I'm dealing with Keras functional API.
Specifically for my experiments, I'm using Keras resnet50 model obtained with:
model = resnet50.ResNet50(weights='imagenet')
Obviously, to get the final output of the network we need to feed a value to the placeholder input_1.
My question is, can I somehow start inferencing this graph from the relu layer which is depicted at the bottom of the picture below, provided that I feed a value of the appropriate dimensions into it?
I tried to achieve this with Keras functions. Something like:
self.inp = model.input
self.outputs = [layer.output for layer in model.layers]
self.functor = K.function([self.inp, K.learning_phase()], [self.outputs[6], self.outputs[17]])
But this approach will not work, because again to inference any output I need to feed value into tensor.
Is recreating graph from scratch my best option here?
Thanks
If I got you right, you can just specify input and output nodes
base_model = tf.keras.applications.ResNet50(weights='imagenet')
inference_model = tf.keras.Model(inputs=base_model.input, outputs=base_model.get_layer('any_layer_name').output)
You can set the output to any layer name
Related
I want to remove the last layer of 'faster_rcnn_nas_lowproposals_coco' model which downloaded from https://github.com/tensorflow/models/blob/master/research/object_detection/g3doc/detection_model_zoo.md.
I know I in Keras we can use model.layers.pop() to remove the last layer.
But I searched in the Internet and there are no equivalent function in tensorflow.
If there are no equivalent function in tensorflow, are there anyone can tell me how to load trained Model zoo by Keras?
You don't need to "pop" a layer, you just have to not load it:
For the example of Mobilenet (but put your downloaded model here) :
model = mobilenet.MobileNet()
x = model.layers[-2].output
The first line load the entire model, the second load the outputs of the before the last layer.
You can change layer[-x] with x being the outputs of the layer you want. So, for loading the model without the last layer, x should be equal to -2.
Then it's possible to use it like this :
x = Dense(256)(x)
predictions = Dense(15, activation = "softmax")(x)
model = Model(inputs = model.input, outputs = predictions)
I'm using a pre-trained deeplab model (from here) to obtain segmentations for an input image. I'm able to obtain the sematic labels (i.e. SemanticPredictions) which is argmax applied to logits (link).
I was wondering if there is an easy way to obtain the logits before argmax? I was hoping to find the output tensor name and simply pass it into my tfsession
as in the following:
tf_session.run(
self.OUTPUT_TENSOR_NAME,
feed_dict={self.INPUT_TENSOR_NAME: [np.asarray(input_image)]})
But I have not been able to locate such tensor name in the code that reveals the logits, or softmax outputs.
For a model trained from MobileNet_V2 setting self.OUTPUT_TENSOR_NAME = 'ResizeBilinear_2:0' retrieves the logits before the argmax is performed.
I suspect this is the same for xception, but have not verified it.
I arrived at this answer by loading my model in tensorflow. Then, printing the name of all layers in the loaded graph. Finally, I took the name of the final output layer before the last 'ArgMax' layer and ran some inferencing using that.
Here is a link to a stackoverflow question on printing the names of the layers in a graph. I found the answer by Ted to be most helpful.
By the way, the output layers of DeeplabV3 models does not apply SoftMax. So you cannot simply take the raw value of the elements of output vectors as a confidence.
Let's say I have a model with one input and two outputs. And I want the output of the third layer of my model to be the y_true in my cost function for my second output.
I've tried this:
model.fit(x, [y, model.layers[3].output], ...)
But got the error:
'Tensor' object has no attribute 'ndim'
Which I believe is referring to the second y_true I gave the fit method.
Is it possible to do something like this in Keras? If so, how?
I managed to this by changing only the cost function, like:
def custom_euclidean_distance_loss(layer_output):
from keras import backend as K
def wrap(y_true, y_pred):
return K.mean(K.square(y_pred - layer_output))
return wrap
And since I do not use any previously known y_true I just fed a dummy one to fit. Note that the printed metrics from Keras won't be correct this way but the model will train with no problem.
If you do know of a better way (like actually feeding the layer output to fit) please let me know
There are many examples about how to do fine-tuning with tensorflow. Almost all these examples are try to resize our images to the specified size that the existing model needs. Like for example, 224×224 is the input size that vgg19 needs. However, in keras, we can change the input size by setting the include_top to false:
base_model = VGG19(include_top=False, weights="imagenet", input_shape=(input_size, input_size, input_channels))
Then we do not have to fix the image size to be 224×224 anymore. Can we do such kind of fine-tuning by using official pre-trained models in tensorflow? I cannot find the solutions up till now, anyone help me?
Yes, it is possible to do this kind of fine-tuning. You would just have to ensure that you also fine-tune some of the first few layers (to account for changed input) of the original network in addition to the last few layers (to account for changed output).
I work with TensorFlow using Keras. If you are open to that, then there is a code snippet that shows the general fine-tuning flow here:
https://keras.io/applications/
Specifically, I had to write the following code to make it work for my case:
#img_width,img_height is the size of your new input, 3 is the number of channels
input_tensor = Input(shape=(img_width, img_height, 3))
base_model =
keras.applications.vgg19.VGG19(include_top=False,weights='imagenet', input_tensor=input_tensor)
#instantiate whatever other layers you need
model = Model(inputs=base_model.inputs, outputs=predictions)
#predictions is the new logistic layer added to account for new classes
Hope this helps.
I wanna draw the weights of tf.layers.dense in tensorboard histogram, but it not show in the parameter, how could I do that?
The weights are added as a variable named kernel, so you could use
x = tf.dense(...)
weights = tf.get_default_graph().get_tensor_by_name(
os.path.split(x.name)[0] + '/kernel:0')
You can obviously replace tf.get_default_graph() by any other graph you are working in.
I came across this problem and just solved it. tf.layers.dense 's name is not necessary to be the same with the kernel's name's prefix. My tensor is "dense_2/xxx" but it's kernel is "dense_1/kernel:0". To ensure that tf.get_variable works, you'd better set the name=xxx in the tf.layers.dense function to make two names owning same prefix. It works as the demo below:
l=tf.layers.dense(input_tf_xxx,300,name='ip1')
with tf.variable_scope('ip1', reuse=True):
w = tf.get_variable('kernel')
By the way, my tf version is 1.3.
The latest tensorflow layers api creates all the variables using the tf.get_variable call. This ensures that if you wish to use the variable again, you can just use the tf.get_variable function and provide the name of the variable that you wish to obtain.
In the case of a tf.layers.dense, the variable is created as: layer_name/kernel. So, you can obtain the variable by saying:
with tf.variable_scope("layer_name", reuse=True):
weights = tf.get_variable("kernel") # do not specify
# the shape here or it will confuse tensorflow into creating a new one.
[Edit]: The new version of Tensorflow now has both Functional and Object-Oriented interfaces to the layers api. If you need the layers only for computational purposes, then using the functional api is a good choice. The function names start with small letters for instance -> tf.layers.dense(...). The Layer Objects can be created using capital first letters e.g. -> tf.layers.Dense(...). Once you have a handle to this layer object, you can use all of its functionality. For obtaining the weights, just use obj.trainable_weights this returns a list of all the trainable variables found in that layer's scope.
I am going crazy with tensorflow.
I run this:
sess.run(x.kernel)
after training, and I get the weights.
Comes from the properties described here.
I am saying that I am going crazy because it seems that there are a million slightly different ways to do something in tf, and that fragments the tutorials around.
Is there anything wrong with
model.get_weights()
After I create a model, compile it and run fit, this function returns a numpy array of the weights for me.
In TF 2 if you're inside a #tf.function (graph mode):
weights = optimizer.weights
If you're in eager mode (default in TF2 except in #tf.function decorated functions):
weights = optimizer.get_weights()
in TF2 weights will output a list in length 2
weights_out[0] = kernel weight
weights_out[1] = bias weight
the second layer weight (layer[0] is the input layer with no weights) in a model in size: 50 with input size: 784
inputs = keras.Input(shape=(784,), name="digits")
x = layers.Dense(50, activation="relu", name="dense_1")(inputs)
x = layers.Dense(50, activation="relu", name="dense_2")(x)
outputs = layers.Dense(10, activation="softmax", name="predictions")(x)
model = keras.Model(inputs=inputs, outputs=outputs)
model.compile(...)
model.fit(...)
kernel_weight = model.layers[1].weights[0]
bias_weight = model.layers[1].weights[1]
all_weight = model.layers[1].weights
print(len(all_weight)) # 2
print(kernel_weight.shape) # (784,50)
print(bias_weight.shape) # (50,)
Try to make a loop for getting the weight of each layer in your sequential network by printing the name of the layer first which you can get from:
model.summary()
Then u can get the weight of each layer running this code:
for layer in model.layers:
print(layer.name)
print(layer.get_weights())