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我被困住了，我正在使用 Python 和 Pygame 制作游戏项目。（屏幕为 770 x 485）它涉及在随机生成的矩形下拉时将一个圆圈向左或向右移动通过 5 个垂直车道。我有一个 Wall 类来随机生成它们，但我不知道如何检测矩形的每个实例和我的圆之间的碰撞。

就在带有 z 和 c 列表的位中，我尝试尝试使它们发生碰撞，但它只与第一个矩形发生碰撞，并且它没有检测到任何其他下降的矩形（它们是在彼此之间 100 个滴答声之后生成的) 谁能指出我正确的方向？我真的不确定 z 变量、c 列表和 v 变量：\

提前致谢！

class Wall:
    def __init__(self,colour,x,y,life):
        self.colour = colour
        self.x = x
        self.y = y
        self.life = life
    def drop(self):
        pygame.draw.rect(screen, self.colour, (self.x, self.y, 154, 30))
        self.y += 1.25
        self.life -= 1
        if self.life < 0:
            wall.remove(self)

def playMode(): #assume everything is indented properly
global wall, points
check   = True
left    = False
right   = False
circlex = 385
degrees = 0
health  = 3
wall = []
x = 2
points = 0
n = 0
c = []
starttime = time.time()
while True:
    runningtime = time.clock()
    screen.fill((255,255,255))
    x -= 1
    if x == 1:
      wall.append(Wall((random.randint(1,255),random.randint(1,255),random.randint(1,255)),
random.choice([0, 154, 308, 462, 616]), -30 , 450))
        x = 100
    for i in wall:
        i.drop()
    z = wall[-1]
    c.append(z)
    v = c[(n)]
    if 445 >= v.y >= 340 and v.x == (circlex - 77):           
        health -= 1
        points -= 5
        v.y = 485 #moves this instance offscreen so it doesn't make the hp go to 0 immediately
        n += 1
        print v.y

    circle = pygame.draw.circle(screen, colour,(circlex, 409), 50)

    for event in pygame.event.get():
        if event.type == KEYDOWN:
            if event.key == K_ESCAPE:
                showPauseScreen()
            elif event.key == K_p:
                showPauseScreen()
            elif event.key == K_LEFT:
                if circlex > 77:
                    circlex = circlex - 154
            elif event.key == K_RIGHT:
                if circlex < 616:
                    circlex = circlex + 154
        elif event.type == QUIT:
            terminate()

    if health == 0:
        check = False
        return #goes to gameover screen basically
    pygame.display.update()
    clock.tick(100)
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1 回答 1

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你可以看看 pygame 文档，但基本上你想使用这个函数，它将返回一个精灵是否与另一个精灵发生碰撞的布尔值：

pygame.sprite.spritecollide(<sprite>,<group>,False)     #in almost all circumstances the third argument should be False. This will check for collision between a sprite and group.

有关更多信息，请查看 pygame 文档：

http://www.pygame.org/docs/ref/sprite.html#pygame.sprite.spritecollide

于 2013-11-22T02:42:56.573 回答