我想在屏幕上随机绘制矩形,但是如果已经绘制的矩形数超过3,那么我想开始删除“较旧的矩形”,这意味着正在绘制的矩形屏幕最多。
import pygame
import random
import time
pygame.init()
x_axis = [500, 650, 350, 400]
y_axis = [100, 50, 450, 300]
sq_width = [10, 15, 20, 25, 30]
sec = [0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5]
sq_display = []
class Squares:
def __init__(self, x, y, width, height):
self.x = x
self.y = y
self.width = width
self.height = width
def draw(self):
a = pygame.draw.rect(window, (255, 0, 0), (random.choice(x_axis), random.choice(y_axis), self.width, self.height))
sq_display.append(a)
if len(sq_display) > 3:
sq_display.remove(sq_display[0])
time.sleep(random.choice(sec))
return
我尝试每次将矩形存储到变量中并将其附加到列表中,以为以后可以删除它。好吧,它没有用。所以我想知道我的问题是否有解决方案。
答案 0 :(得分:2)
从draw
方法中删除新平方的生成:
class Squares:
def __init__(self, x, y, width):
self.x = x
self.y = y
self.width = width
self.height = width
def draw(self):
pygame.draw.rect(window, (255, 0, 0), (self.x, self.y, self.width, self.height))
您必须在应用程序循环的每一帧中重新绘制整个场景。在绘制正方形之前先清除显示,然后在列表中绘制所有正方形,最后更新显示:
while run:
# [...]
a = Squares(random.choice(x_axis), random.choice(y_axis),
random.choice(sq_width), random.choice(sq_width))
sq_display.append(a)
if len(sq_display) > 3:
sq_display.remove(sq_display[0])
window.fill(0)
for r in sq_display:
r.draw()
pygame.display.flip()
如果要保持应用程序响应,则不能将应用程序循环延迟time.sleep
。使用pygame.time.get_ticks()
获取自pygame.init()
以来的当前毫秒数,并在随机时间过去后创建一个新的平方:
next_square_time = 0
while run:
# [...]
current_time = pygame.time.get_ticks()
if next_square_time <= current_time:
next_square_time += random.choice(sec) * 1000
# create new square
# [...]
最小示例:
import pygame
import random
pygame.init()
x_axis = [500, 650, 350, 400]
y_axis = [100, 50, 450, 300]
sq_width = [10, 15, 20, 25, 30]
sec = [0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5]
sq_display = []
class Squares:
def __init__(self, x, y, width):
self.x = x
self.y = y
self.width = width
self.height = width
def draw(self):
pygame.draw.rect(window, (255, 0, 0), (self.x, self.y, self.width, self.height))
window = pygame.display.set_mode((800, 600))
next_square_time = 0
run = True
while run:
for event in pygame.event.get():
if event.type == pygame.QUIT:
run = False
current_time = pygame.time.get_ticks()
if next_square_time <= current_time:
next_square_time += random.choice(sec) * 1000
a = Squares(random.choice(x_axis), random.choice(y_axis), random.choice(sq_width))
sq_display.append(a)
if len(sq_display) > 3:
sq_display.remove(sq_display[0])
window.fill(0)
for r in sq_display:
r.draw()
pygame.display.flip()
答案 1 :(得分:0)
存储所有当前绘制的矩形的队列,每当列表长度超过3时,只需使用存储的坐标绘制一个白色矩形(或背景的任何颜色)即可。
答案 2 :(得分:-1)
只需将矩形移出显示屏即可。
例如。如果您的显示尺寸为800x600像素,则将矩形X,Y坐标更改为1000,1000。它在屏幕上将不再可见