如果您无法通过使用或组合操作(例如旋转,然后应用EMBOSS滤镜,重新旋转)(或增强对比度然后进行压印)来实现目标,则可以诉诸于更改(或创建自己的)滤镜矩阵。
在ImageFilter.py中,您将找到此类:
##
# Embossing filter.
class EMBOSS(BuiltinFilter):
name = "Emboss"
filterargs = (3, 3), 1, 128, (
-1, 0, 0,
0, 1, 0,
0, 0, 0
)
将-1放置在矩阵的另一个角会改变方位角,使其变为-2 可能 会达到您想要的效果。
逐像素应用矩阵。矩阵中的每个元素对应于当前像素和周围像素;代表当前像素的中心值。新的变换后的当前像素将创建为所有9个像素的组合,并按矩阵中的值加权。例如,中心为全零且为1的矩阵不会更改图像。
附加参数为scale
和offset
。对于内置的EMBOSS,值是1(比例)和128(偏移)。更改这些将改变结果的整体强度。
从ImageFilter.py:
# @keyparam scale Scale factor. If given, the result for each
# pixel is divided by this value. The default is the sum
# of the kernel weights.
# @keyparam offset Offset. If given, this value is added to the
# result, after it has been divided by the scale factor.
由于我不熟悉GIMP的“深度”参数的影响,因此无法确定哪种方法最有可能满足您的要求。
您还可以使矩阵具有不同的大小。将(3,3)替换为(5,5),然后创建25个元素的矩阵。
要在不重新保存源代码的情况下对过滤器进行临时更改,只需执行以下操作:
ImageFilter.EMBOSS.filterargs=((3, 3), 1, 128, (-1, 0, 0, 0, 1, 0, 0, 0, 0))
from PIL import Image
import numpy
# defining azimuth, elevation, and depth
ele = numpy.pi/2.2 # radians
azi = numpy.pi/4. # radians
dep = 10. # (0-100)
# get a B&W version of the image
img = Image.open('daisy.jpg').convert('L')
# get an array
a = numpy.asarray(img).astype('float')
# find the gradient
grad = numpy.gradient(a)
# (it is two arrays: grad_x and grad_y)
grad_x, grad_y = grad
# getting the unit incident ray
gd = numpy.cos(ele) # length of projection of ray on ground plane
dx = gd*numpy.cos(azi)
dy = gd*numpy.sin(azi)
dz = numpy.sin(ele)
# adjusting the gradient by the "depth" factor
# (I think this is how GIMP defines it)
grad_x = grad_x*dep/100.
grad_y = grad_y*dep/100.
# finding the unit normal vectors for the image
leng = numpy.sqrt(grad_x**2 + grad_y**2 + 1.)
uni_x = grad_x/leng
uni_y = grad_y/leng
uni_z = 1./leng
# take the dot product
a2 = 255*(dx*uni_x + dy*uni_y + dz*uni_z)
# avoid overflow
a2 = a2.clip(0,255)
# you must convert back to uint8 /before/ converting to an image
img2 = Image.fromarray(a2.astype('uint8'))
img2.save('daisy2.png')
我希望这有帮助。现在,我明白为什么您对PIL的结果感到失望了。Wolfram Mathworld是向量代数复习的好资源。