基于python的图像最大似然分割算法
我想使用python中实现的最大似然算法执行图像分割。 类的平均向量和协方差矩阵是已知的,通过迭代图像(非常大…5100X7020),我们可以计算每个像素成为给定类一部分的概率 简单地用Python编写基于python的图像最大似然分割算法,python,algorithm,performance,numpy,Python,Algorithm,Performance,Numpy,我想使用python中实现的最大似然算法执行图像分割。 类的平均向量和协方差矩阵是已知的,通过迭代图像(非常大…5100X7020),我们可以计算每个像素成为给定类一部分的概率 简单地用Python编写 import numpy as np from numpy.linalg import inv from numpy.linalg import det ... probImage1 = [] probImage1Vector = [] norm = 1.0 / (np.power((2*n
import numpy as np
from numpy.linalg import inv
from numpy.linalg import det
...
probImage1 = []
probImage1Vector = []
norm = 1.0 / (np.power((2*np.pi), i/2) * np.sqrt(np.linalg.det(covMatrixClass1)))
covMatrixInverz = np.linalg.inv(covMatrixClass1)
for x in xrange(x_img):
for y in xrange(y_img):
X = realImage[x,y]
pixelValueDifference = X - meanVectorClass1
mult1 = np.multiply(-0.5,np.transpose(pixelValueDifference))
mult2 = np.dot(covMatrixInverz,pixelValueDifference)
multMult = np.dot(mult1,mult2)
expo = np.exp(multMult)
probImage1Vector.append(np.multiply(norm,expo))
probImage1.append(probImage1Vector)
probImage1Vector = []
这个代码在大图像上执行时非常慢的问题。。。
向量减法和乘法之类的计算耗费大量时间,即使它们只是1X3个向量
你能给我一个提示如何加速这个代码吗?我将非常感激。对不起,如果我不清楚的话,我仍然是python的初学者 仔细看看:
mult1 = np.multiply(-0.5,np.transpose(pixelValueDifference))
mult2 = np.dot(covMatrixInverz,pixelValueDifference)
multMult = np.dot(mult1,mult2)
我们看到,这一行动基本上是:
A.T (d) C (d) A # where `(d)` is the dot-product
这三个步骤可以很容易地用一个字符串表示,如下所示-
np.einsum('k,lk,l->',pA,covMatrixInverz,-0.5*pA)
在迭代器i(=x)
和j(=y)
中执行此操作,我们将得到一个完全矢量化的表达式-
np.einsum('ijk,lk,ijl->ij',pA,covMatrixInverz,-0.5*pA))
或者,我们可以使用np.tensordot
-
mult2_vectorized = np.tensordot(pA, covMatrixInverz, axes=([2],[1]))
output = np.einsum('ijk,ijk->ij',-0.5*pA, mult2_vectorized)
标杆管理 将所有方法列为函数-
# Original code posted by OP to return array
def org_app(meanVectorClass1, realImage, covMatrixInverz, norm):
probImage1 = []
probImage1Vector = []
x_img, y_img = realImage.shape[:2]
for x in xrange(x_img):
for y in xrange(y_img):
X = realImage[x,y]
pixelValueDifference = X - meanVectorClass1
mult1 = np.multiply(-0.5,np.transpose(pixelValueDifference))
mult2 = np.dot(covMatrixInverz,pixelValueDifference)
multMult = np.dot(mult1,mult2)
expo = np.exp(multMult)
probImage1Vector.append(np.multiply(norm,expo))
probImage1.append(probImage1Vector)
probImage1Vector = []
return np.asarray(probImage1).reshape(x_img,y_img)
def vectorized(meanVectorClass1, realImage, covMatrixInverz, norm):
pA = realImage - meanVectorClass1
mult2_vectorized = np.tensordot(pA, covMatrixInverz, axes=([2],[1]))
return np.exp(np.einsum('ijk,ijk->ij',-0.5*pA, mult2_vectorized))*norm
def vectorized2(meanVectorClass1, realImage, covMatrixInverz, norm):
pA = realImage - meanVectorClass1
return np.exp(np.einsum('ijk,lk,ijl->ij',pA,covMatrixInverz,-0.5*pA))*norm
时间安排-
In [19]: # Setup inputs
...: meanVectorClass1 = np.array([23.96000000, 58.159999, 61.5399])
...:
...: covMatrixClass1 = np.array([[ 514.20040404, 461.68323232, 364.35515152],
...: [ 461.68323232, 519.63070707, 446.48848485],
...: [ 364.35515152, 446.48848485, 476.37212121]])
...: covMatrixInverz = np.linalg.inv(covMatrixClass1)
...:
...: norm = 0.234 # Random float number
...: realImage = np.random.rand(1000,2000,3)
...:
In [20]: out1 = org_app(meanVectorClass1, realImage, covMatrixInverz, norm )
...: out2 = vectorized(meanVectorClass1, realImage, covMatrixInverz, norm )
...: out3 = vectorized2(meanVectorClass1, realImage, covMatrixInverz, norm )
...: print np.allclose(out1, out2)
...: print np.allclose(out1, out3)
...:
True
True
In [21]: %timeit org_app(meanVectorClass1, realImage, covMatrixInverz, norm )
1 loops, best of 3: 27.8 s per loop
In [22]: %timeit vectorized(meanVectorClass1, realImage, covMatrixInverz, norm )
1 loops, best of 3: 182 ms per loop
In [23]: %timeit vectorized2(meanVectorClass1, realImage, covMatrixInverz, norm )
1 loops, best of 3: 275 ms per loop
看起来完全矢量化的einsum+tensordot
混合解决方案做得很好
为了进一步提高性能,还可以研究如何在大型阵列上加速指数计算。作为第一步,我将消除不必要的函数调用,如转置、点和乘法。这些都是简单的计算,你应该在线进行。当您能够真正看到自己在做什么时,就可以更容易地理解性能问题,而不是将内容隐藏在函数中 这里的基本问题是,这似乎至少是一个四次复杂度的操作。您可能希望简单地乘以您在所有循环中执行的操作数。是5亿,20亿,3500亿?有多少
要控制性能,您需要了解您正在执行的指令数量。一台现代计算机每秒可以执行约10亿条指令,但如果涉及内存移动,速度可能会大大降低。谢谢,非常感谢!我要试试看@user2955708运气好吗?请看下面我的评论@user2955708在您的实际用例中,
meanVectorClass1
、realImage
和covMatrixInverz
的形状是什么?请查找这些形状!('meanVectorClass1',(3L,)('covMatrixClass1',(3L,3L))('realImage',(7020L,5100L,3L))