2 回答
TA贡献1848条经验 获得超2个赞
关键是将所有东西参数化为合适大小的立方体,然后将填充代码变成可以调用来填充这些立方体的子程序。下面是沿着这些行重新编写的代码,可以将任何奇数 3 或更大的数作为输入处理:
from graphics import *
UNIT = 100
def patchWork(win, size):
blocks = size // 2
def hatch_box(x, y):
for n in range(0, UNIT, UNIT//5):
line = Line(Point(n + x * UNIT, y * UNIT), Point((x + 1) * UNIT, UNIT - n + y * UNIT))
line.draw(win)
for n in range(UNIT, 0, -UNIT//5):
line = Line(Point(n + x * UNIT, (y + 1) * UNIT), Point(x * UNIT, UNIT - n + y * UNIT))
line.draw(win)
for n in range(UNIT, 0, -UNIT//5):
line = Line(Point((x + 1) * UNIT, n + y * UNIT), Point(n + x * UNIT, (y + 1) * UNIT))
line.setFill('red')
line.draw(win)
for n in range(0, UNIT, UNIT//5):
line = Line(Point(x * UNIT, UNIT - n + y * UNIT), Point(UNIT - n + x * UNIT, y * UNIT))
line.setFill('red')
line.draw(win)
for y in range(blocks):
for x in range(blocks):
hatch_box(x, y)
for y in range(blocks + 1, 2 * blocks + 1):
for x in range(blocks + 1, 2 * blocks + 1):
hatch_box(x, y)
def draw_circles(x, y):
for o in range(0, UNIT, UNIT // 10):
for c in range(0, UNIT, UNIT // 10):
circle = Circle(Point(c + UNIT // 20 + x * UNIT, o + UNIT // 20 + y * UNIT), UNIT // 20)
circle.setFill('blue')
circle.draw(win)
for y in range(blocks):
for x in range(blocks + 1, 2 * blocks + 1):
draw_circles(x, y)
draw_circles(y, x)
def draw_cube(x, y):
cube = Rectangle(Point(x * UNIT, y * UNIT), Point((x + 1) * UNIT, (y + 1) * UNIT))
cube.setFill('yellow')
cube.draw(win)
x = blocks
for y in range(0, 2 * blocks + 1):
draw_cube(x, y)
draw_cube(y, x)
def layout(size):
win = GraphWin('Patchwork', size * UNIT, size * UNIT)
for i in range(0, size * UNIT, UNIT):
Line(Point(i, 0), Point(i, size * UNIT)).draw(win)
Line(Point(0, i), Point(size * UNIT, i)).draw(win)
return win
def main():
patchWorkSize = int(input("What patchwork size would you like? "))
if patchWorkSize % 2 == 1 and patchWorkSize > 2:
win = layout(patchWorkSize)
patchWork(win, patchWorkSize)
win.getMouse() # pause for click in window
win.close()
else:
print('Only odd sizes 3 or greater are available')
main()
你应该能够UNIT在合理范围内改变。
TA贡献1810条经验 获得超4个赞
我不会为您编写所有内容,但下面的代码显示了如何根据每个区域在拼凑网格 (i和j) 中的位置为每个区域选择颜色。颜色在 中确定color_func()。
希望这足以让您弄清楚如何应用所示的相对简单的编码模式来绘制所需的图形。
提示:根据(或相对于)这些相同的位置值计算每个图形元素的坐标(即不要为每个可能的拼凑尺寸编写单独的函数)。
这段代码对我来说似乎仍然过于重复,可能可以做得更简洁,但我也会把它留给你...... ;¬)
def main():
# size = int(input("What patchwork size would you like?"))
# Hardcoded for testing.
#
# if size % 2 == 0:
# raise RuntimeError('patchwork size must be odd')
patch_work(5)
print()
patch_work(7)
def patch_work(n):
colors = '0', '1', '2'
mid = n // 2
# Call color_func() for every possible position in patchwork and
# prints the results in rows.
for i in range(n):
row = []
for j in range(n):
row.append(color_func(n, i, j, colors))
print(''.join(row))
def color_func(n, i, j, colors):
mid = n // 2
if i == mid:
index = 2
elif i < mid:
if j < mid:
index = 0
elif j == mid:
index = 2
elif j > mid:
index = 1
elif i > mid:
if j < mid:
index = 1
elif j == mid:
index = 2
elif j > mid:
index = 0
return colors[index]
if __name__ == '__main__':
main()
输出:
00211
00211
22222
11200
11200
0002111
0002111
0002111
2222222
1112000
1112000
1112000
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