如何在除了冲压之外的Python龟图形中绘制椭圆？

2 回答

• 2

  在测试@ moomoomoo309的椭圆代码并发现问题（打印错误的地方，宽度和高度与参数不匹配，忽略龟头，因此无法打印倾斜的椭圆， Headers 不跟踪绘图，不会使笔处于原始状态等我决定尝试自己写 .

  我选择使用 turtle.circle() 作为相对于现有龟位置和 Headers 绘制椭圆的位置的模型，允许用户更改步骤（即制作其他不规则多边形），保留笔状态和开始位置，这就是我提出的（我使用了 self 而不是 turtle 或 pen ，因为我打算将其作为方法安装）：

  import turtle
  import math
  
  def ellipse(self, x_radius, y_radius, steps=60):
  
      down = self.isdown()  # record pen state for restoration later
  
      if not down:
          self.pendown()
  
      heading_radians = math.radians(self.heading())
      theta_radians = -math.pi / 2
      extent_radians = 2 * math.pi
      step_radians = extent_radians / steps
      extent_radians += theta_radians
      x_center, y_start = self.position()
      y_center = y_start + y_radius
  
      cos_heading, sin_heading = math.cos(heading_radians), math.sin(heading_radians)
  
      while True:
          x, y = x_center + math.cos(theta_radians) * x_radius, y_center + math.sin(theta_radians) * y_radius
          # readjust x & y to set the angle of the ellipse based on the original heading of the turtle
          x, y = x - x_center, y - y_start
          x, y = x * cos_heading - y * sin_heading, x * sin_heading + y * cos_heading
          x, y = x + x_center, y + y_start
  
          self.setheading(self.towards(x, y))  # turtle faces direction in which ellipse is drawn
          self.goto(x, y)
  
          if theta_radians == extent_radians:
              break
  
          theta_radians = min(theta_radians + step_radians, extent_radians)  # don't overshoot our starting point
  
      self.setheading(self.towards(x_center, y_start))  # set correct heading for the next thing we draw
  
      if not down:  # restore pen state on return
          self.penup()
  

  （可选）根据Adding a Method to an Existing Object Instance将此方法添加到我们的龟中：

  from functools import partial
  
  yertle = turtle.Turtle()
  yertle.ellipse = partial(ellipse, yertle)
  

  演示代码以显示我们可以使用 turtle.ellipse() 绘制的所有新形状：

  if __name__ == "__main__":
  
      from functools import partial
  
      yertle = turtle.Turtle()
      yertle.ellipse = partial(ellipse, yertle)
  
      import random
  
      yertle.speed("fastest")
      yertle.hideturtle()
      yertle.penup()
  
      screen = turtle.Screen()
  
      for _ in range(75):
  
          radius = random.randint(10, 50)
  
          yertle.setheading(random.randint(0, 360))
          yertle.setx(random.randint(-screen.window_width()/2 + radius * 2, screen.window_width()/2 - radius * 2))
          yertle.sety(random.randint(-screen.window_height()/2 + radius + 2, screen.window_height()/2 - radius * 2))
          yertle.color((random.random(), random.random(), random.random()), (random.random(), random.random(), random.random()))
  
          flag = random.choice([True, False, False])
  
          if flag:
              yertle.begin_fill()
  
          yertle.ellipse(radius, radius / 0.5 + random.random() * 3, steps=random.choice([3, 4, 5, 6, 7, 8, 60, 60, 60]))
  
          if flag:
              yertle.end_fill()
  
      screen.exitonclick()
  

  EXAMPLE OUTPUT

  

  我试图实现 extent a la turtle.circle() 但是无法使其适当地在任意范围内工作（即以这样的方式，你可以以相同的范围调用 turtle.ellipse() 两次，并让它继续它停止的曲线）所以我已经离开了那一天 .

  将我的答案带回OP的原始问题，我们现在可以做到：

  import turtle
  import math
  
  def ellipse(self, x_radius, y_radius, steps=60):
  
      # ...
  
  def draw_O():
      # Draw an O
  
      turtle.penup()
      turtle.forward(letter_height/4)
      turtle.pendown()
  
      ellipse(turtle, letter_width, letter_height)
  
      turtle.penup()
      turtle.forward(space_width + letter_height/4)
      turtle.pendown()
  
  letter_width = 10
  letter_height = 170
  
  space_width = 5
  
  turtle.onkey(draw_O, "o")
  
  turtle.listen()
  turtle.done()
  

  要生成OP所需的基于椭圆的骨骼字母O：

  

  回复于 2024-04-28T09:20:04+08:00
• 0

  我很确定这会起作用，但宽度为180/180和高度/ 180可能会关闭 .

  from math import sin,cos,pi
  def ellipse(pen, x, y, width, height):
      pen.penup()
      pen.goto(x + width / 2, height)
      pen.pendown()
      penX, penY = pen.pos()
      for i in range(0, 360):
          penX += cos(i*pi/180)*width/180
          penY += sin(i*pi/180)*height/180
          pen.goto(penX, penY)
      pen.penup()
  

  回复于 2024-04-28T09:20:04+08:00