在这个例子中我想做的是将图像包裹在一个圆圈上,如下所示。
为了包装图像,我简单地使用 trig 计算了 x,y 坐标。
问题是计算出的 X 和 Y 位置被四舍五入以使其成为整数。这会导致上面的包裹图像中出现空白像素。 x,y 位置必须是整数,因为它们是列表中的位置。
我在下面的代码中再次执行了此操作,但没有任何图像以使事情更容易查看。我所做的就是创建两个具有二进制值的数组,一个数组是黑色的,另一个是白色的,然后将一个数组包裹在另一个数组上。
代码的输出是。
import math as m
from PIL import Image # only used for showing output as image
width = 254.0
height = 24.0
Ro = 40.0
img = [[1 for x in range(int(width))] for y in range(int(height))]
cir = [[0 for x in range(int(Ro * 2))] for y in range(int(Ro * 2))]
def shom_im(img): # for showing data as image
list_image = [item for sublist in img for item in sublist]
new_image = Image.new("1", (len(img[0]), len(img)))
new_image.putdata(list_image)
new_image.show()
increment = m.radians(360 / width)
rad = Ro - 0.5
for i, row in enumerate(img):
hyp = rad - i
for j, column in enumerate(row):
alpha = j * increment
x = m.cos(alpha) * hyp + rad
y = m.sin(alpha) * hyp + rad
# put value from original image to its position in new image
cir[int(round(y))][int(round(x))] = img[i][j]
shom_im(cir)
我后来发现了中点圆算法,但结果更糟
from PIL import Image # only used for showing output as image
width, height = 254, 24
ro = 40
img = [[(0, 0, 0, 1) for x in range(int(width))]
for y in range(int(height))]
cir = [[(0, 0, 0, 255) for x in range(int(ro * 2))] for y in range(int(ro * 2))]
def shom_im(img): # for showing data as image
list_image = [item for sublist in img for item in sublist]
new_image = Image.new("RGBA", (len(img[0]), len(img)))
new_image.putdata(list_image)
new_image.show()
def putpixel(x0, y0):
global cir
cir[y0][x0] = (255, 255, 255, 255)
def drawcircle(x0, y0, radius):
x = radius
y = 0
err = 0
while (x >= y):
putpixel(x0 + x, y0 + y)
putpixel(x0 + y, y0 + x)
putpixel(x0 - y, y0 + x)
putpixel(x0 - x, y0 + y)
putpixel(x0 - x, y0 - y)
putpixel(x0 - y, y0 - x)
putpixel(x0 + y, y0 - x)
putpixel(x0 + x, y0 - y)
y += 1
err += 1 + 2 * y
if (2 * (err - x) + 1 > 0):
x -= 1
err += 1 - 2 * x
for i, row in enumerate(img):
rad = ro - i
drawcircle(int(ro - 1), int(ro - 1), rad)
shom_im(cir)
有人能建议一种消除空白像素的方法吗?