。这段代码产生与上面相同的输出：
def draw_board():
v = '| | | |'
h = ' ____ ____ ____ '
for i in range(0,10):
if i%3==0:
print(h)
else:
print(v)
draw_board()
输出：
____ ____ ____
| | | |
| | | |
____ ____ ____
| | | |
| | | |
____ ____ ____
| | | |
| | | |
____ ____ ____
答案 5 :(得分：0)
如果您不想使用变量/函数/循环，而想要基于打印命令的简单的一线解决方案：
print("__|__|__", "__|__|__", " | | ", sep='\n')
答案 6 :(得分：0)
您可以尝试以下操作： 在下面找到端到端互动井字棋盘游戏的python代码。 代码看起来很冗长，可以优化，但它可以完美地用作交互式井字棋盘游戏。
#Function code to clear the output space (screen)
from IPython.display import clear_output
#code to display just board-
def ttt_borad(board):
cl = clear_output()
print('Your Tic-Tac-Toe board now:\n')
print(board[1] + "|" + board[2] + "|" + board[3])
print("________")
print(board[4] + "|" + board[5] + "|" + board[6])
print("________")
print(board[7] + "|" + board[8] + "|" + board[9])
#function code to accept player key choices-
def player_key():
player_choice = ''
play1 = ''
play2 = ''
while player_choice not in ('Y', 'N'):
player_choice = input("Player-1 would like to go first ? Enter Y/N: ")
player_choice = player_choice.upper()
if player_choice not in ('Y', 'N'):
print("Invalid Key")
else:
pass
if player_choice == 'Y':
while play1 not in ('X', 'O'):
play1 = input("Select your Key for Player-1 X or O: ")
play1 = play1.upper()
if play1 not in ('X', 'O'):
print("Invalid Key")
else:
pass
else:
while play2 not in ('X', 'O'):
play2 = input("Select your Key for Player-2 X or O: ")
play2 = play2.upper()
if play2 not in ('X', 'O'):
print("Invalid Key")
else:
pass
if play1 == 'X':
play2 = 'O'
elif play1 == 'O':
play2 = 'X'
elif play2 == 'X':
play1 = 'O'
elif play2 == 'O':
play1 = 'X'
print(f'Key for Player-1 is: {play1} and Key for Player-2 is: {play2}')
return play1, play2
#function code to accept key strokes to play game
def enter_key(key, bp):
play1, play2 = key
ind = ['1', '2', '3', '4', '5', '6', '7', '8', '9']
i = 1
while i < 10:
j = 0
k = 0
print(f'Game Move: {i}')
while j not in ind:
j = input("Player-1: Select position (1-9) for your Move: ")
if j not in ind:
print("Invalid Key or Position already marked")
else:
pass
x = ind.index(j)
ind.pop(x)
j = int(j)
bp[j] = play1
ttt_borad(bp)
i = i + 1
tf = game_winner(key, bp)
if tf == 1:
print("The Winner is: Player-1 !!")
break
print(f'Game Move: {i}')
if i == 10:
break
while k not in ind:
k = input("Player-2: Select position (1-9) for your Move: ")
if k not in ind:
print("Invalid Key or Position already marked")
else:
pass
y = ind.index(k)
ind.pop(y)
k = int(k)
bp[k] = play2
ttt_borad(bp)
i = i + 1
ft = game_winner(key, bp)
if ft == 2:
print("The Winner is: Player-2 !!")
break
return bp
#function code to calculate and display winner of the game-
def game_winner(key, game):
p1, p2 = key
p = 0
if game[1] == game[2] == game[3] == p1:
p = 1
return p
elif game[1] == game[4] == game[7] == p1:
p = 1
return p
elif game[1] == game[5] == game[9] == p1:
p = 1
return p
elif game[2] == game[5] == game[8] == p1:
p = 1
return p
elif game[3] == game[6] == game[9] == p1:
p = 1
return p
elif game[4] == game[5] == game[6] == p1:
p = 1
return p
elif game[3] == game[5] == game[7] == p1:
p = 1
return p
elif game[1] == game[2] == game[3] == p2:
p = 2
return p
elif game[1] == game[4] == game[7] == p2:
p = 2
return p
elif game[1] == game[5] == game[9] == p2:
p = 2
return p
elif game[2] == game[5] == game[8] == p2:
p = 2
return p
elif game[3] == game[6] == game[9] == p2:
p = 2
return p
elif game[4] == game[5] == game[6] == p2:
p = 2
return p
elif game[3] == game[5] == game[7] == p2:
p = 2
return p
else:
p = 3
return p
#Function code to call all functions in order to start and play game-
def game_play():
clear_output()
entry = ['M', '1', '2', '3', '4', '5', '6', '7', '8', '9']
ttt_borad(entry)
plk = player_key()
new_board = enter_key(plk, entry)
tie = game_winner(plk, new_board)
if tie == 3:
print("Game Tie !!! :-( ")
print('Would you like to play again? ')
pa = input("Enter Y to continue OR Enter any other key to exit game: ")
pa = pa.upper()
if pa == 'Y':
game_play()
else:
pass
game_play()
＃在任何Python3编辑器中尝试整个代码，并让我知道您的反馈。 我已附上示例板代码显示方式。Sample Tic-Tac-Toe board by code
谢谢
