用张量流中的线性代数帮助新手(3阶张量)

根据操作员的意见进行编辑

可以将输入的特征向量展平为形状 [-1, n_type * n_features] ,应用精心选择的矩阵乘法,并从 [-1, n_type * n_neurons] 到 [-1, n_type, n_neurons]

运算张量是块对角线 [n_type * n_features, n_type * n_neurons] 一个,每个块是 n_type 张量 weights .

为了构建一个块对角矩阵,我使用了另一个答案(来自 here )

这个看起来像

inputs = tf.placeholder("float", shape=[None, n_type, n_features])
inputs = tf.reshape(inputs, shape=[-1, n_type * n_features])

weights = tf.Variable(FNN_weight_initializer([n_type, n_features, n_neurons]))

split_weights = tf.split(weights, num_or_size_splits=n_type, axis=1)
# each element of split_weights is a tensor of shape : [1, n_features, n_neurons] -> need to squeeze
split_weights = tf.map_fn(lambda elt : tf.squeeze(elt, axis=0), split_weights)

block_matrix = block_diagonal(split_weights) # from the abovementioned reference

Hidden1 = tf.matmul(inputs, block_matrix)
# shape : [None, n_type * n_neurons]

Hidden1 = tf.reshape(Hidden1, [-1, n_type, n_neurons])
# shape : [None, n_type, n_neurons]

原始答案

根据文件 tf.matmul ( reference )你要乘的张量必须是同一个等级。

当等级是 >2 只有最后两个维度需要矩阵乘法兼容,第一个其他维度需要完全匹配。

那么,对于这个问题,“能不能用tf.matmul把rank3张量相乘?”答案是“是的,这是可能的,但从概念上讲,它仍然是第二级乘法”。

因此,需要进行一些整形:

inputs = tf.placeholder("float", shape=[None, n_type, n_features])

inputs = tf.reshape(inputs, shape=[-1, n_type, 1, n_features])

weights = tf.Variable(FNN_weight_initializer([n_type, n_features, n_neurons]))

weights = tf.expand_dims(weights, 0)
# shape : [1, n_type, n_features, n_neurons]

weights = tf.tile(weights, [tf.shape(inputs)[0], 1, 1, 1])
# shape : [None, n_type, n_features, n_neurons]

Hidden1 = tf.matmul(inputs, weights)
# shape : [None, n_type, 1, n_neurons]

Hidden1 = tf.reshape(Hidden1, [-1, n_type, n_neurons])
# shape : [None, n_type, n_neurons]