Python 在'中使用标准tensorflow层；单元格'-相貌

我的问题与*有关

是否有可能将标准的tensorflow层转换成“细胞”，与之一起用于构建递归神经网络

因此，新的“单元”应该存储参数（权重，…），并且能够在不同的输入上调用。大概是这样的：

from tf.nn import batch_normalization, conv2d
from tf.contrib.rnn import MultiRNNCell, LSTMCell

bn_cell = cell_creation_fun(batch_normalization, otherparams) # batch norm cell
conv_cell = cell_creation_fun(conv2d, otherparams )           # non-rnn conv cell
# or `conv_cell = cell_creation_fun(tf.layers.Conv2D, otherparams )` # using tf.layers  

因此，它们可以像这样使用：

multi_cell = MultiRNNCell([LSTMCell(...), conv_cell, bn_cell])

h = ...
conv_h, _ = conv_cell(h, state=None)
normed_h, _ = bn_cell(h, state=None)

或者像这样：

multi_cell = MultiRNNCell([LSTMCell(...), conv_cell, bn_cell])

h = ...
conv_h, _ = conv_cell(h, state=None)
normed_h, _ = bn_cell(h, state=None)

---

我唯一能想到的就是为我想使用的每一层手动编写这样一个“单元”，子类化。但是，如果在创建过程中不能传递“输入”参数，那么使用现有函数（如Conv2D）似乎并不简单。（我管理时将发布代码。）

---

---

*也许以更有针对性的方式提问有可能得到答案。

好的，以下是我到目前为止得到的答案：

class LayerCell(rnn_cell_impl.RNNCell):

    def __init__(self, tf_layer, **kwargs):
        ''' :param tf_layer: a tensorflow layer, e.g. tf.layers.Conv2D or 
            tf.keras.layers.Conv2D. NOT tf.layers.conv2d !'''
        self.layer_fn = tf_layer(**kwargs)

    def __call__(self, inputs, state, scope=None):
        ''' Every `RNNCell` must implement `call` with
          the signature `(output, next_state) = call(input, state)`.  The optional
          third input argument, `scope`, is allowed for backwards compatibility
          purposes; but should be left off for new subclasses.'''
        return (self.layer_fn(inputs), state)

    def __str__(self):
            return "Cell wrapper of " + str(self.layer_fn)

    def __getattr__(self, attr):
        '''credits to https://stackoverflow.com/questions/1382871/dynamically-attaching-a-method-to-an-existing-python-object-generated-with-swig/1383646#1383646'''
        return getattr(self.layer_fn, attr)

    @property
    def state_size(self):
        """size(s) of state(s) used by this cell.

        It can be represented by an Integer, a TensorShape or a tuple of Integers
        or TensorShapes.
        """
        return  (0,) 

    @property
    def output_size(self):
        """Integer or TensorShape: size of outputs produced by this cell."""
        # use with caution; could be uninitialized
        return self.layer_fn.output_shape

---

（当然，不要与重复层一起使用，因为状态保持将被破坏。）

似乎适用于：tf.layers.Conv2D、tf.keras.layers.Conv2D、tf.keras.layers.Activation、tf.layers.BatchNormalization

不适用于：tf.keras.layers.BatchNormalization。 至少我在tf.while循环中使用它时失败了；抱怨组合来自不同帧的变量，类似于。也许凯拉斯用

---

用益：

cell0 = tf.contrib.rnn.ConvLSTMCell(conv_ndims=2, input_shape=[40, 40, 3], output_channels=16, kernel_shape=[5, 5])
cell1 = LayerCell(tf.keras.layers.Conv2D, filters=8, kernel_size=[5, 5], strides=(1, 1), padding='same')
cell2 = LayerCell(tf.layers.BatchNormalization, axis=-1)

inputs =  np.random.rand(10, 40, 40, 3).astype(np.float32)
multicell = tf.contrib.rnn.MultiRNNCell([cell0, cell1, cell2])
state = multicell.zero_state(batch_size=10, dtype=tf.float32)

output = multicell(inputs, state)
print("Yippee!")