Question

我想在Python上绘制一个棋盘，但我只得到一个黑色方块。你能帮我解决一下这个程序吗？

import turtle
def filled_square(size, color, x, y):
    turtle.setpos(x, y)
    turtle.color(color)
    turtle.begin_fill()
    for i in range(4):
       angle = 90
       turtle.fd(size)
       turtle.lt(angle)
    turtle.end_fill()
    turtle.up()
import sys
n = int(sys.argv[1])
s = int(sys.argv[2])
square_size = s//n
y=0
for i in range(n):
    x = 0
    for j in range(n):
        if (i+j)%2==0:
           filled_square(square_size, "red", x, y)
        else:
           filled_square(square_size, "black", x, y)
    x+=square_size
turtle.down()
turtle.done()

Answer 1

y=0
for i in range(n):
    x = 0
    for j in range(n):
        if (i+j)%2==0:
           filled_square(square_size, "red", x, y)
        else:
           filled_square(square_size, "black", x, y)
    x+=square_size

在这里遇到问题。

在下一次迭代中将x重置为零时，增加x的值没有多大意义，所以初始赋值应该高于for循环。
你永远不会更新y的值。

x = 0
for i in range(n):
    y=0
    for j in range(n):
        if (i+j)%2==0:
           filled_square(square_size, "red", x, y)
        else:
           filled_square(square_size, "black", x, y)
        y+=square_size
    x+=square_size

现在你应该得到你想要的棋盘形状。

替代解决方案：您可以通过根本不具备这些值来避免记账x和y的值的问题。您可以直接从i和j推导出方块的坐标。

for i in range(n):
    for j in range(n):
        if (i+j)%2==0:
            filled_square(square_size, "red", i*square_size, j*square_size)
        else:
            filled_square(square_size, "black", i*square_size, j*square_size)

在if和else块中合并公共逻辑也可能很好，只能区分实际更改的值（即颜色）

for i in range(n):
    for j in range(n):
        if (i+j)%2==0:
            color = "red"
        else:
            color = "black"
        filled_square(square_size, color, i*square_size, j*square_size)

Answer 2

您需要增加x和y。另请注意，x应在内循环中递增。这是工作代码，

import turtle
def filled_square(size, color, x, y):
    turtle.setpos(x, y)
    turtle.color(color)
    turtle.begin_fill()
    for i in range(4):
        angle = 90
        turtle.fd(size)
        turtle.lt(angle)
    turtle.end_fill()
    turtle.up()
import sys
n = int(sys.argv[1])
s = int(sys.argv[2])
square_size = s//n
y=0
for i in range(n):
    x = 0
    for j in range(n):
        if (i+j)%2==0:
            filled_square(square_size, "red", x, y)
        else:
            filled_square(square_size, "black", x, y)
        x+=square_size
    y+=square_size
turtle.down()
turtle.done()

Answer 3

我建议使用这样的功能编程进行电路板/游戏工作。它将使一切变得更容易维护，并使实现新功能变得更容易。

import turtle

def filled_square(size, color, x, y):
    turtle.up()
    turtle.setpos(x, y)
    turtle.color(color)
    turtle.begin_fill()
    for i in range(4):
        angle = 90
        turtle.fd(size)
        turtle.lt(angle)
    turtle.end_fill()

def board(length, size, x_pos=0, y_pos=0):
    for y in range(length):
        for x in range(length):
            if (x+y)%2==0:
                filled_square(
                    size,  "red", (x * square_size) - x_pos, (y * square_size) - y_pos)
            else:
                filled_square(
                    size, "black", (x * square_size) - x_pos, (y * square_size) - y_pos)
    turtle.done()

关键字参数也很棒。如果您决定添加更多颜色，那么就像添加两个参数一样简单，

def board(length, size, x_pos=0, y_pos=0, color1="red", color2="black"):
    for y in range(length):
        for x in range(length):
            if (x+y)%2==0:
                filled_square(
                    size,  color1, (x * square_size) - x_pos, (y * square_size) - y_pos)
            else:
                filled_square(
                    size, color2, (x * square_size) - x_pos, (y * square_size) - y_pos)

在Python上绘制Checkerboard

3 个答案: