绘制然后擦除正方形

Question

我正在尝试使用 python 图形绘制一个正方形，然后在 3 秒后将其擦除。我有以下代码：

import threading as th
import turtle
import random

thread_count = 0

def draw_square():
    turtle.goto(random.randint(-200,200), random.randint(-200,200))
    for i in range(4):
        turtle.forward(100)
        turtle.left(90)

def erase_square(x,y):
    turtle.pencolor('white')
    turtle.fillcolor('white')
    turtle.goto(x,y)
    turtle.begin_fill()
    for i in range(4):
        turtle.forward(target_size)
        turtle.left(90)
    turtle.end_fill()

def square_timer():
    global thread_count
    if thread_count <= 10: 
        print("Thread count", thread_count)
        draw_square()
        T = th.Timer(3, square_timer)
        T.start()

square_update() 函数将绘制 10 个方块然后停止。我试图让它画一个正方形，然后使用 erase_square() 清除它，方法是在它上面涂上白色，然后再画一个并重复这个过程，当它达到 10 时停止。我不知道如何制作擦除功能转到绘制正方形的随机位置并 'erase' 它。如何让它绘制和擦除正方形？

Answer 1

清除绘图的最直接方法是使用 clear() 方法，而不是尝试在图像上绘制白色：

from turtle import Screen, Turtle
from random import randint

def draw_square():
    turtle.goto(randint(-200, 200), randint(-200, 200))

    turtle.pendown()

    for _ in range(4):
        turtle.forward(100)
        turtle.left(90)

    turtle.penup()

def erase_square():
    turtle.clear()
    screen.ontimer(square_update)  # no time, ASAP

def square_update():
    draw_square()
    screen.ontimer(erase_square, 3000)  # 3 seconds in milliseconds

screen = Screen()

turtle = Turtle()
turtle.hideturtle()
turtle.speed('fastest')
turtle.penup()

square_update()

screen.exitonclick()

您可以按照@JonathanDrukker 的建议使用undo() 方法，但您必须撤消每个步骤。即使对于像正方形这样的简单形状，这也需要一点思考，但对于复杂的形状，这就很困难了。添加到撤消缓冲区的操作数并不总是与您对 turtle 的调用相匹配。但是我们可以从 turtle 本身得到这个计数，只要我们在绘制和擦除之间不做任何其他 turtle 动作：

def draw_square():
    turtle.goto(randint(-200, 200), randint(-200, 200))

    count = turtle.undobufferentries()
    turtle.pendown()

    for _ in range(4):
        turtle.forward(100)
        turtle.left(90)

    turtle.penup()

    return turtle.undobufferentries() - count

def erase_square(undo_count):
    for _ in range(undo_count):
        turtle.undo()

    screen.ontimer(square_update)

def square_update():
    undo_count = draw_square()
    screen.ontimer(lambda: erase_square(undo_count), 3000)

但我要做的是让乌龟 自己成为正方形 然后移动它而不是擦除和（重新）绘制正方形：

from turtle import Screen, Turtle
from random import randint

CURSOR_SIZE = 20

def draw_square():
    turtle.goto(randint(-200, 200), randint(-200, 200))

def square_update():
    turtle.hideturtle()
    draw_square()
    turtle.showturtle()

    screen.ontimer(square_update, 3000)

screen = Screen()

turtle = Turtle()
turtle.hideturtle()
turtle.shape('square')
turtle.shapesize(100 / CURSOR_SIZE)
turtle.fillcolor('white')
turtle.penup()

square_update()

screen.exitonclick()

绘制然后擦除正方形

Draw then erase square

python

turtle-graphics

python-turtle