我最近在Pygame从事游戏。它会加载两个不同的背景(以供将来进行视差滚动),然后将其放置在可以用箭头键控制并且可以左右移动的精灵中。在我运行它之前,所有负载都可以正常运行,直到尝试控制精灵为止。它的响应速度很慢,我必须继续按箭头键才能使其移动。我用来移动精灵的代码与我之前在另一个程序中使用的代码相同,该程序只是在后台和精灵中加载并控制了精灵的移动。对于我的新程序,我刚刚添加了屏幕调整大小和全屏选项。那是当它变慢的时候。是什么在减慢速度?我需要更好地优化代码吗?
如果有帮助,我还注意到当我快速按下另一个键以及箭头键时,它会以正确的速度移动。就像这样更新它。是否需要进行某种更新?另外,我已经在if语句中使用了“ Pygame.display.update()”来控制精灵键。
import pygame
from pygame.locals import *
pygame.init()
#two backgrounds
bg = pygame.image.load('/Users/luke.redwine/Documents/Python/PyGame/Llama game/Llama imaging/backgrounds concepts/Mountains/PPP_BG1.png')
fg = pygame.image.load('/Users/luke.redwine/Documents/Python/PyGame/Llama game/Llama imaging/backgrounds concepts/Mountains/PPP_fg1.png')
clock = pygame.time.Clock()
def Sresize():
#screen resize:
pygame.event.pump()
event=pygame.event.wait()
if event.type==QUIT:
pygame.quit()
elif event.type==VIDEORESIZE:
sprite.screen=pygame.display.set_mode(event.dict['size'],HWSURFACE|DOUBLEBUF|RESIZABLE)
sprite.screen.blit(pygame.transform.scale(bg,event.dict['size']),(0,0))
sprite.screen.blit(pygame.transform.scale(fg,event.dict['size']),(0,0))
pygame.display.update()
#PLAYER CLASS
class player(object):
def __init__(self,x, y, width, height):
self.x = x
self.y = y
self.width = width
self.height = height
self.velocity = 15
self.isJump = False
self.jumpCount = 7
self.right = False
self.left = False
self.walkCount = 0
self.screen = pygame.display.set_mode((1280, 800), HWSURFACE|DOUBLEBUF|RESIZABLE)
self.walkRight = [pygame.image.load('/Users/luke.redwine/Documents/Python/PyGame/Examples/sprite-Game/sprite characters main/R %s.png' % frame) for frame in range(1, 9)]
self.walkLeft = [pygame.image.load('/Users/luke.redwine/Documents/Python/PyGame/Examples/sprite-Game/sprite characters main/L %s.png' % frame) for frame in range(14, 21)]
self.char = pygame.image.load('/Users/luke.redwine/Documents/Python/PyGame/Examples/sprite-Game/sprite characters main/character 25.png')
def draw(self, screen):
if self.walkCount + 1 >= 8:
self.walkCount = 0
if self.left:
self.screen.blit(self.walkLeft[self.walkCount//1], (self.x, self.y))
self.walkCount += 1
elif self.right:
self.screen.blit(self.walkRight[self.walkCount//1], (self.x, self.y))
self.walkCount += 1
else:
self.screen.blit(self.char, (self.x, self.y))
sprite = player(4, 480, 16, 16)
def redrawGamescreen():
sprite.screen.blit(pygame.transform.scale(bg,(1280, 800)),(0,0))
sprite.screen.blit(pygame.transform.scale(fg,(1280, 800)),(0,0))
sprite.draw(sprite.screen)
pygame.display.update()
#mainloop:-----------------------------------------------------------------
run = True
while run:
keys = pygame.key.get_pressed()
Sresize()
clock.tick(20)
#left
if keys[pygame.K_LEFT] and sprite.x > sprite.velocity:
sprite.x -= sprite.velocity
sprite.left = True
sprite.right = False
#right
elif keys[pygame.K_RIGHT] and sprite.x < 1280 - sprite.width - sprite.velocity:
sprite.x += sprite.velocity
sprite.right = True
sprite.left = False
#otherwise, stand still
else:
sprite.right = False
sprite.left = False
sprite.walkCount = 0
#jumping
if not(sprite.isJump):
if keys[pygame.K_SPACE]:
sprite.isJump = True
sprite.right = False
sprite.left = False
else:
if sprite.jumpCount >= -7:
neg = 1
pygame.display.update()
if sprite.jumpCount < 0:
neg = -1
sprite.y -= (sprite.jumpCount ** 2) * 0.5 * neg
sprite.jumpCount -= 1
else:
sprite.isJump = False
sprite.jumpCount = 7
sprite.velocity = 15
redrawGamescreen()
pygame.quit()
答案 0 :(得分:2)
问题是调用event=pygame.event.wait()
。注意[pygame.event.wait()
](等待队列中的单个事件)等待队列中的单个事件。如果队列为空,此函数将等待直到创建一个。这会使您的游戏滞后。由于pygame.event.pump()
之前被调用,这一事实使问题更加复杂。
pygame.event.pump() event=pygame.event.wait()
改为使用pygame.event.get()
:
events=pygame.event.get()
pygame.event.get()
获取会从事件队列中删除所有消息,并且。
例如:
def HandleEvents():
events = pygame.event.get()
for event in events:
if event.type==QUIT:
pygame.quit()
elif event.type==VIDEORESIZE:
sprite.screen=pygame.display.set_mode(event.dict['size'],HWSURFACE|DOUBLEBUF|RESIZABLE)
sprite.screen.blit(pygame.transform.scale(bg,event.dict['size']),(0,0))
sprite.screen.blit(pygame.transform.scale(fg,event.dict['size']),(0,0))
pygame.display.update()
注意,为了获得良好的控制流程,您必须在应用程序循环中的事件循环上实现。事件循环必须处理所有用户输入。