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我的python代码如下......它需要永远。一定有一些我可以使用的 numpy 技巧？我正在分析的图片很小而且是灰度的......

def gaussian_probability(x,mean,standard_dev):
        termA = 1.0 / (standard_dev*np.sqrt(2.0*np.pi))
        termB = np.exp(-((x - mean)**2.0)/(2.0*(standard_dev**2.0)))
        g = (termA*termB)
        return g
def sum_of_gaussians(x):
        return sum([self.mixing_coefficients[i] * 
                    gaussian_probability(x, self.means[i], self.variances[i]**0.5) 
                    for i in range(self.num_components)])
def expectation():
        dim = self.image_matrix.shape
        rows, cols = dim[0], dim[1]
        responsibilities = []
        for i in range(self.num_components):
            gamma_k = np.zeros([rows, cols])
            for j in range(rows):
                for k in range(cols):
                    p = (self.mixing_coefficients[i] *      
                         gaussian_probability(self.image_matrix[j,k],
                                              self.means[i], 
                                              self.variances[i]**0.5))
                    gamma_k[j,k] = p / sum_of_gaussians(self.image_matrix[j,k])
            responsibilities.append(gamma_k)
        return responsibilities

我只包括了期望步骤，因为虽然最大化步骤循环遍历矩阵责任数组的每个元素，但它似乎走得相对较快（所以瓶颈可能是所有 gaussian_probability 计算？）

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1 回答 1

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您可以通过做两件事来大大加快计算速度：

  • 不要在每个循环中计算归一化！如目前所写，对于具有 M 个组件的 NxN 图像，您正在计算每个相关的计算N * N * M时间，从而产生一个O[N^4 M^2]算法！相反，您应该计算一次所有元素，然后除以总和，即O[N^2 M].

  • 使用 numpy 矢量化而不是显式循环。这可以按照您设置代码的方式非常简单地完成。

本质上，您的expectation函数应如下所示：

def expectation(self):
    responsibilities = (self.mixing_coefficients[:, None, None] *
                        gaussian_probability(self.image_matrix,
                                             self.means[:, None, None],
                                             self.variances[:, None, None] ** 0.5))
    return responsibilities / responsibilities.sum(0)

你没有提供一个完整的例子，所以我不得不即兴创作一些来检查和基准测试，但这里有一个快速的例子：

import numpy as np

def gaussian_probability(x,mean,standard_dev):
    termA = 1.0 / (standard_dev*np.sqrt(2.0*np.pi))
    termB = np.exp(-((x - mean)**2.0)/(2.0*(standard_dev**2.0)))
    return termA * termB

class EM(object):
    def __init__(self, N=5):
        self.image_matrix = np.random.rand(20, 20)
        self.num_components = N
        self.mixing_coefficients = 1 + np.random.rand(N)
        self.means = 10 * np.random.rand(N)
        self.variances = np.ones(N)

    def sum_of_gaussians(self, x):
        return sum([self.mixing_coefficients[i] * 
                    gaussian_probability(x, self.means[i], self.variances[i]**0.5) 
                    for i in range(self.num_components)])

    def expectation(self):
        dim = self.image_matrix.shape
        rows, cols = dim[0], dim[1]
        responsibilities = []
        for i in range(self.num_components):
            gamma_k = np.zeros([rows, cols])
            for j in range(rows):
                for k in range(cols):
                    p = (self.mixing_coefficients[i] *      
                         gaussian_probability(self.image_matrix[j,k],
                                              self.means[i], 
                                              self.variances[i]**0.5))
                    gamma_k[j,k] = p / self.sum_of_gaussians(self.image_matrix[j,k])
            responsibilities.append(gamma_k)
        return responsibilities

    def expectation_fast(self):
        responsibilities = (self.mixing_coefficients[:, None, None] *
                            gaussian_probability(self.image_matrix,
                                                 self.means[:, None, None],
                                                 self.variances[:, None, None] ** 0.5))
        return responsibilities / responsibilities.sum(0)

现在我们可以实例化对象并比较期望步骤的两种实现：

em = EM(5)
np.allclose(em.expectation(),
            em.expectation_fast())
# True

查看时间，对于具有 5 个组件的 20x20 图像，我们的速度提高了大约 1000 倍：

%timeit em.expectation()
10 loops, best of 3: 65.9 ms per loop

%timeit em.expectation_fast()
10000 loops, best of 3: 74.5 µs per loop

这种改进将随着图像大小和组件数量的增加而增长。祝你好运！

于 2015-11-14T13:37:33.003 回答