在网格中处理光标移动时如何提高性能？ （pygame的）

Question

好的，用鼠标移动一些矩形很容易。然而，当你想在一个＆＃34;网格中移动它时，事情就不那么简单了。

我的游戏中的分辨率默认为1024 | 768，并且整个屏幕都填充了（1024/16）|（768/16）的图块。这意味着64 | 48。每块瓷砖可以是墙壁或玩家，敌人或物品或任何东西。

所以光标是一个＆＃34;选择光标＆＃34;提供有关图块的信息（它按块移动，而不是按像素移动）。

我知道怎么做，但是有一个问题：

def move(self,pos):

    mouseX = pos[0]
    mouseY = pos[1]

    for i in range(0,16*Wall.BRICK_HEIGHT,Wall.BRICK_HEIGHT): // Brick Height = 48
        for j in range(0,16*Wall.BRICK_WIDTH,Wall.BRICK_WIDTH): // Brick Width = 64
            if (((mouseX > j) and (mouseX < (j + Wall.BRICK_WIDTH)) and (mouseY > i) and (mouseY < (i + Wall.BRICK_HEIGHT)))):
                self.x = j
                self.y = i

我在主游戏循环中运行此程序：

while not e.Global.DONE:

game.runLanguageMenu()
game.runIntroduction()
game.runMainMenu()
game.runDungeonMenu()
game.loadStuff()
game.runPauseMenu()

event = pygame.event.poll()
if event.type == pygame.QUIT:
    e.Global.DONE = True
elif event.type == pygame.MOUSEMOTION:
    if e.Global.gameScreen == "GAME":
        game.cursor.move(event.pos) // <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

if e.Global.gameScreen == "GAME":
    game.player.move(pygame.key.get_pressed())
    game.runGame()

pygame.display.flip()

这意味着在EACH FRAME中，它会执行二维循环，以便将鼠标选择光标移动到切片中。结果？当然是滞后。 当鼠标位于其中时，光标会更改。为了检查它，必须循环遍历当前正在绘制的所有图块。你知道怎么做而不使用二维回路来节省一些处理能力吗？

Answer 1

您可以通过为块/切片提供世界坐标（它只是表示网格中块的列和行）来简化解决方案，而不是每次都手动计算它。

然后，您可以在将鼠标位置转换为世界坐标后简单地检查一下。

此外，您可以创建一个字典，将每个列/行映射到块/磁贴对象（您没有显示其余的代码，所以我不知道您实际存储的方式tile;也许是列表列表？如果是，那么你只需要用索引检索正确的tile。

那会让你摆脱循环。

---

此外，我认为使用pygame.event.poll()代替pygame.event.get()可能也是一个问题。

使用poll时，您只能从事件队列中获取一个事件。这意味着您每帧只能处理一个事件。也许这会导致你的滞后？

您也应该检查在主循环中运行的其他功能。我会运行一个分析器来查看游戏的瓶颈在哪里（有时结果令人惊讶）。

---

这是一个使用dict在鼠标光标下找到图块的简单示例。评论中有更多解释。

import pygame
import random

pygame.init()

BRICK_HEIGHT = 48
BRICK_WIDTH = 64

class Block(object):
    def __init__(self, x=0, y=0, num=0, width=0, color=(122, 122, 122)):
        # w_x and w_y represents the world coordinates
        # that means the 'column' and 'row' in the 'grid'
        self.w_x, self.w_y = x, y
        # to calculate the absolute (screen) position,
        # simply multiply x, y with BRICK_WIDTH, BRICK_HEIGHT
        self.rect = pygame.rect.Rect(x*BRICK_WIDTH, y*BRICK_HEIGHT, BRICK_WIDTH, BRICK_HEIGHT)
        self.width = width
        self.color = color

    def draw(self, surface):
        pygame.draw.rect(surface, self.color, self.rect, self.width)

    def move(self, global_pos):
        # to translate the absolute (screen) position
        # back to the world coordinates, simply divide
        # with BRICK_WIDTH, BRICK_HEIGHT
        x, y = global_pos
        self.w_x = (x / BRICK_WIDTH)
        self.w_y = (y / BRICK_HEIGHT)
        # recalculate the absolute (screen) position,
        # so the cursor "snaps" to the grid
        self.rect.x = self.w_x * BRICK_WIDTH
        self.rect.y = self.w_y * BRICK_HEIGHT

screen = pygame.display.set_mode((10*BRICK_WIDTH, 10*BRICK_HEIGHT))
clock = pygame.time.Clock()

c = Block(width=4, color=(200, 255, 200))

def r_color():
    return (random.randint(30, 255), 
            random.randint(30, 255),
            random.randint(30, 255))

# blocks maps (column, row) to a block
# note that this keeps the information of the
# position of a block at *two* places, which needs
# to be in sync. That's a drawback, but the
# lookup is fast!
blocks = {}
for x in xrange(10):
    for y in xrange(10):
        if random.randint(0, 100) < 33:
            blocks[(x, y)] = Block(x, y, color=r_color())

while True:
    pos = pygame.mouse.get_pos()
    c.move(pos)

    for e in pygame.event.get(): 
        if e.type == pygame.QUIT:
            raise Exception()
        if e.type == pygame.MOUSEBUTTONDOWN:
            # finding the tile under the mouse is as easy as:
            block = blocks.get((c.w_x, c.w_y))
            # and since it's a dict, the lookup is very fast.
            print 'Selected block color: {}'.format(block.color) if block else 'No block selected'

    screen.fill((0, 30, 30))
    for b in blocks.values():
        # The blocks calculated their absolute position by themself.
        # That may or may not what you want. Another way is to calculate
        # their absolute position here, so the blocks only needs to 
        # know about their world coordinates
        b.draw(screen)
    c.draw(screen)

    # run at 60 FPS
    clock.tick(60)
    pygame.display.flip()