我试图在InceptionResNetV2 (或任何其他可以通过tf.keras.applications导入的预先培训的网络)中添加层。我知道可以将对象添加到顺序模型或函数模型中。但是,当我这样做时,我将无法访问来自各层的单个输出,以便在Grad或类似的应用程序中使用它们。
我现在正在使用下面的模型结构。它有效,它可以训练。但是,它不允许我访问关于特定输入和特定输出的InceptionResNetV2的最后一个卷积层的输出。
from tensorflow.keras import layers, models
InceptionResNetV2 = tf.keras.applications.inception_resnet_v2.InceptionResNetV2
def get_base():
conv_base = InceptionResNetV2(weights=None, include_top=False, input_shape=(224, 224, 3))
conv_base.trainable = False
return(conv_base)
def get_model():
base = get_base()
inputs = tf.keras.Input(shape=(224, 224, 3))
x = base(inputs, training=False)
x = layers.Flatten()(x)
x = layers.Dense(512, "relu")(x)
x = layers.Dropout(0.25)(x)
x = layers.Dense(256, "relu")(x)
x = layers.Dropout(0.25)(x)
dims = layers.Dense(2, name="Valence_Arousal")(x)
expression = layers.Dense(2, name="Emotion_Category")(x)
model = models.Model(inputs=[inputs], outputs=[expression, dims])
return(model)
print(get_model().summary())发布于 2022-09-15 04:16:58
在创建嵌套模型之后,扩展它们是很困难的。将input_tensor参数传递给预先训练的模型,可以得到预期的结果。
def get_model():
inputs = tf.keras.Input(shape=(224, 224, 3))
conv_base = InceptionResNetV2(weights=None, include_top=False, input_tensor = inputs)
conv_base.trainable = False
x = layers.Flatten()(conv_base.output)
x = layers.Dense(512, "relu")(x)
x = layers.Dropout(0.25)(x)
x = layers.Dense(256, "relu")(x)
x = layers.Dropout(0.25)(x)
dims = layers.Dense(2, name="Valence_Arousal")(x)
expression = layers.Dense(2, name="Emotion_Category")(x)
model = models.Model(inputs=[inputs], outputs=[expression, dims])
return(model)示范摘要:
input_1 (InputLayer) [(None, 224, 224, 3 0 []
)]
conv2d (Conv2D) (None, 111, 111, 32 864 ['input_1[0][0]']
)
...https://stackoverflow.com/questions/73723176
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