When I run my loss function and it will be occur this error or warning.
I really can not figure out what cause it.
I guess that maybe I didn't use the origin input,for example:
def loss(predict,label):
#because some reason I need to extract some values in predict
predictProcessed = process(predict)
#predictProcessed is a subset of predict
loss = tf.square(predict - label)
return loss
My guess is right or not?
And I also use double for-loop in this code,Should the code use less for for-loop?thanks
Related
I'm trying to set up a DNN for classification and at one point I want to take the tensor product of a vector with itself. I'm using the Keras functional API at the moment but it isn't immediately clear that there is a layer that does this already.
I've been attempting to use a Lambda layer and numpy in order to try this, but it's not working.
Doing a bit of googling reveals
tf.linalg.LinearOperatorKronecker, which does not seem to work either.
Here's what I've tried:
I have a layer called part_layer whose output is a single vector (rank one tensor).
keras.layers.Lambda(lambda x_array: np.outer(x_array, x_array),) ( part_layer) )
Ideally I would want this to to take a vector of the form [1,2] and give me [[1,2],[2,4]].
But the error I'm getting suggests that the np.outer function is not recognizing its arguments:
AttributeError: 'numpy.ndarray' object has no attribute '_keras_history
Any ideas on what to try next, or if there is a simple function to use?
You can use two operations:
If you want to consider the batch size you can use the Dot function
Otherwise, you can use the the dot function
In both case the code should look like this:
dot_lambda = lambda x_array: tf.keras.layers.dot(x_array, x_array)
# dot_lambda = lambda x_array: tf.keras.layers.Dot(x_array, x_array)
keras.layers.Lambda(dot_lamda)( part_layer)
Hope this help.
Use tf.tensordot(x_array, x_array, axes=0) to achieve what you want. For example, the expression print(tf.tensordot([1,2], [1,2], axes=0)) gives the desired result: [[1,2],[2,4]].
Keras/Tensorflow needs to keep an history of operations applied to tensors to perform the optimization. Numpy has no notion of history, so using it in the middle of a layer is not allowed. tf.tensordot performs the same operation, but keeps the history.
I am using TF tensorboard to monitor the training progress for a model. I am getting a bit confused because I am seeing the two points that represent the validation loss value showing a different direction:
Time=13:30 Smoothed=18.33 Value=15.41..........
Time=13:45 Smoothed=17.76 Value=16.92
In this case, is the validation loss increasing or decreasing? thanks!
As I cannot put figures in the comments, have a look at this graph.
If you watch the falling slope between x = 50 and x = 100, you will see that locally, the real values increase at some points (usually after downward spikes). So you could conclude that your function values are increasing. But at a larger scope you will see that the function values are decreasing. The smoothing helps you to get make the interpretation easier, but does not return exact values.
Coming back to the local example, it would give you the insight that the overall trend is a decreasing function, but it does not provide accurate loss values.
I am trying to minimize x*x with adagrad optimiser. I expect to get x=0 as result, but I get value x, close to initial value.
import tensorflow as tf
x=tf.Variable(-2.)
sq=x*x
o = tf.train.AdamOptimizer(1e-1).minimize(sq)
with tf.Session() as sess:
init = tf.global_variables_initializer()
sess.run([init])
sess.run([o])
r=sess.run([x])
print("done",r)
I get -1.9 as a result, instead of expected 0.
Do I understand correctly that -2 is initial value here, or is it something else? Does AdamOptimiser perform just one step or is it possible to launch it for continious optimisation? How do I get x=0 as result?
sess.run([0]) runs only a single step. To perform a full optimization, you need to run many steps, which can be done by repeating the single step in a loop.
Thus, you can replace sess.run([o]) with:
for i in range(1000):
sess.run([o])
This yields the results 3.4735016e-23, very close to the expected 0.
In my experience, people usually run many optimization steps just as I demonstrated, with a for loop. If you are interested in implementing the loop as a TensorFlow operation, and then running this operation only once, this can be done, but it is not recommended. The reasons are: (a) I don't think you will gain any "elegance" in your code by doing this. (b) If you want to run 1000 steps, you will need to add 1000 sets of operations to your graph, and group them as one. Contrast this to needing only one set of operations.
You can see more relevant information in this question.
I am using this function of tensorflow to get my function jacobian. Came across two problems:
The tensorflow documentation is contradicted to itself in the following two paragraph if I am not mistaken:
gradients() adds ops to the graph to output the partial derivatives of ys with respect to xs. It returns a list of Tensor of length len(xs) where each tensor is the sum(dy/dx) for y in ys.
Blockquote
Blockquote
Returns:
A list of sum(dy/dx) for each x in xs.
Blockquote
According to my test, it is, in fact, return a vector of len(ys) which is the sum(dy/dx) for each x in xs.
I do not understand why they designed it in a way that the return is the sum of the columns(or row, depending on how you define your Jacobian).
How can I really get the Jacobian?
4.In the loss, I need the partial derivative of my function with respect to input (x), but when I am optimizing with respect to the network weights, I define x as a placeholder whose value is fed later, and weights are variable, in this case, can I still define the symbolic derivative of function with respect to input (x)? and put it in the loss? ( which later when we optimize with respect to weights will bring second order derivative of the function.)
I think you are right and there is a typo there, it was probably meant to be "of length len(ys)".
For efficiency. I can't explain exactly the reasoning, but this seems to be a pretty fundamental characteristic of how TensorFlow handles automatic differentiation. See issue #675.
There is no straightforward way to get the Jacobian matrix in TensorFlow. Take a look at this answer and again issue #675. Basically, you need one call to tf.gradients per column/row.
Yes, of course. You can compute whatever gradients you want, there is no real difference between a placeholder and any other operation really. There are a few operations that do not have a gradient because it is not well defined or not implemented (in which case it will generally return 0), but that's all.
Question: What is the most efficient way to get the delta of my weights in the most efficient way in a TensorFlow network?
Background: I've got the operators hooked up as follows (thanks to this SO question):
self.cost = `the rest of the network`
self.rmsprop = tf.train.RMSPropOptimizer(lr,rms_decay,0.0,rms_eps)
self.comp_grads = self.rmsprop.compute_gradients(self.cost)
self.grad_placeholder = [(tf.placeholder("float", shape=grad[1].get_shape(), name="grad_placeholder"), grad[1]) for grad in self.comp_grads]
self.apply_grads = self.rmsprop.apply_gradients(self.grad_placeholder)
Now, to feed in information, I run the following:
feed_dict = `training variables`
grad_vals = self.sess.run([grad[0] for grad in self.comp_grads], feed_dict=feed_dict)
feed_dict2 = `feed_dict plus gradient values added to self.grad_placeholder`
self.sess.run(self.apply_grads, feed_dict=feed_dict2)
The command of run(self.apply_grads) will update the network weights, but when I compute the differences in the starting and ending weights (run(self.w1)), those numbers are different than what is stored in grad_vals[0]. I figure this is because the RMSPropOptimizer does more to the raw gradients, but I'm not sure what, or where to find out what it does.
So back to the question: How do I get the delta on my weights in the most efficient way? Am I stuck running self.w1.eval(sess) multiple times to get the weights and calc the difference? Is there something that I'm missing with the tf.RMSPropOptimizer function.
Thanks!
RMSprop does not subtract the gradient from the parameters but use more complicated formula involving a combination of:
a momentum, if the corresponding parameter is not 0
a gradient step, rescaled non uniformly (on each coordinate) by the square root of the squared average of the gradient.
For more information you can refer to these slides or this recent paper.
The delta is first computed in memory by tensorflow in the slot variable 'momentum' and then the variable is updated (see the C++ operator).
Thus, you should be able to access it and construct a delta node with delta_w1 = self.rmsprop.get_slot(self.w1, 'momentum'). (I have not tried it yet.)
You can add the weights to the list of things to fetch each run call. Then you can compute the deltas outside of TensorFlow since you will have the iterates. This should be reasonably efficient, although it might incur an extra elementwise difference, but to avoid that you might have to hack around in the guts of the optimizer and find where it puts the update before it applies it and fetch that each step. Fetching the weights each call shouldn't do wasteful extra evaluations of part of the graph at least.
RMSProp does complicated scaling of the learning rate for each weight. Basically it divides the learning rate for a weight by a running average of the magnitudes of recent gradients of that weight.