使用Python龟制作没有圆圈功能的圆圈

时间:2014-10-27 03:58:18

标签: python turtle-graphics

我有这个学校作业:

建立雪人没有海龟circle功能

雪人应该在蓝色背景上,并且应该用白色填充。

雪人的轮廓应该是黑色的。

雪人的身体应该由3个圆圈组成。

每个圆圈的轮廓应为3像素宽。

底部圆的半径应为100像素。

中间圆的半径应为70像素。

顶部圆的半径应为40像素。

每个圆圈应位于其下方的圆圈上方(底部圆圈除外,可以位于任何位置)。

圈子之间应该没有差距。

给雪人一个嘴,眼睛和一个鼻子(帽子是可选的)。

确保每只手上都包含两个手臂和至少两个手指。

到目前为止,我创造了这个,但在我继续之前,似乎无法让圆圈正确。 此外,不知道圈子中如何着色为眼睛做点。请帮帮我,第一次编码。

import turtle                               # allows us to use turtle library
wn = turtle.Screen()                        # allows us to create a graphics window
wn.bgcolor("blue")                          # sets gtaphics windows background color to blue
import math                                 # allows us to use math functions
quinn = turtle.Turtle()                     # sets up turtle quinn
quinn.setpos(0,0)
quinn.pensize(3)
quinn.up()

# drawing first circle middle
quinn.forward(70)
quinn.down()
quinn.left(90)

# calculation of cicumference of a circle
a = (math.pi*140.00/360)

#itineration for first circle
for i in range (1,361,1):
    quinn.left(a)
    quinn.forward (1)

# drawing second circle bottom
quinn.up()
quinn.home()
quinn.right(90)
quinn.forward(70)
quinn.left(90)
quinn.down()

b = (math.pi*200.00/360)

for i in range (1,361,1):
    quinn.right(b)
    quinn.forward(1)

# drawing third circle head top

quinn.up ()
quinn.goto(0,70)
quinn.right(90)
quinn.down()

c =(math.pi*80/360)

for i in range (1,361,1):
    quinn.left(c)
    quinn.forward(1)

wn.exitonclick()

6 个答案:

答案 0 :(得分:2)

以下是绘制填充蓝色圆圈的示例函数:

def draw_circle(radius):    
    turtle.up()
    turtle.goto(0,radius) # go to (0, radius)
    turtle.begin_fill() # start fill
    turtle.down() # pen down
    turtle.color('blue')
    times_y_crossed = 0
    x_sign = 1.0
    while times_y_crossed <= 1:
        turtle.forward(2*math.pi*radius/360.0) # move by 1/360
        turtle.right(1.0)
        x_sign_new = math.copysign(1, turtle.xcor())        
        if(x_sign_new != x_sign):
            times_y_crossed += 1
        x_sign = x_sign_new
    turtle.up() # pen up
    turtle.end_fill() # end fill.
    return

然后你可以修改上面的函数,为圆心的位置(x,y)添加参数:

def draw_circle(radius, x, y):    
    turtle.up()
    turtle.goto(x,y+radius) # go to (x, y + radius)
    turtle.begin_fill() # start fill
    turtle.down() # pen down
    turtle.color('blue')
    times_y_crossed = 0
    x_sign = 1.0
    while times_y_crossed <= 1:
        turtle.forward(2*math.pi*radius/360.0) # move by 1/360
        turtle.right(1.0)
        x_sign_new = math.copysign(1, turtle.xcor())        
        if(x_sign_new != x_sign):
            times_y_crossed += 1
        x_sign = x_sign_new
    turtle.up() # pen up
    turtle.end_fill() # end fill.
    return

您可以轻松添加点,例如:

turtle.goto(-20,10)
turtle.color('red')
turtle.dot(20)
turtle.goto(40,10)
turtle.dot(20)

汇总:

import turtle
import math

def draw_circle(radius, x, y):    
    turtle.up()
    turtle.goto(x,y+radius) # go to (0, radius)
    turtle.begin_fill() # start fill
    turtle.down() # pen down
    turtle.color('blue')
    times_y_crossed = 0
    x_sign = 1.0
    while times_y_crossed <= 1:
        turtle.forward(2*math.pi*radius/360.0) # move by 1/360
        turtle.right(1.0)
        x_sign_new = math.copysign(1, turtle.xcor())        
        if(x_sign_new != x_sign):
            times_y_crossed += 1
        x_sign = x_sign_new
    turtle.up() # pen up
    turtle.end_fill() # end fill.
    return


draw_circle(100, 10, 10)
turtle.goto(-20,10)
turtle.color('red')
turtle.dot(20)
turtle.goto(40,10)
turtle.dot(20)
turtle.pen(shown=False)
turtle.done()

您应该尝试自己完成作业的剩余部分..;)

答案 1 :(得分:1)

很抱歉没有给出解释。 第一部分是Ramanujan对pi的近似,但不是一个非常好的,因为它只能在循环的300,000次迭代之后达到pi的近似,并且它仅精确到5位小数。这将是这一部分:

    r += (1 / k) * (-1)**i
    pi = (4 * (1 - r))

然后我使用圆周:

t.forward(2*5*pi)

最后,我只是在20岁时将乌龟行走时钟。

import turtle 
t = turtle.Turtle()
t.right(90)
t.penup()
t.goto(100, 0)
t.pendown()

i = 0
r = 0
k = 3

while i <= 360:
    r += (1 / k) * (-1)**i
    pi = (4 * (1 - r))
    t.write(pi)
    t.forward(2*5*pi)
    t.right(20)

    i += 1
    k += 2

turtle.done()

答案 2 :(得分:1)

“没有龟圈功能”的大多数解决方案都涉及编写自己的等同于龟圈功能。但是还有另外两种方法可以用乌龟绘制概述的圆圈。

一个是你可以使用同心

turtle.color('black')
turtle.dot(100)
turtle.color('white')
turtle.dot(100 - 6)

请记住,dot()取直径而circle()取半径:

enter image description here

但是,我更喜欢使用 stamping 来解决这些问题:

''' Build a Snowman without turtle circle function '''

from turtle import Turtle, Screen

# The snowman’s body should be made of 3 filled circles.

# The bottom circle should have a radius of 100 pixels.
# The middle circle should have a radius of 70 pixels.
# The top circle should have a radius of 40 pixels.

RADII = (100, 70, 40)

STAMP_SIZE = 20

# The snowman should be on a blue background
screen = Screen()
screen.bgcolor('blue')

quinn = Turtle('circle')
quinn.setheading(90)
quinn.up()

# The outline of the snowman should be in black, and should be drawn filled with white.
quinn.color('black', 'white')

for not_first, radius in enumerate(RADII):

    if not_first:
        quinn.forward(radius)

    # The outline of each circle should be 3 pixels wide.
    quinn.shapesize((radius * 2) / STAMP_SIZE, outline=3)

    quinn.stamp()

    # Each circle should be centered above the one below it
    # There should be no gap between the circles.
    quinn.forward(radius)

# Give the snowman eyes

quinn.shapesize(15 / STAMP_SIZE)
quinn.color('black')
quinn.backward(3 * RADII[-1] / 4)

for x in (-RADII[-1] / 3, RADII[-1] / 3):
    quinn.setx(x)
    quinn.stamp()

# Give the snowman a mouth, and a nose (a hat is optional).

pass

# Make sure to include two stick-arms and at least two fingers on each hand.

pass

quinn.hideturtle()

screen.exitonclick()

你的想法是将乌龟光标本身扭曲成你需要的东西,在屏幕上制作它的快照,然后将它扭曲到你需要绘制的下一个东西。

enter image description here

答案 3 :(得分:0)

您可以创建一个接受参数fd并左移的函数。

这是我创建的。

 private void runCommand(String commandName,
                        String[] commandArgs) {

    try (CommandContext commandContext = createCommandContext()) {
        // find and run the command
        SparkCommand command = commandContext.findCommand(commandName);

        checkSparkResource(command.context.sc());
        command.main(commandArgs);

    } catch (Exception e) {
        logger.error(e.getMessage(), e);

        String message = "Something wrong~";
        String title = "Run Job on Dataproc:" + commandName + " Fail";
        String text = e.getMessage();

        SlackNotifier.instance()
                     .error(message, title, text);
    }
}

这是计算:范围中的迭代除以fd + left。 那是我的近似值。因此,您应该能够创建类似的功能。

答案 4 :(得分:0)

您可以尝试一下,希望对您有所帮助!

def polygon(length, sides):
    for i in range(sides):
        turtle.forward(length)
        turtle.left(360.0/sides)

polygon(1, 360)

答案 5 :(得分:0)

从数学的角度来看,您可以使用math中的函数sincos绘制圆。

绘制圆后,请使用turtle.begin_fill()turtle.end_fill()方法来填充圆(尽管我确实同意在编程中,this方法更实用)

from turtle import Turtle
from math import sin, cos, radians

def draw_circle(radius, x, y, color="light blue", line_width=3):
    c = Turtle(visible=False)
    c.width(3)
    c.penup()
    c.goto(x + radius, y)
    c.pendown()
    c.color("black", color)
    c.begin_fill()
    # Circle drawing starts here
    for i in range(1, 361):
        c.goto(radius * cos(radians(i)) + x,
               radius * sin(radians(i)) + y)
    # Circle drawing ends here
    c.end_fill()
draw_circle(100, 0, -100)

正如this答案中指出的那样,您可以使用turtle.dot()方法在屏幕上绘制一个点, 笔的宽度就是点的直径。

还有另一种方法,但是与dot方法相比,这是不切实际的。 无论如何,我都会把它扔出去,只是为了说明解决方法有很多可能性:

from turtle import Turtle

def draw_circle(radius, x, y, color="light bue", line_width=3):
    c = Turtle(visible=False)
    c.penup()
    c.goto(x, y)
    c.pendown()
    # Circle drawing starts here
    c.width(radius * 2 + line_width)
    c.forward(0)
    c.color(color)
    c.width(radius * 2 - line_width)
    c.forward(0)
    # Circle drawing ends here

draw_circle(100, 0, -100)

因此turtle.dot()等效于turtle.forward(0) (以及turtle.backward(0)turtle.goto(turtle.pos()), turtle.setpos(turtle.pos())等,等等)

>

输出:

enter image description here