Tensorflow keras多项式特征层

Tensorflow keras多项式特征层,tensorflow,machine-learning,keras,keras-layer,Tensorflow,Machine Learning,Keras,Keras Layer,我想要创建多项式特征乘以X.T*X的图层: class QuadraticLayer(Layer): def __init__(self, **kwargs): super(QuadraticLayer, self).__init__(**kwargs) def build(self, input_shape): assert isinstance(input_shape, tuple) print(input_shape) self.in_shape = i

我想要创建多项式特征乘以X.T*X的图层:

class QuadraticLayer(Layer):

def __init__(self, **kwargs):
    super(QuadraticLayer, self).__init__(**kwargs)

def build(self, input_shape):
    assert isinstance(input_shape, tuple)
    print(input_shape)
    self.in_shape = input_shape[1]
    self.out_shape = input_shape[1] ** 2
    super(QuadraticLayer, self).build(input_shape)  # Be sure to call this at the end

def call(self, x):
    print(x.shape)
    tf.reshape(x, (self.in_shape, 1, -1), name=None)
    x = tf.matmul(x, x, transpose_a=True)
    return tf.reshape(x, (-1, self.out_shape))

def compute_output_shape(self, input_shape):
    return (None, self.out_shape)
我的问题是,在
call
tensor
x
中,是一个批量张量-如何编写每个训练示例都能工作的层,而不是整个批量张量?

这是一个想法:

def call(self, x):
    x = K.backend.batch_dot(tf.reshape(x, (-1, 1, self.in_shape)), tf.reshape(x, (-1, self.in_shape, 1)), axes=[1,2])
    return tf.reshape(x, (-1, self.out_shape))
但问题是有更好的解决方案吗