model.summary() 在使用子类模型时无法打印输出形状

Posted 2023-02-16

【中文标题】model.summary() 在使用子类模型时无法打印输出形状

【英文标题】：model.summary() can't print output shape while using subclass model

【发布时间】：2019-08-09 15:19:49

【问题描述】：

这是创建keras模型的两种方法，但是两种方法的总结结果output shapes不一样。显然，前者打印的信息更多，更容易检查网络的正确性。

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)


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()

输出：

_________________________________________________________________
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
_________________________________________________________________
_________________________________________________________________
Layer (type)                 Output Shape              Param #   
=================================================================
conv2d_1 (Conv2D)            multiple                  784       
=================================================================
Total params: 784
Trainable params: 784
Non-trainable params: 0
_________________________________________________________________

那么，我应该如何使用子类方法在summary()中获取output shape？

【参考方案1】：

我已经用这个方法解决了这个问题，不知道有没有更简单的方法。

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()

【讨论】：

您能解释一下为什么会这样吗？尤其是outputs=self.call(x) 部分。

@Samuel 通过评估outputs=self.call(x)，调用subclass.call(self, x) 方法。这会触发封装实例中的形状计算。此外，Model 的返回实例还会计算在.summary() 中报告的自己的形状。这种方法的主要问题是输入形状是恒定的shape=(24, 24, 3)，所以如果你需要一个动态的解决方案，这是行不通的。

您能解释一下... 中的内容吗？这是一个通用的解决方案，还是您需要在这些调用中使用特定于模型的东西？

@GuySoft ... in init 实例化您的层，而 ... in call 连接构建网络的不同层。它对所有子类 keras 模型都是通用的。

【参考方案2】：

我猜关键点是Network类中的_init_graph_network方法，它是Model的父类。如果在调用__init__ 方法时指定了inputs 和outputs 参数，则将调用_init_graph_network。

所以会有两种可能的方法：

手动调用_init_graph_network方法构建模型图。 使用输入层和输出重新初始化。

这两种方法都需要输入层和输出层（self.call 需要）。

现在调用summary 将给出准确的输出形状。但是它会显示Input 层，它不是子类化模型的一部分。

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
_________________________________________________________________

【参考方案3】：

我解决问题的方式与 Elazar 提到的非常相似。覆盖类subclass 中的函数summary()。然后可以在使用模型子类化的同时直接调用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()

【参考方案4】：

我分析了 Adi Shumely 的答案：

应该不需要添加 Input_shape，因为您在 build() 中将其设置为参数 添加 Input 层对模型没有任何作用，而是作为 call() 方法的参数引入 添加所谓的输出不是我看到的方式。它所做的唯一也是最重要的事情就是调用 call() 方法。

所以我提出了这个解决方案，它不需要对模型进行任何修改，只需要改进模型，因为它是在调用 summary() 方法之前构建的，方法是添加对调用的调用具有输入张量的模型的 () 方法。 我尝试了我自己的模型以及此提要中提供的三个模型，并且到目前为止它工作正常。

来自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()

来自第二条动态

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()

从提要的最后一个帖子开始

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()

【参考方案5】：

遇到同样的问题 - 通过 3 个步骤修复它：

在 _ init _ 中添加 input_shape 添加输入层 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

【参考方案6】：

我已经使用这种方法解决了在 tensorflow 2.1 和 tensorflow 2.4.1 上测试的这个问题。用model.inputs_layer声明InputLayer

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
_________________________________________________________________

【参考方案7】：

Gary's answer 有效。然而，为了更方便，我想从我的自定义类对象中透明地访问keras.Model 的summary 方法。

这可以通过实现内置的__getattr__ 方法轻松完成，如下所示：

from tensorflow.keras import Input, layers, Model

class MyModel(Model):
    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()