所以我正在网上做一个python课程，我被赋予了一个做一个tic-tac-toe游戏的分配。我完成了所有工作，但是当占位符被用户选择的块中的O或X替换时，我仍然坚持更换列表元素。我已经尝试了很多方法来永久地替换列表元素，并花了好几个小时来找到我可以正确运行它的方式，但它仍然打印出相同的原始tic-tac-toe板。
# 2 players should be able to play the game (both sitting at the same computer)
# The board should be printed out every time a player makes a move
# You should be able to accept input of the player position and then place a symbol on the board
# x1 | y1 | z1
# ---- | ---- | ----
# x2 | y2 | z2
# ---- | ---- | ----
# x3 | y3 | z3
x1= 'top-left'
y1= 'top-mid'
z1= 'top-right'
x2= 'mid-left'
y2= 'mid-mid'
z2= 'mid-right'
x3= 'btm-left'
y3= 'btm-mid'
z3= 'btm-right'
e = [x1,y1,z1,x2,y2,z2,x3,y3,z3]
board = e[0]+ ' | '+ e[1]+ ' | '+ e[2]+'\n'+ \
'--------|------------|--------'+'\n'+\
e[3]+ ' | '+ e[4]+ ' | '+ e[5]+'\n'+ \
'--------|------------|--------'+'\n'+\
e[6]+ ' | '+ e[7]+ ' | '+ e[8]
def instructions():
print('Instuctions: \n')
print('Game: tic-tac-toe \n')
print('Player 1: O and Player 2: X \n\n')
print(' top-left | top-mid | top-right')
print(' -------- | ------- | ----------')
print(' mid-left | mid-mid | mid-right')
print(' -------- | ------- | ----------')
print(' btm-left | btm-mid | btm-right')
print('\n\n')
print('Choose your input when your turn comes.\n\n')
def player_one(inputvalue):
global e
for element in e:
if element == inputvalue:
e[e.index(element)]=' O ' #element that i wanna replace
break
query =checkSuccess()
if (query == True):
print('Tic-Tac-Toe ! Player1 has won the game! ')
elif (query == False):
print(board) # this still prints the original, old board.
def player_two(input):
global e
for element in e:
if element== input:
e[e.index(element)] =' X '
break
query =checkSuccess()
if (query == True):
print('Tic-Tac-Toe ! Player2 has won the game! ')
elif (query == False):
print(board)
ticTacToe()
board = e[0]+ ' | '+ e[1]+ ' | '+ e[2]+'\n'+ \
e[3]+ ' | '+ e[4]+ ' | '+ e[5]+'\n'+ \
e[6]+ ' | '+ e[7]+ ' | '+ e[8]
def checkSuccess():
if(e[0]==e[1]==e[2] or e[3]==e[4]==e[5] or e[6]==e[7]==e[8] or e[0] == e[3] == e[6] or
e[1]==e[4]==e[7] or e[2]==e[5]==e[8] or e[0]==e[4]==e[8] or e[6]==e[4]==e[2]):
return True
else:
return False
def ticTacToe():
p1 = input("Player1, Enter your choice \n")
player_one(p1)
p2 = input("Player2, Enter your choice \n")
player_two(p2)
instructions()
print(board)
ticTacToe()
我得到的输出：
top-left | top-mid | top-right
mid-left | mid-mid | mid-right
btm-left | btm-mid | btm-right
Player1, Enter your choice
mid-mid
top-left | top-mid | top-right
mid-left | mid-mid | mid-right
btm-left | btm-mid | btm-right
Player2, Enter your choice
btm-right
top-left | top-mid | top-right
mid-left | mid-mid | mid-right
btm-left | btm-mid | btm-right
Player1, Enter your choice
o / p仍然相同。有没有办法替换元素并打印列表的新值？
答案 0 :(得分：1)
Board变量只定义一次。它是一个字符串，它是不可变的，当你更改e列表的元素时它不会神奇地更新。正如Danilo中建议answer一样，您可以创建一个函数，并在每次打印时都调用它：
from beautifultable import BeautifulTable
def draw_board():
board = BeautifulTable()
board.append_row(e[:3])
board.append_row(e[3:6])
board.append_row(e[6:])
print(board)
您可以使用BeautifulTable格式化您的主板。但是，您需要先安装它：
pip install beautifultable
示例：
>>> e[0] = 'X'
>>> draw_board()
+----------+---------+-----------+
| X | top-mid | top-right |
+----------+---------+-----------+
| mid-left | mid-mid | mid-right |
+----------+---------+-----------+
| btm-left | btm-mid | btm-right |
+----------+---------+-----------+
答案 1 :(得分：-1)
好的这是一堆代码，请将来查看堆栈溢出策略的How to ask a good question。如果您使用列表get和set方法
首先，您要更改e列表而非更改电路板变量。董事会不可靠于电子清单。您已使用1变量创建另一个变量，现在创建了第二个变量，对1变量的任何更改都不会影响第二个变量。
Soo，只需将一块板作为一个函数，而不是采用全局参数e，它就可以工作。
def board():
global e
print( e[0]+ ' | '+ e[1]+ ' | '+ e[2]+'\n'+\
'--------|------------|--------'+'\n'+\
e[3]+ ' | '+ e[4]+ ' | '+ e[5]+'\n'+\
'--------|------------|--------'+'\n'+\
e[6]+ ' | '+ e[7]+ ' | '+ e[8] )
之后你将()括号for loop添加到你的每个变量中，因为它现在是一个函数而且是瞧。
每次想要更改变量时，您都不需要if input_string in e: e[ e.index(input_string) ] = "\tX\t"
else: print("this space is taken, take a hike!!") # or something
，如果变量具有特定值（在列表中也没有重复），您可以使用下一个方法：
putIp(ip, data): any {
const url = `${this.IpUrl}/${ip}`;
return this.authHttp.put(url, data, {headers: this.headers});}