model.summary() 在使用子类模型时无法打印输出形状

我已经使用了此方法来解决此问题，我不知道是否有更简单的方法。

class subclass(Model):
    def __init__(self):
        ...
    def call(self, x):
        ...
    def model(self):
        x = Input(shape=(24, 24, 3))
        return Model(inputs=[x], outputs=self.call(x))

if __name__ == '__main__':
    sub = subclass()
    sub.model().summary()

我解决问题的方式与Elazar的评估非常相似。覆盖subclass类中的函数摘要（）。然后，您可以在使用模型子类时直接调用summary（）：

class subclass(Model):
    def __init__(self):
        ...
    def call(self, x):
        ...
    def summary(self):
        x = Input(shape=(24, 24, 3))
        model = Model(inputs=[x], outputs=self.call(x))
        return model.summary()
if __name__ == '__main__':
    sub = subclass()
    sub.summary()

我想关键点是Network类中的_init_graph_network方法，即Model的父类。如果您在调用__init__方法时指定inputs和outputs参数，则将调用_init_graph_network。

因此，将有两种可能的方法：

1. 手动调用_init_graph_network方法来构建模型的图。
2. 用输入层和输出重新初始化。

，两种方法都需要输入层和输出（self.call要求）。

现在调用summary将给出确切的输出形状。但是，它将显示Input层，这不是子分类模型的一部分。

from tensorflow import keras
from tensorflow.keras import layers as klayers
class MLP(keras.Model):
    def __init__(self, input_shape=(32), **kwargs):
        super(MLP, self).__init__(**kwargs)
        # Add input layer
        self.input_layer = klayers.Input(input_shape)
        self.dense_1 = klayers.Dense(64, activation='relu')
        self.dense_2 = klayers.Dense(10)
        # Get output layer with `call` method
        self.out = self.call(self.input_layer)
        # Reinitial
        super(MLP, self).__init__(
            inputs=self.input_layer,
            outputs=self.out,
            **kwargs)
    def build(self):
        # Initialize the graph
        self._is_graph_network = True
        self._init_graph_network(
            inputs=self.input_layer,
            outputs=self.out
        )
    def call(self, inputs):
        x = self.dense_1(inputs)
        return self.dense_2(x)
if __name__ == '__main__':
    mlp = MLP(16)
    mlp.summary()

输出将是：

Model: "mlp_1"
_________________________________________________________________
Layer (type)                 Output Shape              Param #   
=================================================================
input_1 (InputLayer)         [(None, 16)]              0         
_________________________________________________________________
dense (Dense)                (None, 64)                1088      
_________________________________________________________________
dense_1 (Dense)              (None, 10)                650       
=================================================================
Total params: 1,738
Trainable params: 1,738
Non-trainable params: 0
_________________________________________________________________

我分析了adi shumely的答案：

• 不需要添加input_shape，因为您将其设置为build（）为参数
• 添加输入层对模型无济于事，它是作为call（）方法的参数带来的
• 添加所谓的输出不是我看到的。它做的唯一也是最重要的是调用call（）方法。

因此，我提出并提出了该解决方案，该解决方案不需要任何模型中的任何修改，只需要在呼叫摘要（）方法之前就需要改进模型，然后将呼叫添加到呼叫中（）具有输入张量的模型方法。我尝试了自己的模型，并在此提要中提出的三个模型上尝试了。

此提要的第一篇文章：

import tensorflow as tf
from tensorflow.keras import Input, layers, Model
class subclass(Model):
    def __init__(self):
        super(subclass, self).__init__()
        self.conv = layers.Conv2D(28, 3, strides=1)
    def call(self, x):
        return self.conv(x)
if __name__ == '__main__':
    sub = subclass()
    sub.build(input_shape=(None, 24, 24, 3))
    # Adding this call to the call() method solves it all
    sub.call(Input(shape=(24, 24, 3)))
    # And the summary() outputs all the information
    sub.summary()

feed的第二篇文章

from tensorflow import keras
from tensorflow.keras import layers as klayers
class MLP(keras.Model):
    def __init__(self, **kwargs):
        super(MLP, self).__init__(**kwargs)
        self.dense_1 = klayers.Dense(64, activation='relu')
        self.dense_2 = klayers.Dense(10)
    def call(self, inputs):
        x = self.dense_1(inputs)
        return self.dense_2(x)
if __name__ == '__main__':
    mlp = MLP()
    mlp.build(input_shape=(None, 16))
    mlp.call(klayers.Input(shape=(16)))
    mlp.summary()

从提要的最后一篇文章

import tensorflow as tf
class MyModel(tf.keras.Model):
    def __init__(self, **kwargs):
        super(MyModel, self).__init__(**kwargs) 
        self.dense10 = tf.keras.layers.Dense(10, activation=tf.keras.activations.softmax)    
        self.dense20 = tf.keras.layers.Dense(20, activation=tf.keras.activations.softmax)
    
    def call(self, inputs):
        x =  self.dense10(inputs)
        y_pred =  self.dense20(x)
        return y_pred
model = MyModel()
model.build(input_shape = (None, 32, 32, 1))
model.call(tf.keras.layers.Input(shape = (32, 32, 1)))
model.summary()

有相同的问题 - 通过3个步骤修复它：

1. 在_ init _
中添加input_shape
2. 添加一个input_layer
3. 添加图层

class MyModel(tf.keras.Model):
    
    def __init__(self,input_shape=(32,32,1), **kwargs):
        super(MyModel, self).__init__(**kwargs) 
        self.input_layer = tf.keras.layers.Input(input_shape)
        self.dense10 = tf.keras.layers.Dense(10, activation=tf.keras.activations.softmax)    
        self.dense20 = tf.keras.layers.Dense(20, activation=tf.keras.activations.softmax)
        self.out = self.call(self.input_layer)    
    
    def call(self, inputs):
        x =  self.dense10(inputs)
        y_pred =  self.dense20(x)
     
        return y_pred
model = MyModel()
model(x_test[:99])
print('x_test[:99].shape:',x_test[:10].shape)
model.summary()

输出：

x_test[:99].shape: (99, 32, 32, 1)
Model: "my_model_32"
_________________________________________________________________
Layer (type)                 Output Shape              Param #   
=================================================================
dense_79 (Dense)             (None, 32, 32, 10)        20        
_________________________________________________________________
dense_80 (Dense)             (None, 32, 32, 20)        220       
=================================================================
Total params: 240
Trainable params: 240
Non-trainable params: 0

我已经使用此方法解决了在Tensorflow 2.1和Tensorflow 2.4.1上测试的此问题。用model.inputs_layer

class Logistic(tf.keras.models.Model):
    def __init__(self, hidden_size = 5, output_size=1, dynamic=False, **kwargs):
        '''
        name: String name of the model.
        dynamic: (Subclassed models only) Set this to `True` if your model should
            only be run eagerly, and should not be used to generate a static
            computation graph. This attribute is automatically set for Functional API
            models.
        trainable: Boolean, whether the model's variables should be trainable.
        dtype: (Subclassed models only) Default dtype of the model's weights (
            default of `None` means use the type of the first input). This attribute
            has no effect on Functional API models, which do not have weights of their
            own.
        '''
        super().__init__(dynamic=dynamic, **kwargs)
        self.inputs_ = tf.keras.Input(shape=(2,), name="hello")
        self._set_input_layer(self.inputs_)
        self.hidden_size = hidden_size
        self.dense = layers.Dense(hidden_size, name = "linear")
        self.outlayer = layers.Dense(output_size, 
                        activation = 'sigmoid', name = "out_layer")
        self.build()
        
    def _set_input_layer(self, inputs):
        """add inputLayer to model and display InputLayers in model.summary()
        Args:
            inputs ([dict]): the result from `tf.keras.Input`
        """
        if isinstance(inputs, dict):
            self.inputs_layer = {n: tf.keras.layers.InputLayer(input_tensor=i, name=n) 
                                    for n, i in inputs.items()}
        elif isinstance(inputs, (list, tuple)):
            self.inputs_layer = [tf.keras.layers.InputLayer(input_tensor=i, name=i.name) 
                                    for i in inputs]
        elif tf.is_tensor(inputs):
            self.inputs_layer = tf.keras.layers.InputLayer(input_tensor=inputs, name=inputs.name)
    
    def build(self):
        super(Logistic, self).build(self.inputs_.shape if tf.is_tensor(self.inputs_) else self.inputs_)
        _ = self.call(self.inputs_)
    
    def call(self, X):
        X = self.dense(X)
        Y = self.outlayer(X)
        return Y
model = Logistic()
model.summary()

Model: "logistic"
_________________________________________________________________
Layer (type)                 Output Shape              Param #   
=================================================================
hello:0 (InputLayer)         [(None, 2)]               0         
_________________________________________________________________
linear (Dense)               (None, 5)                 15        
_________________________________________________________________
out_layer (Dense)            (None, 1)                 6         
=================================================================
Total params: 21
Trainable params: 21
Non-trainable params: 0
_________________________________________________________________

我在您的代码中仅添加了一行（下）。

self.call(Input(shape=(24, 24, 3)))

我的代码是

import tensorflow as tf
from tensorflow.keras import Input, layers, Model
class subclass(Model):
    def __init__(self):
        super(subclass, self).__init__()
        self.conv = layers.Conv2D(28, 3, strides=1)
    
        # add this code
        self.call(Input(shape=(24, 24, 3)))
    def call(self, x):
        return self.conv(x)

def func_api():
    x = Input(shape=(24, 24, 3))
    y = layers.Conv2D(28, 3, strides=1)(x)
    return Model(inputs=[x], outputs=[y])
if __name__ == '__main__':
    func = func_api()
    func.summary()
    sub = subclass()
    sub.build(input_shape=(None, 24, 24, 3))
    sub.summary()

结果

Model: "model"
_________________________________________________________________
Layer (type)                 Output Shape              Param #
=================================================================
input_1 (InputLayer)         [(None, 24, 24, 3)]       0
_________________________________________________________________
conv2d (Conv2D)              (None, 22, 22, 28)        784
=================================================================
Total params: 784
Trainable params: 784
Non-trainable params: 0
_________________________________________________________________
Model: "subclass"
_________________________________________________________________
Layer (type)                 Output Shape              Param #
=================================================================
conv2d_1 (Conv2D)            (None, 22, 22, 28)        784
=================================================================
Total params: 784
Trainable params: 784
Non-trainable params: 0                
_______________________________________________________________

加里的答案有效。但是，为了更方便，我想从我的自定义类对象透明地访问keras.Model的summary方法。

这可以通过实现内置__getattr__方法（可以在Python官方文档中找到更多信息）来轻松完成：

from tensorflow.keras import Input, layers, Model
class MyModel():
    def __init__(self):
        self.model = self.get_model()
    def get_model(self):
        # here we use the usual Keras functional API
        x = Input(shape=(24, 24, 3))
        y = layers.Conv2D(28, 3, strides=1)(x)
        return Model(inputs=[x], outputs=[y])
    def __getattr__(self, name):
        """
        This method enables to access an attribute/method of self.model.
        Thus, any method of keras.Model() can be used transparently from a MyModel object
        """
        return getattr(self.model, name)

if __name__ == '__main__':
    mymodel = MyModel()
    mymodel.summary()  # underlyingly calls MyModel.model.summary()