Tru*_*ude 3 pygame python-3.x pygame-surface
在 的文档页面上pygame.display.update(),它说您可以将 rect 传递到方法中以更新部分屏幕。但是,我看到的所有示例都只是从程序中的图像或形状传递了一个现有的矩形。如何告诉它直接更新屏幕上的区域?例如,在绘制矩形时,我可以使用 rect 参数(100,200,30,40)。这将绘制一个矩形,顶部为 200,左侧为 100,宽度为 30,高度为 40。如何将类似的参数传递给pygame.display.update()?我试过pygame.display.update((100,200,30,40)),但这会更新整个窗口。
只需定义一个矩形并将其传递给以pygame.display.update()仅更新显示的此特定区域。您还可以传递一个矩形列表。
import random
import pygame as pg
from pygame.math import Vector2
# A simple sprite, just to have something moving on the screen.
class Ball(pg.sprite.Sprite):
def __init__(self, screen_rect):
super().__init__()
radius = random.randrange(5, 31)
self.image = pg.Surface((radius*2, radius*2), pg.SRCALPHA)
pg.draw.circle(self.image, pg.Color('dodgerblue1'), (radius, radius), radius)
pg.draw.circle(self.image, pg.Color('dodgerblue3'), (radius, radius), radius-2)
self.rect = self.image.get_rect(center=screen_rect.center)
self.vel = Vector2(random.uniform(-2, 2), random.uniform(-2, 2))
self.pos = Vector2(self.rect.center)
self.screen_rect = screen_rect
self.lifetime = 350
def update(self):
self.pos += self.vel
self.rect.center = self.pos
self.lifetime -= 1
if not self.screen_rect.contains(self.rect) or self.lifetime <= 0:
self.kill()
def main():
screen = pg.display.set_mode((800, 600))
screen.fill((20, 40, 70))
pg.display.update()
screen_rect = screen.get_rect()
clock = pg.time.Clock()
all_sprites = pg.sprite.Group()
# Pass this rect to `pg.display.update` to update only this area.
update_rect = pg.Rect(50, 50, 500, 400)
done = False
while not done:
for event in pg.event.get():
if event.type == pg.QUIT:
done = True
all_sprites.add(Ball(screen_rect))
all_sprites.update()
screen.fill((20, 50, 90))
all_sprites.draw(screen)
# Update only the area that we specified with the `update_rect`.
pg.display.update(update_rect)
clock.tick(60)
if __name__ == '__main__':
pg.init()
main()
pg.quit()
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