Python 用Theano实现LeCun局部对比度归一化
我试图使用我找到的代码来实现LeCun局部对比度标准化,但结果不正确。代码是Python语言,使用Python 用Theano实现LeCun局部对比度归一化,python,machine-learning,theano,Python,Machine Learning,Theano,我试图使用我找到的代码来实现LeCun局部对比度标准化,但结果不正确。代码是Python语言,使用theano库 def lecun_lcn(input, img_shape, kernel_shape, threshold=1e-4): """ Yann LeCun's local contrast normalization Orginal code in Theano by: Guillaume Desjardins """ input = inpu
theano
库
def lecun_lcn(input, img_shape, kernel_shape, threshold=1e-4):
"""
Yann LeCun's local contrast normalization
Orginal code in Theano by: Guillaume Desjardins
"""
input = input.reshape(input.shape[0], 1, img_shape[0], img_shape[1])
X = T.matrix(dtype=theano.config.floatX)
X = X.reshape(input.shape)
filter_shape = (1, 1, kernel_shape, kernel_shape)
filters = gaussian_filter(kernel_shape).reshape(filter_shape)
convout = conv.conv2d(input=X,
filters=filters,
image_shape=(input.shape[0], 1, img_shape[0], img_shape[1]),
filter_shape=filter_shape,
border_mode='full')
# For each pixel, remove mean of 9x9 neighborhood
mid = int(np.floor(kernel_shape / 2.))
centered_X = X - convout[:, :, mid:-mid, mid:-mid]
# Scale down norm of 9x9 patch if norm is bigger than 1
sum_sqr_XX = conv.conv2d(input=centered_X ** 2,
filters=filters,
image_shape=(input.shape[0], 1, img_shape[0], img_shape[1]),
filter_shape=filter_shape,
border_mode='full')
denom = T.sqrt(sum_sqr_XX[:, :, mid:-mid, mid:-mid])
per_img_mean = denom.mean(axis=[1, 2])
divisor = T.largest(per_img_mean.dimshuffle(0, 'x', 'x', 1), denom)
divisor = T.maximum(divisor, threshold)
new_X = centered_X / divisor
new_X = new_X.dimshuffle(0, 2, 3, 1)
new_X = new_X.flatten(ndim=3)
f = theano.function([X], new_X)
return f(input)
以下是测试代码:
x_img_origin = plt.imread("..//data//Lenna.png")
x_img = plt.imread("..//data//Lenna.png")
x_img_real_result = plt.imread("..//data//Lenna_Processed.png")
x_img = x_img.reshape(1, x_img.shape[0], x_img.shape[1], x_img.shape[2])
for d in range(3):
x_img[:, :, :, d] = tools.lecun_lcn(x_img[:, :, :, d], (x_img.shape[1], x_img.shape[2]), 9)
x_img = x_img[0]
pylab.subplot(1, 3, 1); pylab.axis('off'); pylab.imshow(x_img_origin)
pylab.gray()
pylab.subplot(1, 3, 2); pylab.axis('off'); pylab.imshow(x_img)
pylab.subplot(1, 3, 3); pylab.axis('off'); pylab.imshow(x_img_real_result)
pylab.show()
结果如下:
(从左到右:原点、我的结果、预期结果)
有人能告诉我代码的错误吗?我想这两行在矩阵轴上可能有一些错误:
per_img_mean = denom.mean(axis=[1, 2])
divisor = T.largest(per_img_mean.dimshuffle(0, 'x', 'x', 1), denom)
应该改写为:
per_img_mean = denom.mean(axis=[2, 3])
divisor = T.largest(per_img_mean.dimshuffle(0, 1, 'x', 'x'), denom)
下面是我如何实现Jarrett等人()中报告的局部对比度标准化。可以将其用作单独的层 我在theano的LeNet教程中的代码上测试了它,在该教程中,我将LCN应用于输入和每个卷积层,从而产生稍好的结果 您可以在此处找到完整的代码:
我曾经做过类似的事情,但不幸的是无法运行您的代码,因为它不是自包含的。你能让它可以运行,从而可以调试吗?除此之外,您还需要指定您使用的高斯滤波函数。嗨,艾肯伯格,这是代码,请根据需要更改图像路径。这是图片:如果你还需要什么,请告诉我。我认为门槛是罪魁祸首。如果我增加阈值,它会变得更接近预期结果。如果您还添加了一个如何调用函数的示例,那将非常酷。我对这些方法非常陌生,调用LecunCn得到的结果是
Elemwise{true\u div,no\u inplace}.0
,我不知道它是否正确以及如何从中提取图像。对于像我这样对theano一无所知的人,函数的输出是一个对象,还需要通过计算该对象来编译。因此,如果要提取卷积图像,必须编写LecunLCN(X).output.eval()[0][0]
class LecunLCN(object):
def __init__(self, X, image_shape, threshold=1e-4, radius=9, use_divisor=True):
"""
Allocate an LCN.
:type X: theano.tensor.dtensor4
:param X: symbolic image tensor, of shape image_shape
:type image_shape: tuple or list of length 4
:param image_shape: (batch size, num input feature maps,
image height, image width)
:type threshold: double
:param threshold: the threshold will be used to avoid division by zeros
:type radius: int
:param radius: determines size of Gaussian filter patch (default 9x9)
:type use_divisor: Boolean
:param use_divisor: whether or not to apply divisive normalization
"""
# Get Gaussian filter
filter_shape = (1, image_shape[1], radius, radius)
self.filters = theano.shared(self.gaussian_filter(filter_shape), borrow=True)
# Compute the Guassian weighted average by means of convolution
convout = conv.conv2d(
input=X,
filters=self.filters,
image_shape=image_shape,
filter_shape=filter_shape,
border_mode='full'
)
# Subtractive step
mid = int(numpy.floor(filter_shape[2] / 2.))
# Make filter dimension broadcastable and subtract
centered_X = X - T.addbroadcast(convout[:, :, mid:-mid, mid:-mid], 1)
# Boolean marks whether or not to perform divisive step
if use_divisor:
# Note that the local variances can be computed by using the centered_X
# tensor. If we convolve this with the mean filter, that should give us
# the variance at each point. We simply take the square root to get our
# denominator
# Compute variances
sum_sqr_XX = conv.conv2d(
input=T.sqr(centered_X),
filters=self.filters,
image_shape=image_shape,
filter_shape=filter_shape,
border_mode='full'
)
# Take square root to get local standard deviation
denom = T.sqrt(sum_sqr_XX[:,:,mid:-mid,mid:-mid])
per_img_mean = denom.mean(axis=[2,3])
divisor = T.largest(per_img_mean.dimshuffle(0, 1, 'x', 'x'), denom)
# Divisise step
new_X = centered_X / T.maximum(T.addbroadcast(divisor, 1), threshold)
else:
new_X = centered_X
self.output = new_X
def gaussian_filter(self, kernel_shape):
x = numpy.zeros(kernel_shape, dtype=theano.config.floatX)
def gauss(x, y, sigma=2.0):
Z = 2 * numpy.pi * sigma ** 2
return 1. / Z * numpy.exp(-(x ** 2 + y ** 2) / (2. * sigma ** 2))
mid = numpy.floor(kernel_shape[-1] / 2.)
for kernel_idx in xrange(0, kernel_shape[1]):
for i in xrange(0, kernel_shape[2]):
for j in xrange(0, kernel_shape[3]):
x[0, kernel_idx, i, j] = gauss(i - mid, j - mid)
return x / numpy.sum(x)