如何在python中绘制填充矩形
因此,我正在创建一个程序,它将创造某种景观,我正在尝试制作草。我想把光标放在300300,然后画一个巨大的矩形,然后把它填成绿色。到目前为止,我无法使矩形工作,或使其覆盖整个屏幕的下半部分。有人能帮我吗?和你的另一个问题一样,你的如何在python中绘制填充矩形,python,turtle-graphics,Python,Turtle Graphics,因此,我正在创建一个程序,它将创造某种景观,我正在尝试制作草。我想把光标放在300300,然后画一个巨大的矩形,然后把它填成绿色。到目前为止,我无法使矩形工作,或使其覆盖整个屏幕的下半部分。有人能帮我吗?和你的另一个问题一样,你的penup()函数毫无意义。下面是我如何用绿色填充半个屏幕: import turtle def penup(): for x in drawings: x.penup() penup() def drawgrass(): for x
penup()
函数毫无意义。下面是我如何用绿色填充半个屏幕:
import turtle
def penup():
for x in drawings:
x.penup()
penup()
def drawgrass():
for x in range(10):
grass.goto(300,300)
grass.color("green")
grass.begin_fill()
grass.forward(200)
grass.left(300)
grass.forward(200)
grass.left(300)
grass.end_fill()
penup()
drawgrass()
但我真正要做的是盖章:
可能重复的left()
仅用于旋转(即90度)-不移动(即300像素)。要创建矩形,需要4次forward()
和4次left()
(或right()
)。
from turtle import Screen, Turtle
def drawgrass():
grass.color("green")
width, height = screen.window_width(), screen.window_height()
grass.penup()
grass.goto(-width/2, 0)
grass.pendown()
grass.begin_fill()
for _ in range(2):
grass.forward(width)
grass.right(90)
grass.forward(height/2)
grass.right(90)
grass.end_fill()
screen = Screen()
grass = Turtle()
grass.speed('fastest') # because I have no patience
drawgrass()
grass.hideturtle()
screen.exitonclick()
from turtle import Screen, Turtle
CURSOR_SIZE = 20
def drawgrass():
grass.color('green')
grass.shape('square')
width, height = screen.window_width(), screen.window_height()
grass.penup()
grass.goto(0, -height/4)
grass.shapesize(height/2 / CURSOR_SIZE, width / CURSOR_SIZE)
grass.stamp()
screen = Screen()
grass = Turtle()
grass.hideturtle()
drawgrass()
screen.exitonclick()