python - python：简单的方形网格

Question

最近，我一直在尝试找出一个 20 x 20 的方格。到目前为止，我只知道如何绘制 4，无论我将什么作为我的 # of。如果有人能找出我在代码中的缺陷，将不胜感激！

我想要达到的目标

我的代码

import turtle
turtle.hideturtle()
t = turtle.Turtle()
t.hideturtle()

def draw_square(t, size, num, angle):

for i in range(num):
    for x in range(4):
        turtle.forward(size)
        turtle.left(90)
    turtle.right(angle)

draw_square(t, 25, 4, 90)

标签： pythongridturtle-graphics

Answer 1

您的代码有一对嵌套循环。但是，考虑到您绘制网格的方式，您确实需要三个嵌套循环。试图找出从你所在的地方到你想去的地方的最少代码量，我想出了以下内容：

from turtle import Screen, Turtle

def draw_square(turtle, size, num):
    for y in range(num):
        for x in range(num):
            for _ in range(4):
                turtle.forward(size)
                turtle.left(90)

            turtle.forward(size)

        parity = y % 2 == 0
        turn = turtle.left if parity else turtle.right

        turn(90)
        turtle.forward(size * 2 * parity)
        turn(90)

screen = Screen()

yertle = Turtle(visible=False)
yertle.speed('fastest')  # because I have no patience

draw_square(yertle, 25, 20)

screen.exitonclick()

此代码在绘图方面效率低下，因为要重新绘制相同的行，这是要避免的。对于这个问题，我个人最喜欢的解决方案是使用生成器的Tholian Web方法：

from turtle import Turtle, Screen

UNIT_SIZE, GRID_SQUARES = 25, 20

GRID_SIZE = GRID_SQUARES * UNIT_SIZE

def half_grid(turtle):
    speed = turtle.speed()

    for brick in range(GRID_SQUARES):
        direction = [turtle.right, turtle.left][brick % 2 == 1]

        for _ in range(0, GRID_SIZE, speed):
            turtle.forward(speed)
            yield(0)

        direction(90)

        for _ in range(0, UNIT_SIZE, speed):
            turtle.forward(speed)
            yield(0)

        direction(90)

    for _ in range(0, GRID_SIZE, speed):
        turtle.forward(speed)
        yield(0)

heckle = Turtle(shape='arrow')
heckle.speed(5)  # speed needs to be a factor of UNIT_SIZE
heckle.penup()
heckle.goto(-GRID_SIZE / 2, -GRID_SIZE / 2)
heckle.pendown()
heckle.left(90)

jeckle = Turtle(shape='arrow')
jeckle.speed(5)
jeckle.penup()
jeckle.goto(GRID_SIZE / 2, -GRID_SIZE / 2)
jeckle.pendown()
jeckle.left(180)

generator1, generator2 = half_grid(heckle), half_grid(jeckle)

while (next(generator1, 1) + next(generator2, 1) < 2):
    pass

heckle.hideturtle()
jeckle.hideturtle()

screen = Screen()
screen.exitonclick()

但这对于您的目的来说可能是矫枉过正...