我有这个简单的移动脚本。
if x > 0:
if key[pygame.K_a] or key[pygame.K_LEFT]:
rect_player.move_ip(-1 * speed, 0)
if x < SCREEN_WIDTH - 110:
if key[pygame.K_d] or key[pygame.K_RIGHT]:
rect_player.move_ip(speed, 0)
if y > 0:
if key[pygame.K_w] or key[pygame.K_UP]:
rect_player.move_ip(0, -1 * speed)
if y < SCREEN_HEIGHT - 110:
if key[pygame.K_s] or key[pygame.K_DOWN]:
rect_player.move_ip(0, speed)
当玩家朝一个方向移动时,一切都会顺利进行。但当它们沿对角线行进时,速度会更快。有什么办法可以解决这个问题吗?
pygame.Rect
应该代表屏幕上的一个区域,因此 pygame.Rect
对象只能存储整数数据(当您分配浮点数时,小数位会丢失)。
矩形对象的坐标都是整数。 [...]
pygame.math.Vector2
来解决这个问题。将对象的位置存储在 pygame.math.Vector2
对象中:
pos = pygame.math.Vector2(start_x, start_y)
根据按键设置方向向量。使用
speed
将向量缩放到
scale_to_length
的长度。移动对象并更新rect_player
:
key = pygame.key.get_pressed()
up = key[pygame.K_w] or key[pygame.K_UP]
down = key[pygame.K_s] or key[pygame.K_DOWN]
left = key[pygame.K_a] or key[pygame.K_LEFT]
right = key[pygame.K_d] or key[pygame.K_RIGHT]
move = pygame.math.Vector2(right - left, down - up)
if move.length_squared() > 0:
move.scale_to_length(speed)
pos += move
rect_player.topleft = round(pos.x), round(pos.y)
如果您不关心浮点精度,只需移动矩形即可:
key = pygame.key.get_pressed()
up = key[pygame.K_w] or key[pygame.K_UP]
down = key[pygame.K_s] or key[pygame.K_DOWN]
left = key[pygame.K_a] or key[pygame.K_LEFT]
right = key[pygame.K_d] or key[pygame.K_RIGHT]
move = pygame.math.Vector2(right - left, down - up)
if move.length_squared() > 0:
move.scale_to_length(speed)
rect_player.move_ip(round(move.x), round(move.y))
最小示例:
import pygame
pygame.init()
window = pygame.display.set_mode((500, 500))
clock = pygame.time.Clock()
rect_player = pygame.Rect(0, 0, 20, 20)
rect_player.center = window.get_rect().center
speed = 5
run = True
while run:
for event in pygame.event.get():
if event.type == pygame.QUIT:
run = False
key = pygame.key.get_pressed()
up = key[pygame.K_w] or key[pygame.K_UP]
down = key[pygame.K_s] or key[pygame.K_DOWN]
left = key[pygame.K_a] or key[pygame.K_LEFT]
right = key[pygame.K_d] or key[pygame.K_RIGHT]
move = pygame.math.Vector2(right - left, down - up)
if move.length_squared() > 0:
move.scale_to_length(speed)
rect_player.move_ip(round(move.x), round(move.y))
window.fill(0)
pygame.draw.rect(window, (255, 0, 0), rect_player)
pygame.display.flip()
clock.tick(60)
pygame.quit()
exit()
如果您通过
math.sqrt(2)
导出速度,但继续使用 pygame.Rect
,它仍然不会导致准确的运动。 pygame.Rect 对象 (rect_player
) 只能存储整数值(当您分配浮点数时,小数位会丢失)。因此,如果种子值很小(例如 1),这只能近似起作用并且完全失败。该解决方案真正有效的唯一值是 speed=10
,因为 10/math.sqrt(2)
是 7.071 (~7)。
感谢@Rabbid76 帮助我解决这个问题,但这对我来说很有效:
count = 0
if x > 0:
if key[pygame.K_a] or key[pygame.K_LEFT]:
count += 1
if x < SCREEN_WIDTH - 110:
if key[pygame.K_d] or key[pygame.K_RIGHT]:
count += 1
if y > 0:
if key[pygame.K_w] or key[pygame.K_UP]:
count += 1
if y < SCREEN_HEIGHT - 110:
if key[pygame.K_s] or key[pygame.K_DOWN]:
count += 1
if count >= 2:
speed = speed/math.sqrt(2)
if x > 0:
if key[pygame.K_a] or key[pygame.K_LEFT]:
rect_player.move_ip(-1 * speed, 0)
if x < SCREEN_WIDTH - 110:
if key[pygame.K_d] or key[pygame.K_RIGHT]:
rect_player.move_ip(speed, 0)
if y > 0:
if key[pygame.K_w] or key[pygame.K_UP]:
rect_player.move_ip(0, -1 * speed)
if y < SCREEN_HEIGHT - 110:
if key[pygame.K_s] or key[pygame.K_DOWN]:
rect_player.move_ip(0, speed)