如何在Python3中一次移動2個海龜

Question

我想在python中創建一個分形樹。我已經制作了樹，但我想要有2只或更多的海龜來一次繪製我的分形樹。有沒有辦法做到這一點？我尋找解決方案，但沒有一個是我真正想要的。這裏是我的代碼：

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) 

我使用python 3，請讓我知道如果你知道一個解決方案。謝謝！！

Answer 1

使用模塊threading與turtle的關鍵是不能讓額外的線程來操縱海龜 - 他們，而不是排隊的龜請求，並讓他們的主線程處理：我們正在使用

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() 

用於線程安全通信的隊列模塊。我重寫了您的fractalright()和fractalleft()函數，以儘量減少它們所需的各種圖形操作。

如果一切正常，你應該看到在同一時間獨立地繪製樹的淺綠色和深綠色的部分。您的計算機需要至少有幾個硬件線程可用。

Answer 2

蒸餾掉你如何繪製樹的細節，基本的任務是執行一些繪製函數的兩個實例並行使用不同的參數來指定一個「左樹」和一個「右樹」。這種基本結構可實現如下：

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() 

在你的情況，我對fractalleft和fractalright之間的區別有點不清楚，因爲他們似乎是相同的，但邏輯應該成爲你的draw_function的基礎。您應該爲draw_function的每次執行創建一個單獨的烏龜，但請注意，您不需要重新導入烏龜。