tictactoe in python for 2 players

tictactoe.py

"""tictactoe game for 2 players
from blogpost: http://thebillington.co.uk/blog/posts/writing-a-tic-tac-toe-game-in-python by  BILLY REBECCHI,
slightly improved by Horst JENS"""
from __future__ import print_function

choices = []

for x in range (0, 9) :
    choices.append(str(x + 1))

playerOneTurn = True
winner = False

def printBoard() :
    print( '\n -----')
    print( '|' + choices[0] + '|' + choices[1] + '|' + choices[2] + '|')
    print( ' -----')
    print( '|' + choices[3] + '|' + choices[4] + '|' + choices[5] + '|')
    print( ' -----')
    print( '|' + choices[6] + '|' + choices[7] + '|' + choices[8] + '|')
    print( ' -----\n')

while not winner :
    printBoard()

    if playerOneTurn :
        print( "Player 1:")
    else :
        print( "Player 2:")

    try:
        choice = int(input(">> "))
    except:
        print("please enter a valid field")
        continue
    if choices[choice - 1] == 'X' or choices [choice-1] == 'O':
        print("illegal move, plase try again")
        continue

    if playerOneTurn :
        choices[choice - 1] = 'X'
    else :
        choices[choice - 1] = 'O'

    playerOneTurn = not playerOneTurn

    for x in range (0, 3) :
        y = x * 3
        if (choices[y] == choices[(y + 1)] and choices[y] == choices[(y + 2)]) :
            winner = True
            printBoard()
        if (choices[x] == choices[(x + 3)] and choices[x] == choices[(x + 6)]) :
            winner = True
            printBoard()

    if((choices[0] == choices[4] and choices[0] == choices[8]) or 
       (choices[2] == choices[4] and choices[4] == choices[6])) :
        winner = True
        printBoard()

print ("Player " + str(int(playerOneTurn + 1)) + " wins!\n")

You might want to define the numbers that draw the game. Also might be nice to ask if players want to play again

For the following cases, printBoard is called twice -

X X X
X O O
X O O

X O O
X X O
X O X

The conditions on lines 52 and 56 should include (not winner) and

A very efficient way to create Tic Tac Toe game in python.
tic_tic_toe.py

same code with "Lost" condition

from future import print_function

choices = []
checkturns = []
for x in range(0, 9):
choices.append(str(x + 1))

playerOneTurn = True
winner = False

def printBoard():
print('\n -----')
print('|' + choices[0] + '|' + choices[1] + '|' + choices[2] + '|')
print(' -----')
print('|' + choices[3] + '|' + choices[4] + '|' + choices[5] + '|')
print(' -----')
print('|' + choices[6] + '|' + choices[7] + '|' + choices[8] + '|')
print(' -----\n')

while not winner:
printBoard()

if (len(checkturns) == 9 and winner == False):
    print("Both lost")
    break

if playerOneTurn:
    print("Player 1:")
else:
    print("Player 2:")

try:
    choice = int(input(">> "))
    checkturns.append(choice)
except:
    print("please enter a valid field")
    continue
if choices[choice - 1] == 'X' or choices[choice-1] == 'O':
    print("illegal move, plase try again")
    continue

if playerOneTurn:
    choices[choice - 1] = 'X'
else:
    choices[choice - 1] = 'O'

playerOneTurn = not playerOneTurn

for x in range(0, 3):
    y = x * 3
    if (choices[y] == choices[(y + 1)] and choices[y] == choices[(y + 2)]):
        winner = True
        printBoard()
    if (choices[x] == choices[(x + 3)] and choices[x] == choices[(x + 6)]):
        winner = True
        printBoard()

if((choices[0] == choices[4] and choices[0] == choices[8]) or
   (choices[2] == choices[4] and choices[4] == choices[6])):
    winner = True
    printBoard()

if winner == True:
print("Player " + str(int(playerOneTurn + 1)) + " wins!\n")

tnx............. bro

Amazing code but it doesn't have a way to show the game is draw

I kinda changed the code to solve that
link: https://gist.github.com/vivek1339/a35edf19740bc5c58bb729b1bbf690ca/eac36b2e6f90139c33332b04e7f5c26c5de2198b

Doesn't stop when game draws...