如何同时移动多只乌龟？

Question

我有一个作业，要求将两只乌龟放在赛道上（大小相同但赛道不同）。我可以让它们移动，但只有当第一个移动了一半轨道时，第二个才会移动。我不知道如何让乌龟同时移动。这是我的代码：

import turtle
import random
import time

wn = turtle.Screen()
wn.bgcolor("lightgreen")

t = turtle.Turtle()
t.shape('turtle')
t.color('red')

t2 = turtle.Turtle()
t2.shape('turtle')
t2.color('blue')

#user input function

p = float(input('please insert the perimeter:'))

#set the track
def drawTrack(p,r):
    shortside = (p/2.0)/(r+1)
    longside = r*shortside
    turtle.setup((shortside*2)+60, longside +40)
    t.penup()
    t2.penup()
    t.setposition(-shortside-10, -longside/2)
    t2.setposition(10, -longside/2)   
    for i in range (2):
        #first track
        t.speed(1)
        t.pendown()
        t.forward(shortside)
        t.left(90)
        t.forward(longside)
        t.left(90)
        
        #second track
        t2.speed(1)
        t2.pendown()
        t2.forward(shortside)
        t2.left(90)
        t2.forward(longside)
        t2.left(90) 

drawTrack(p,2)

wn.exitonclick()

Answer 1

有几种方法可以解决这个问题。

使用

screen.ontimer()

事件的一种方法（请参阅海龟文档）。这种方法很好，因为您可以将海龟调整为实际时钟时间，并且这可以在海龟事件循环内运行，以便其他事件也可以发生（如

exitonclick()

）。

我下面使用的方法是将海龟运动分解为 Python generator 中的微小步骤，它在每次运动后都会产生结果。这使我们能够在海龟之间交替。比赛发生在海龟事件循环之前，因此

exitonclick()

在比赛结束之前无效。

为了提供速度差异，我使用海龟绘制速度作为运动计算的一部分，因此如果您说

turtle1.speed("fast")

，它会比

turtle2.speed("slow")

移动得更快。还有其他方法可以使用随机和/或变化的速度来解决此问题。

我已将周长提示更改为对话框，并进行了各种样式调整：

from turtle import Turtle, Screen

screen = Screen()
screen.bgcolor("lightgreen")

turtle1 = Turtle(shape='turtle')
turtle1.color('red')
turtle1.speed("slow")  # = 3
turtle1.penup()

turtle2 = Turtle(shape='turtle')
turtle2.color('blue')
turtle2.speed(4)  # "slow" (3) < 4 < "normal" (6)
turtle2.penup()

# user input function

perimeter = screen.numinput("Track Perimeter", "Please enter the perimeter:", default=2000, minval=500, maxval=3000)

def full_track_crawl(turtle, shortside, longside):
    speed = turtle.speed()
    turtle.pendown()

    for j in range (2):
        for i in range(0, int(shortside), speed):
            turtle.forward(speed)
            yield(0)
        turtle.left(90)
        for i in range(0, int(longside), speed):
            turtle.forward(speed)
            yield(0)
        turtle.left(90)

    turtle.penup()

# set the track
def drawTrack(perimeter, ratio):
    shortside = (perimeter / 2.0) / (ratio + 1)
    longside = ratio * shortside

    screen.setup(shortside * 2 + 60, longside + 40)

    turtle1.setposition(-shortside - 10, -longside / 2)
    turtle2.setposition(10, -longside / 2)   

    generator1 = full_track_crawl(turtle1, shortside, longside)
    generator2 = full_track_crawl(turtle2, shortside, longside)

    while (next(generator1, 1) + next(generator2, 1) < 2):
        pass

drawTrack(perimeter, 2)

screen.exitonclick()

快乐赛车！

Answer 2

我让每只海龟重复移动一个微小的增量，这样它们之间的最大距离就只有 1 个像素：

# Setup 
import turtle

screen = turtle.Screen()
screen.bgcolor("white")
screen.title("Turtle Movement")
screen.setup(width=500, height=500)

# Object 1 set up
obj1 = turtle.Turtle()
obj1.color("red") # Sets obj1's colour
obj1.speed(0) # The drawing speed will go as fast as it can if it is set to zero
obj1.penup() # Don't want it to draw a line as I goes to 0, -50
obj1.goto(0, -50) # goes to 0, -50
obj1.pendown() # Now we want it to draw

# Object 2 set up
obj2 = turtle.Turtle()
obj2.color("blue") # sets obj2's colour
obj2.speed(0) # The drawing speed will go as fast as it can if it is set to zero
obj2.penup() # Don't want it to draw a line as I goes to 0, -50
obj2.goto(0, 50) # goes to 0, 50
obj2.pendown() # Now we want it to draw

# Movement of objects
while True: # infinite loop
    obj1.forward(1) # Moves obj1 forwards by 1
    obj2.forward(1) # Moves obj2 forwards by 1

注意： ‘移动量’越小， 例如

forward(Movement amount is in here)

， 运动变得越顺畅，但你会失去速度。 我鼓励您调整值并进行实验！

Answer 3

以循环方式小增量移动海龟（如本线程其他地方所述）对于许多情况来说是一个很好的解决方案，但更灵活和通用的解决方案是使用

tracer

、

ontimer

禁用海龟对事件循环的控制和

update

：

from random import randint
from turtle import Screen, Turtle


def tick():
    for t in turtles:
        t.forward(1)

        if randint(0, 1):
            t.left(randint(-5, 5))

    screen.update()
    screen.ontimer(tick, 1000 // 30)


screen = Screen()
screen.tracer(0)
turtles = [Turtle() for _ in range(100)]

for t in turtles:
    t.setheading(randint(0, 360))

tick()
screen.exitonclick()

另请参阅：Python 多只海龟（看似）同时移动

Answer 4

解决方案是禁用更新每只海龟的位置，然后在计算出新位置后强制更新整个屏幕。

import turtle

# our two turtle instances
first, second = turtle.Turtle(), turtle.Turtle()

first.tracer(False)  # disable updating view on screen for this turtle!
second.tracer(False)

# make one move - note this will not appear on screen.
first.forward(50)
second.left(20)

# when you are ready to see the whole screen update
turtle.update()