Python乌龟颜色更改和填充

Question

当用户输入要绘制的正方形数量和这些正方形的大小时，此代码将绘制形状。我希望每个绘制的正方形都改变颜色，并填充它。但是代码卡住了：

import turtle
t = turtle.Turtle()

size = int(input("How long do you want the side lengths to be?"))
number = int(input("How many squares do you want in the image?"))
red = 40.0
blue = 30.0
green = 10.0

def square (size):
 count = 0
 while count < 4:
  t.forward(size)
  t.right(90)
  count = count + 1

def drawing(number):
  times = 0
  while times < number:
    t.pencolor(red, blue, green)
    t.fillcolor(red, blue, green)
    t.begin_fill()
    square(size)
    t.right(360/number)
    if t.filling():
      t.pensize(5)
    else:
      t.pensize(3)
    t.color()
    (red + 1, blue + 1, green + 1)
    times = times + 1

drawing(number)

Answer 1

您的代码有几个问题-主要的问题是您使用了错误的颜色模型。默认情况下，乌龟颜色被指定为0.0到1.0之间的浮点值。但是，您想使用代码中的值，我们可以通过colormode(255)进行切换，该值允许从0到255。

接下来，begin_fill()必须具有与之匹配的end_fill()，而您丢失了它。颜色是按(red, green, blue)而不是(red, blue, green)的顺序指定的。另外，如果将pencolor()和fillcolor()都设置为相同的颜色，则只需使用一个调用color()。

最后，这两个语句不执行任何操作：

t.color()
(red + 1, blue + 1, green + 1)

就您的代码而言。您需要重新考虑它们。以下是我对您的代码的重做，该代码解决了上述问题和其他一些细节：

from turtle import Screen, Turtle

def square(size):
    count = 0

    if turtle.filling():
        turtle.pensize(5)
    else:
        turtle.pensize(3)

    while count < 4:
        turtle.forward(size)
        turtle.right(90)
        count += 1

def drawing(number):
    red = 30
    green = 10
    blue = 20

    times = 0

    while times < number:
        turtle.color(red % 255, green % 255, blue % 255)

        turtle.begin_fill()
        square(size)
        turtle.end_fill()

        turtle.right(360 / number)

        red, green, blue = red + 20, green + 30, blue + 10
        times += 1

size = int(input("How long do you want the side lengths to be? "))
number = int(input("How many squares do you want in the image? "))

screen = Screen()
screen.colormode(255)

turtle = Turtle()
turtle.speed('fastest')  # because I have no patience

drawing(number)

screen.exitonclick()