Question

RBM有一个实现。原始RBM实现是针对图像等离散数据，我的数据是二进制数据，代码是否也适用于真实数据？我在某处读到有一个适用于此的典型RBM的高斯版本，它是否也在该模块中实现？

Answer 1

简而言之，RBM只是二分图上的马尔可夫随机场。因此，您可以使用任何概率分布来描述节点之间的关系。

就代码而言，您并不需要明确地复制内容。选择的概率函数将扮演的角色是对比分歧算法。您只需要更改选择样本的方式。需要更改的代码部分将在下面复制。

def sample_h_given_v(self, v0_sample):
    ''' This function infers state of hidden units given visible units '''
    # compute the activation of the hidden units given a sample of
    # the visibles
    pre_sigmoid_h1, h1_mean = self.propup(v0_sample)
    # get a sample of the hiddens given their activation
    # Note that theano_rng.binomial returns a symbolic sample of dtype
    # int64 by default. If we want to keep our computations in floatX
    # for the GPU we need to specify to return the dtype floatX
    h1_sample = self.theano_rng.binomial(size=h1_mean.shape,
                                         n=1, p=h1_mean,
                                         dtype=theano.config.floatX)
    return [pre_sigmoid_h1, h1_mean, h1_sample]

def propdown(self, hid):
    '''This function propagates the hidden units activation downwards to
    the visible units

    Note that we return also the pre_sigmoid_activation of the
    layer. As it will turn out later, due to how Theano deals with
    optimizations, this symbolic variable will be needed to write
    down a more stable computational graph (see details in the
    reconstruction cost function)

    '''
    pre_sigmoid_activation = T.dot(hid, self.W.T) + self.vbias
    return [pre_sigmoid_activation, T.nnet.sigmoid(pre_sigmoid_activation)]

def sample_v_given_h(self, h0_sample):
    ''' This function infers state of visible units given hidden units '''
    # compute the activation of the visible given the hidden sample
    pre_sigmoid_v1, v1_mean = self.propdown(h0_sample)
    # get a sample of the visible given their activation
    # Note that theano_rng.binomial returns a symbolic sample of dtype
    # int64 by default. If we want to keep our computations in floatX
    # for the GPU we need to specify to return the dtype floatX
    v1_sample = self.theano_rng.binomial(size=v1_mean.shape,
                                         n=1, p=v1_mean,
                                         dtype=theano.config.floatX)
    return [pre_sigmoid_v1, v1_mean, v1_sample]

受限制的Boltzman机器用于连续输入

1 个答案: