Question

我想在python中创建一个分形树。我已经制作了这棵树，但我希望有两只或更多的乌龟可以同时绘制我的分形树。有没有办法做到这一点？我已经找到了解决方案，但没有一个是我真正想要的。这是我的代码：

import turtle
tree = turtle.Turtle()
tree.ht()
tree.penup()
tree.sety(-200)
tree.left(90)

import turtle
tree0 = turtle.Turtle()
tree0.ht()
tree0.penup()
tree0.sety(-200)
tree0.left(90)

startx = tree.xcor()
starty = tree.ycor()
startx = tree0.xcor()
starty = tree0.ycor()

def fractalright(angle, length, x, y):
    tree.speed(0)
    tree.setx(x)
    tree.sety(y)
    tree.pendown()
    tree.forward(length)
    tree.right(angle)
    length = length - 20
    x = tree.xcor()
    y = tree.ycor()
    if length < 0:
        return
    tree.penup()
    fractalright(angle, length, x, y)
    tree.penup()
    tree.setx(x)
    tree.sety(y)
    tree.left(angle)
    fractalright (-angle, length, x, y)
def fractalleft(angle, length, x, y):
    tree0.speed(0)
    tree0.setx(x)
    tree0.sety(y)
    tree0.pendown()
    tree0.forward(length)
    tree0.right(angle)
    length = length - 20
    x = tree0.xcor()
    y = tree0.ycor()
    if length < 0:
        return
    tree0.penup()
    fractalleft(angle, length, x, y)
    tree0.penup()
    tree0.setx(x)
    tree0.sety(y)
    tree0.left(angle)
    fractalleft (-angle, length, x, y)

我正在使用python 3，如果您知道解决方案，请告诉我。谢谢！

Answer 1

提取出如何绘制树的细节，基本任务是与不同的参数并行执行一些绘制函数的两个实例，以指定一个“左树”和一个“右树”。这个基本结构可以实现如下：

from multiprocessing.dummy import Pool
import time


def draw_function(current_tree):
    # Replace the following lines with what you need to do with turtles
    print("Drawing tree {}".format(current_tree))
    time.sleep(1)


# Replace this with list of tuples of arguments to draw_function specifying
# angle, length, x, y, left vs right, etc.
list_of_trees = ["left_tree", "right_tree"]

my_pool = Pool(2)

results = my_pool.map(draw_function, list_of_trees)

my_pool.close()
my_pool.join()

在您的情况下，我对fractalleft和fractalright之间的区别有点不清楚，因为它们看起来相同，但逻辑应构成draw_function的基础。你应该为draw_function的每次执行创建一个单独的乌龟，但请注意，你不需要重新导入乌龟。

Answer 2

将模块threading与turtle一起使用的关键是不允许其他线程操纵海龟 - 他们改为排队他们的乌龟请求并让主线程处理它们：

import queue
from threading import Thread, active_count
from turtle import Turtle, Screen

def forward(turtle, distance):
    graphics.put((turtle.forward, distance))

def right(turtle, angle):
    graphics.put((turtle.right, angle))

def left(turtle, angle):
    graphics.put((turtle.left, angle))

def fractalright(turtle, angle, length):
    forward(turtle, length)

    if length - 20 > 0:
        right(turtle, angle)
        fractalright(turtle, angle, length - 20)
        left(turtle, angle)
        fractalright(turtle, -angle, length - 20)

    forward(turtle, -length)

def fractalleft(turtle, angle, length):
    forward(turtle, length)

    if length - 20 > 0:
        left(turtle, angle)
        fractalleft(turtle, angle, length - 20)
        right(turtle, angle)
        fractalleft(turtle, -angle, length - 20)

    forward(turtle, -length)

def process_queue():
    while not graphics.empty():
        action, argument = graphics.get()
        action(argument)

    if active_count() > 1:
        screen.ontimer(process_queue, 100)

START_X, START_Y = 0, -200

screen = Screen()
screen.mode('logo')  # make starting direction 0 degrees towards top

tree1 = Turtle(visible=False)
tree1.color('green')
tree1.penup()
tree1.goto(START_X, START_Y)
tree1.pendown()

tree2 = Turtle(visible=False)
tree2.color('dark green')
tree2.penup()
tree2.goto(START_X, START_Y)
tree2.pendown()

graphics = queue.Queue(1)  # size = number of hardware threads you have - 1

def fractal1():
    fractalright(tree1, 30, 100)

def fractal2():
    fractalleft(tree2, 30, 100)

thread1 = Thread(target=fractal1)
thread1.daemon = True  # thread dies when main thread (only non-daemon thread) exits.
thread1.start()

thread2 = Thread(target=fractal2)
thread2.daemon = True  # thread dies when main thread (only non-daemon thread) exits.
thread2.start()

process_queue()

screen.exitonclick()

我们正在使用队列模块进行线程安全通信。我重新编写了fractalright()和fractalleft()函数，以最大限度地减少所需的各种图形操作。

如果一切正常，您应该看到树的浅绿色和深绿色部分是同时独立绘制的。您的计算机至少需要有几个硬件线程。

如何在Python3中一次移动2只乌龟

2 个答案: