我怎样才能随机地在蛇身上打洞?
How can I randomly make holes in snake?
我有一条蛇在他身后转动并绘制轨迹,但我不知道如何在他绘制时像随机停顿一样打洞。我试过了,它不是随机的,但问题是它的暂停速度非常快。这就像传送移动。
这是我的代码:
import pygame
pygame.init()
import random
from random import randint
win = pygame.display.set_mode((1280,720))
background = pygame.Surface(win.get_size(), pygame.SRCALPHA)
background.fill((0, 0, 0, 1))
x = randint(150,1130)
y = randint(150,570)
vel = 0.6
drawing_time = 0
direction = pygame.math.Vector2(vel, 0).rotate(random.randint(0, 360))
run = True
while run:
pygame.time.delay(5)
drawing_time += 1
if drawing_time == 1000:
for pause in range(0, 60):
head = pygame.draw.circle(win, (255,0,0), (round(x), round(y)), 0)
keys = pygame.key.get_pressed()
if keys[pygame.K_LEFT]:
direction.rotate_ip(-0.8)
if keys[pygame.K_RIGHT]:
direction.rotate_ip(0.8)
x += direction.x
y += direction.y
time = 0
keys = pygame.key.get_pressed()
if keys[pygame.K_LEFT]:
direction.rotate_ip(-0.8)
if keys[pygame.K_RIGHT]:
direction.rotate_ip(0.8)
x += direction.x
y += direction.y
win.blit(background, (12020,0))
head= pygame.draw.circle(win, (255,0,0), (round(x), round(y)), 5)
pygame.display.update()
pygame.quit()
我试着给那个头部半径 0,也许这是个问题,但我不知道如何解决它。
切勿在应用程序循环中实现控制游戏的循环。您需要应用程序循环。使用它!
使用pygame.time.get_ticks()
to return the number of milliseconds since pygame.init()
被调用。添加一个变量,指示是否需要绘制蛇 (draw_snake
)。定义一个变量,指示状态需要更改的时间 (next_change_time
)。当时间到期时,更改变量的状态并定义必须再次更改状态的新随机时间。仅在设置变量时绘制蛇。这会导致蛇上出现洞:
next_change_time = 0
draw_snake = False
run = True
while run:
# [...]
current_time = pygame.time.get_ticks()
if current_time > next_change_time:
next_change_time = current_time + random.randint(500, 1000)
draw_snake = not draw_snake
# [...]
if draw_snake:
pygame.draw.circle(win, (255,0,0), (round(x), round(y)), 5)
# [...]
使用pygame.time.Clock
控制每秒帧数,从而控制游戏速度。
方法 tick()
of a pygame.time.Clock
对象,以这种方式延迟游戏,即循环的每次迭代都消耗相同的时间。
这意味着循环:
clock = pygame.time.Clock()
run = True
while run:
clock.tick(60)
每秒运行 60 次。
完整示例:
import pygame
pygame.init()
import random
from random import randint
win = pygame.display.set_mode((1280,720))
background = pygame.Surface(win.get_size(), pygame.SRCALPHA)
background.fill((0, 0, 0, 1))
clock = pygame.time.Clock()
x = randint(150,1130)
y = randint(150,570)
vel = 0.6
drawing_time = 0
next_change_time = 0
draw_snake = False
direction = pygame.math.Vector2(vel, 0).rotate(random.randint(0, 360))
run = True
while run:
clock.tick(60)
for event in pygame.event.get():
if event.type == pygame.QUIT:
run = False
current_time = pygame.time.get_ticks()
if current_time > next_change_time:
next_change_time = current_time + random.randint(500, 1000)
draw_snake = not draw_snake
keys = pygame.key.get_pressed()
if keys[pygame.K_LEFT]:
direction.rotate_ip(-0.8)
if keys[pygame.K_RIGHT]:
direction.rotate_ip(0.8)
x += direction.x
y += direction.y
win.blit(background, (12020,0))
if draw_snake:
pygame.draw.circle(win, (255,0,0), (round(x), round(y)), 5)
pygame.display.update()
pygame.quit()
如果您想要其他尺寸的孔,附加条件和不同的随机时间:
if current_time > next_change_time:
draw_snake = not draw_snake
if draw_snake:
next_change_time = current_time + random.randint(500, 1000)
else:
next_change_time = current_time + random.randint(250, 500)
我有一条蛇在他身后转动并绘制轨迹,但我不知道如何在他绘制时像随机停顿一样打洞。我试过了,它不是随机的,但问题是它的暂停速度非常快。这就像传送移动。 这是我的代码:
import pygame
pygame.init()
import random
from random import randint
win = pygame.display.set_mode((1280,720))
background = pygame.Surface(win.get_size(), pygame.SRCALPHA)
background.fill((0, 0, 0, 1))
x = randint(150,1130)
y = randint(150,570)
vel = 0.6
drawing_time = 0
direction = pygame.math.Vector2(vel, 0).rotate(random.randint(0, 360))
run = True
while run:
pygame.time.delay(5)
drawing_time += 1
if drawing_time == 1000:
for pause in range(0, 60):
head = pygame.draw.circle(win, (255,0,0), (round(x), round(y)), 0)
keys = pygame.key.get_pressed()
if keys[pygame.K_LEFT]:
direction.rotate_ip(-0.8)
if keys[pygame.K_RIGHT]:
direction.rotate_ip(0.8)
x += direction.x
y += direction.y
time = 0
keys = pygame.key.get_pressed()
if keys[pygame.K_LEFT]:
direction.rotate_ip(-0.8)
if keys[pygame.K_RIGHT]:
direction.rotate_ip(0.8)
x += direction.x
y += direction.y
win.blit(background, (12020,0))
head= pygame.draw.circle(win, (255,0,0), (round(x), round(y)), 5)
pygame.display.update()
pygame.quit()
我试着给那个头部半径 0,也许这是个问题,但我不知道如何解决它。
切勿在应用程序循环中实现控制游戏的循环。您需要应用程序循环。使用它!
使用pygame.time.get_ticks()
to return the number of milliseconds since pygame.init()
被调用。添加一个变量,指示是否需要绘制蛇 (draw_snake
)。定义一个变量,指示状态需要更改的时间 (next_change_time
)。当时间到期时,更改变量的状态并定义必须再次更改状态的新随机时间。仅在设置变量时绘制蛇。这会导致蛇上出现洞:
next_change_time = 0
draw_snake = False
run = True
while run:
# [...]
current_time = pygame.time.get_ticks()
if current_time > next_change_time:
next_change_time = current_time + random.randint(500, 1000)
draw_snake = not draw_snake
# [...]
if draw_snake:
pygame.draw.circle(win, (255,0,0), (round(x), round(y)), 5)
# [...]
使用pygame.time.Clock
控制每秒帧数,从而控制游戏速度。
方法 tick()
of a pygame.time.Clock
对象,以这种方式延迟游戏,即循环的每次迭代都消耗相同的时间。
这意味着循环:
clock = pygame.time.Clock() run = True while run: clock.tick(60)
每秒运行 60 次。
完整示例:
import pygame
pygame.init()
import random
from random import randint
win = pygame.display.set_mode((1280,720))
background = pygame.Surface(win.get_size(), pygame.SRCALPHA)
background.fill((0, 0, 0, 1))
clock = pygame.time.Clock()
x = randint(150,1130)
y = randint(150,570)
vel = 0.6
drawing_time = 0
next_change_time = 0
draw_snake = False
direction = pygame.math.Vector2(vel, 0).rotate(random.randint(0, 360))
run = True
while run:
clock.tick(60)
for event in pygame.event.get():
if event.type == pygame.QUIT:
run = False
current_time = pygame.time.get_ticks()
if current_time > next_change_time:
next_change_time = current_time + random.randint(500, 1000)
draw_snake = not draw_snake
keys = pygame.key.get_pressed()
if keys[pygame.K_LEFT]:
direction.rotate_ip(-0.8)
if keys[pygame.K_RIGHT]:
direction.rotate_ip(0.8)
x += direction.x
y += direction.y
win.blit(background, (12020,0))
if draw_snake:
pygame.draw.circle(win, (255,0,0), (round(x), round(y)), 5)
pygame.display.update()
pygame.quit()
如果您想要其他尺寸的孔,附加条件和不同的随机时间:
if current_time > next_change_time:
draw_snake = not draw_snake
if draw_snake:
next_change_time = current_time + random.randint(500, 1000)
else:
next_change_time = current_time + random.randint(250, 500)