不使用鼠標關閉pygame

Question

我做了一個基本的井字游戲，我有 2 個特工隨機玩。 我希望游戲運行 10000 次。 我現在可以這樣做，但只有在我按下 pygame window 上的關閉按鈕后，實例才會重置。 相反，我希望 window 不斷關閉並再次打開，或者讓 window 通過重置游戲板等保持打開狀態 - 兩者都很好。 這是代碼：

import pygame
import numpy as np
import random
import time
import sys

def get_position(mouseX, mouseY):
    if mouseX <= width/3:
        col = 0
    elif mouseX <= 2*width/3:
        col = 1
    else:
        col = 2

    if mouseY <= height/3:
        row = 0
    elif mouseY <= 2*height/3:
        row = 1
    else:
        row = 2

    return (row, col)
def show_board(input, board):
    input.blit(board, (0,0))
    pygame.display.flip()

def initialise_board(input):
    background = pygame.Surface(input.get_size())
    background = background.convert()
    background.fill((250, 250, 250))
    pygame.draw.line(background, (0, 0, 0), (width / 3, 0), (width / 3, height), 2)
    pygame.draw.line(background, (0, 0, 0), (2 * width / 3, 0), (2 * width / 3, height), 2)
    pygame.draw.line(background, (0, 0, 0), (0, height / 3), (width, height / 3), 2)
    pygame.draw.line(background, (0, 0, 0), (0, 2 * height / 3), (width, 2 * height / 3), 2)
    input.blit(background, (0, 0))
    pygame.display.flip()
    return background

def check_board(row, col, array_board):
    if array_board[row][col] != 0:
        return False
    else:
        return True


#def get_mouse():
#    (mouseX, mouseY) = pygame.mouse.get_pos()
#    (row, col) = get_position(mouseX, mouseY)
#    return (row, col)

def check_win(array_board):
    global winner
    for row in range(3):
        if array_board[row][0] == array_board[row][1] == array_board[row][2] != 0:
            winner = array_board[row][0]
            pygame.draw.line(board, (0, 0, 0), (75, (row*round(height/3) + 150)), (825, (row*round(height/3) + 150)), 3)
            break
    for col in range(3):
        if array_board[0][col] == array_board[1][col] == array_board[2][col] != 0:
            winner = array_board[col][0]
            pygame.draw.line(board, (0, 0, 0), (col*round(width/3) + 150, 75), (col*round(width/3)+ 150, 825), 3)
            break

    if array_board[0][0] == array_board[1][1] == array_board[2][2] != 0:
        winner = array_board[0][0]
        pygame.draw.line(board, (0, 0, 0,), (75, 75), (825, 825), 3)
    if array_board[0][2] == array_board[1][1] == array_board[2][0] != 0:
        winner = array_board[0][2]
        pygame.draw.line(board, (0,0,0), (825, 75), (75, 825), 3)
    return winner

def click_board(array_board):
    global team
    (row, col) = get_position(random.randint(1,900), random.randint(1,900))

    centerX = (col * round(width/3)) + 150
    centerY = (row * round(height/3)) + 150
    if team == 2 and check_board(row, col, array_board):
        pygame.draw.circle(board, (0, 0, 0), np.round((centerX, centerY)), round(width/12), 2)
        team = 1
        array_board[row][col] = 2
    elif team == 1 and check_board(row, col, array_board):
        pygame.draw.line(board, (0,0,0), (centerX - width/15, centerY - height/15), (centerX + width/15, centerY + height/15), 2)
        pygame.draw.line(board, (0,0,0), (centerX + width/15, centerY - height/15), (centerX - width/15, centerY + height/15), 2)
        team = 2
        array_board[row][col] = 1


def game(array_board, board):
    global winner, xwins, owins
    winner = check_win(array_board)
    if winner == 0 and team == 1:
        message = "X's turn"
    elif winner == 0 and team == 2:
        message = "O's turn"
    elif winner != 0 and team == 2:
        message = "X wins!"
        xwins += 1
    elif winner != 0 and team == 1:
        message = "O wins!"
        owins += 1

    font = pygame.font.Font(None, 24)
    text = font.render(message, 1, (10,10,10))
    board.fill((250,250,250), (width/2 - 20, 50, width/2 + 20, 50))
    board.blit(text, (width/2,50))

xwins = 0
owins = 0

for i in range(10):
    pygame.init()
    width = 900
    height = 900
    input = pygame.display.set_mode((width,height))
    pygame.display.set_caption('Tic-Tac-Toe')
    array_board = np.zeros([3,3])
    winner = 0
    team = 1 #team 1 is X, team 2 is O
    board = initialise_board(input)



    running = True
    while running:
        for event in pygame.event.get():
                if event.type == pygame.QUIT:
                    running = False
            #elif event.type == pygame.MOUSEBUTTONDOWN and winner == 0:
        if winner == 0:
            print(winner)
            game(array_board, board)
            show_board(input, board)
            check_win(array_board)
            click_board(array_board)
            time.sleep(0.1)

    pygame.display.quit()
    pygame.quit()

Answer 1

有兩個條件結束游戲：

• 獲勝者已設置（ winner != 0 ）
• 或操場上的所有場地都被占用
  ( np.array(np.where(array_board == 0)).size == 0 )

您所要做的就是在您的應用程序中實現else案例並重置所有游戲狀態並重新繪制空的游樂場。
此外，您可以計算游戲數並在 10000 場游戲后break循環。 for循環（ for i in range(10): ）不再需要。

no_of_games = 0
running = True
while running:
    for event in pygame.event.get():
            if event.type == pygame.QUIT:
                running = False
    
    end_of_game = winner != 0 or np.array(np.where(array_board == 0)).size == 0
    if not end_of_game:
        print(winner)
        game(array_board, board)
        show_board(input, board)
        check_win(array_board)
        click_board(array_board)
        time.sleep(0.1)
    else:
        array_board = np.zeros([3,3])
        team = 1
        winner = 0
        time.sleep(1)
        board = initialise_board(input)
        no_of_games += 1

    if no_of_games >= 10000:
        break

pygame.display.quit()
pygame.quit()