如何获取/定义我的图形的输入 output 名称或以正确的形式冻结它

Question

Hi everyone we have been trying to save our model in.bytes format so that we can use it in a c# script. We are using tensorflow 1.7.0 Here is our model:

bsize=16


# define cnn model
def define_model():
    # load model
    model = VGG16(include_top=False, input_shape=(224, 224, 3))
    # mark loaded layers as not trainable
    for layer in model.layers:
        layer.trainable = False
    # add new classifier layers
    flat1 = Flatten()(model.layers[-1].output)
    class1 = Dense(128, activation='relu', kernel_initializer='he_uniform')(flat1)
    output = Dense(2, activation='softmax')(class1)
    # define new model
    model = Model(inputs=model.inputs, outputs=output)
    # compile model
    # compile model
    opt = Adam(lr=0.001)
    model.compile(optimizer=opt, loss='categorical_crossentropy', metrics=['accuracy'])
    return model

Training:

sess=tf.Session()
tf.global_variables_initializer().run(session=sess)
model = define_model()
# create data generator
datagen = ImageDataGenerator(rescale=1.0/255.0)
# prepare iterators
train_it = datagen.flow_from_directory(mpath+'/train',
class_mode='categorical', batch_size=bsize, target_size=(224, 224))
test_it = datagen.flow_from_directory(mpath+'/test',
class_mode='categorical', batch_size=bsize, target_size=(224, 224))
# fit model
history = model.fit_generator(train_it, steps_per_epoch=len(train_it),
validation_data=test_it, validation_steps=len(test_it), epochs=1, verbose=0)
# evaluate model
_, acc = model.evaluate_generator(test_it, steps=len(test_it), verbose=0)
print('> %.3f' % (acc * 100.0))

model.save_weights("weights.h5")

Freeze:

K.clear_session()
K.set_learning_phase(0)
model = define_model()
model.load_weights("weights.h5")

save_dir = "./out"


tf.saved_model.simple_save(K.get_session(),
                           save_dir,
                           inputs={"input": model.inputs[0]},
                           outputs={"output": model.outputs[0]})

freeze_graph.freeze_graph(None,
                          None,
                          None,
                          None,
                          model.outputs[0].op.name,
                          None,
                          None,
                          os.path.join(save_dir, "frozen_model.bytes"),
                          False,
                          "",
                          input_saved_model_dir=save_dir)

Then we try to see input and output names using:

    model.inputs[0].name
    model.ouputs[0].name

Finally we would like to use this graph in c# as:

.AddInput(graph["input_1_1:0"][0], tensor).Fetch(graph["output"][0]);

However, as we understand from the error input and output names are wrong. Moreover whenever we call

model.inputs[0].name
model.ouputs[0].name

they print different input and output names even if we define output name as name="output"

Do you have any suggestions about how to freeze this model, obtain input and output names etc

regards

Answer 1

How about this:

my_input = tf.keras.layers.Input(shape=[224,224,3], name="my_input")
vgg = tf.keras.applications.VGG16(input_tensor=my_input)
my_output = tf.identity(vgg.output, name="my_output")

Now you know the names of your input and output tensors:

In [0]: my_input                                                                                                                                                                    
Out[0]: <tf.Tensor 'my_input:0' shape=(?, 224, 224, 3) dtype=float32>

In [1]: my_output                                                                                                                                                                   
Out[1]: <tf.Tensor 'my_output:0' shape=(?, 1000) dtype=float32>

Answer 2

For those having similar issues:

model.inputs[0].name
model.ouputs[0].name

yield sth like this:

input_1:0
dense_2/Softmax:0

but you shouldnt use them directly in.AddInput so instead of:

.AddInput(graph["input_1:0"][0], tensor).Fetch(graph[dense_2/Softmax:0][0]);

do it this way:

.AddInput(graph["input_1"][0], tensor).Fetch(graph[dense_2/Softmax][0]);

So the code above actually works just omit the:0 part from the name strings and thats all:D