如何通过xcor和ycor判断Python Turtle是否处于指定圆内？

作业要求：

• 绘制一个圆
• 将海龟返回起始位置
• 让用户移动海龟十次
• 判断海龟是否处于圆内并显示胜负提示

我需要通过xcor()和ycor()来完成判断，但不清楚具体计算方法，以下是我当前编写的代码：

# import turtle
import turtle

# declare variables
answer = str()
move_counter = int()
user_turn = str()
user_turn_num = int()
user_move = str()
user_move_num = int()

# establish turtle color and background color
turtle.pencolor("magenta")
turtle.bgcolor("black")

# ask the user what shape they'd like the turtle to be
print("Turtle has many shapes to choose from. Your options are classic, arrow, turtle, circle, square and triangle.")
answer = input("What shape would you like the turtle to be? ")

# if structure for changing the shape depending on the user's answer
if answer == "classic":
  turtle.shape("classic")
elif answer == "turtle":
  turtle.shape("turtle")
elif answer == "arrow":
  turtle.shape("arrow")
elif answer == "circle":
  turtle.shape("circle")
elif answer == "square":
  turtle.shape("square")
elif answer == "triangle":
  turtle.shape("triangle")

# bring turtle to circle location
turtle.penup()
turtle.goto(100,-200)

# have turtle draw the circle
turtle.pendown()
turtle.fillcolor("magenta")
turtle.begin_fill()
turtle.circle(100)
turtle.end_fill()

# return turtle to starting area
turtle.penup()
turtle.goto(0,0)

while move_counter < 10:
  # ask user if they'd like to turn
  user_turn = input("Which direction would you like to turn (left, right, neither)? ")
  
  # create if structure for turning
  if user_turn == "left":
    user_turn_num = int(input("Turn how many degrees to the left? "))
    turtle.left(user_turn_num)
  elif user_turn == "right":
    user_turn_num = int(input("Turn how many degrees to the right?"))
    turtle.right(user_turn_num)
  elif user_turn == "neither":
    print("No turn made.")
  
  # ask user if they'd like to move forward or backward
  user_move = input("Would you like to move forward or backward? ")

  # create if structure for moving forward or backward
  if user_move == "forward":
    # ask user how many spaces they'd like to move forward and move turtle that many spaces
    user_move_num = int(input("How many spaces would you like to move forward?" ))
    turtle.forward(user_move_num)

  elif user_move == "backward":
    # ask user how many spaces they'd like to move backward and move turtle that many spaces
    user_move_num = int(input("How many spaces would you like to move backward?" ))
    turtle.backward(user_move_num)

  # increment counter and let user know how many moves they have left
  move_counter = move_counter + 1
  print("You have", 10 - move_counter, "moves left.")

# determine if turtle is within the circle
xpos = turtle.xcor
ypos = turtle.ycor


# display win message
# display loss message

# make turtle exit on click
turtle.exitonclick()

解决方案 #

判断原理 #

你绘制的圆圆心坐标是(100, -200)，半径为100。判断点是否在圆内，需要计算海龟当前位置到圆心的距离，若距离小于等于半径，说明海龟在圆内（包括边界）。

为了避免开根号的计算开销，直接比较距离的平方和半径的平方即可，结果与比较距离本身一致：
距离平方 = (海龟x坐标 - 圆心x坐标)² + (海龟y坐标 - 圆心y坐标)²
若该值 ≤ 半径²，则海龟在圆内。

代码修改 #

1. 首先修正xcor和ycor的调用：这两个是方法，必须加括号()才能获取坐标值。
2. 添加判断逻辑和胜负提示：

替换原代码中以下部分：

# determine if turtle is within the circle
xpos = turtle.xcor
ypos = turtle.ycor


# display win message
# display loss message

为：

# determine if turtle is within the circle
xpos = turtle.xcor()
ypos = turtle.ycor()

# 定义圆的参数
circle_center_x = 100
circle_center_y = -200
circle_radius = 100

# 计算距离平方与半径平方
distance_squared = (xpos - circle_center_x) ** 2 + (ypos - circle_center_y) ** 2
radius_squared = circle_radius ** 2

# 判断并显示结果
if distance_squared <= radius_squared:
    print("恭喜！你成功将海龟移到圆内，获胜！")
else:
    print("很遗憾，海龟不在圆内，挑战失败！")

额外优化提示 #

• 可以把圆心坐标和radius定义为全局变量，避免硬编码，让代码更易修改维护。
• Python不需要提前声明变量类型，原代码中变量声明部分可简化为直接赋值。

内容的提问来源于stack exchange，提问作者nightstargalaxy