如何在for循环中移动多只海龟？

Question

所以，我是 python 的新手，对于我的 class，我的任务是让 10 只海龟参加一场海龟比赛，这些海龟都在移动并且应该停在终点线。我被指示为海龟列出一个列表，并有一个 while 循环让它们随机移动，还有一个嵌套的 if 循环来检查获胜者。我无法让所有海龟同时移动，当我运行我的代码时，它一个接一个地移动海龟而不是一起移动，有什么想法吗？

import turtle as trtl
import random as rand
zoomers = []
zom = [0,1,2,3,4,5,6,7,8,9]
tloc = -130
trtl.penup()
trtl.goto(-150, 150)
trtl.pendown()
trtl.goto(-150,-140)
trtl.penup()
trtl.goto(180,150)
trtl.pendown()
trtl.goto(180,-140)
trtl.hideturtle()
for z in zom:
  zoom = trtl.Turtle("turtle")
  zoom.penup()
  zoom.goto(-150,-tloc)
  tloc += 25
  robux = rand.randrange(0,20)
  zoomers.append(zoom)
  for n in zoomers:
    cash = 0
    while cash < 100:
      zoom.forward(robux)
      cash = cash + 1
    if zoom.xcor() == 180:
      print("We have a winner!")
      break```

Answer 1

我们甚至可以在不使用 turtle 库的情况下对其进行建模：

import random

twerbles = list(zip(range(10), [0]*10)) #list of turtles all starting at position 0 (numbered 0-10)
finishline = 10 #the threshold they must pass

while True:
    twerbles = [(twerble, position+random.randint(1,2)) for twerble, position in twerbles] #move each turtle up a random amount.

    print(twerbles) #this can be removed if you dont want to see their positions as they go

    winners = [twerble for twerble, position in twerbles if position >= finishline] #list of turtles that passed the threshold (if any)
    if winners: print(winners); break #if there are winners print them and stop looping

[(0, 1), (1, 2), (2, 1), (3, 2), (4, 1), (5, 2), (6, 1), (7, 1), (8, 1), (9, 2)]
[(0, 2), (1, 3), (2, 2), (3, 3), (4, 2), (5, 4), (6, 3), (7, 3), (8, 3), (9, 3)]
[(0, 4), (1, 5), (2, 4), (3, 4), (4, 4), (5, 6), (6, 5), (7, 5), (8, 4), (9, 5)]
[(0, 5), (1, 7), (2, 6), (3, 5), (4, 6), (5, 8), (6, 6), (7, 7), (8, 5), (9, 6)]
[(0, 6), (1, 8), (2, 8), (3, 6), (4, 7), (5, 9), (6, 7), (7, 9), (8, 7), (9, 7)]
[(0, 8), (1, 9), (2, 10), (3, 7), (4, 9), (5, 11), (6, 8), (7, 11), (8, 8), (9, 9)]

[2, 5, 7] #winning turtles

与海龟一起移动

import turtle as trtl
import random

finishline = 30

twerbles = [(trtl.Turtle("turtle"), 0) for i in range(10)]

for i,(twerble,position) in enumerate(twerbles):
  twerble.penup()
  twerble.goto(-150,150 - i*25)

while True:
    twerbles = [(twerble, position+random.randint(1,2)) for twerble, position in twerbles]

    for twerble, position in twerbles:
        twerble.forward(position)

    winners = [twerble for twerble, position in twerbles if position >= finishline]
    if winners:
        break

Answer 2

没有直接的方法可以做到这一点，但这是下一个最好的方法。

创建一只海龟Screen。将其 tracer() 设置为 0，并在循环的每次迭代中更新屏幕。它可能会变得太快，因此从 time 模块中导入 sleep 以减慢速度。

最小示例：

from turtle import Turtle, Screen
from random import randrange
from time import sleep

wn = Screen()
wn.tracer(0)

t1 = Turtle('turtle')
t2 = Turtle('turtle')

t1.penup()
t2.penup()

t1.goto(-100, 50)
t2.goto(-100, -50)

while True:
    sleep(0.1)
    t1.forward(randrange(0, 20))
    t2.forward(randrange(0, 20))
    wn.update()

Answer 3

您需要创建一个海龟集合。为每只海龟循环移动（一步）直到一只海龟完成。

试试这个代码：

import turtle as trtl
import random as rand
zoomers = []
zom = [0,1,2,3,4,5,6,7,8,9]
tloc = -130
trtl.penup()
trtl.goto(-150, 150)
trtl.pendown()
trtl.goto(-150,-140)
trtl.penup()
trtl.goto(180,150)
trtl.pendown()
trtl.goto(180,-140)
trtl.hideturtle()

zoomers = [trtl.Turtle("turtle") for z in zom]

for i,t in enumerate(zoomers):
  t.penup()
  t.goto(-150,150 - i*25)


for i,z in enumerate(zom):
  zoom = zoomers[i]   #trtl.Turtle("turtle")
  tloc += 25
#  zoomers.append(zoom)

running = True
while running:
    for zoom in zoomers:
      robux = rand.randrange(0,20)
      zoom.forward(robux)
      if zoom.xcor() >= 180:
        print("We have a winner!")
        running = False
        break

input('Press enter to exit...')

Answer 4

您正在使用一只乌龟。您可以通过创建多个海龟对象轻松地做到这一点。我用了 3 个乌龟对象，他们有一个小比赛。

试试这个代码，你可以在这个代码中实现你的赢家逻辑。

import turtle as trtl
import random as rand

zoom1 = trtl.Turtle()
zoom2 = trtl.Turtle()
zoom3 = trtl.Turtle()

zoomers = [zoom1, zoom2, zoom3]

for zoom in zoomers:
  zoom.penup()

zoom1.goto(-150, 150)
zoom2.goto(-120, 150)
zoom3.goto(-90, 150)

for zoom in zoomers:
  zoom.pendown()
  zoom.right(90)

for zoom in zoomers:
  robux = rand.randrange(10,50)
  zoom.forward(robux)

编辑：您需要非阻塞代码来实现同时且彼此独立的海龟移动。您可以通过使用多线程来做到这一点。试试 this.

Answer 5

到目前为止，所有其他解决方案都会在固定时间片内将海龟移动一段随机距离。让我们把它反过来，让海龟在随机时间片中移动一个恒定的距离：

from turtle import Turtle, Screen
from random import randrange

def run(turtle):
    turtle.forward(5)

    if turtle.xcor() < half_width:
        screen.ontimer(lambda: run(turtle), randrange(20, 150))

screen = Screen()

half_width = screen.window_width() / 2
lane_width = 20

for order, color in enumerate(['red', 'green', 'blue']):
    turtle = Turtle('turtle')
    turtle.speed('fastest')
    turtle.color(color)
    turtle.penup()

    turtle.goto(-half_width, (order | 1) * lane_width)
    lane_width *= -1
    run(turtle)

screen.exitonclick()

如何在for循环中移动多只海龟？

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