Cache intermediate tensor and update periodically - tensorflow

I have a large tensor that is expensive to calculate, but realistically I only need to recalculate it every 10 iterations or so (during gradient descent). What's the best way to do this?
More specifically:
Suppose I have an intermediate_tensor that is used in the calculation of final_tensor each time the a tf.Session is run. final_tensor is, in my case, a set of modified gradients to use in optimization. It is possible to define a graph that contains both intermediate_tensor and final_tensor. However, running this graph will be inefficient when intermediate_tensor changes slowly. In pseudocode, this is what I'd like to do:
intermediate_tensor = tf.some_operation(earlier_variable)
final_tensor = tf.matmul(intermediate_tensor, other_earlier_variable)
with tf.Session() as sess:
# pretending `partial_run` works like I want it to:
sess.partial_run(intermediate_tensor, feed_dict = {})
for i in range(5):
ft = sess.partial_run(final_tensor, feed_dict = {})
print(ft)
The experimental partial_run feature is almost what I'm looking for. However, partial_run can only be used if I want to evaluate final_tensor just once for each time I evaluate intemediate_tensor. It won't work for a for loop.
My workaround for the moment is to use tf.placeholder. I evaluate intermediate_tensor in one call to sess.run, then feed the result to a new call of sess.run as a placeholder. However, this is very inflexible. It requires that I hardcode the variable shape at compile time, for example. It's also not very good when the number of intermediate variables I'd like to use is very large.
Is there a better way? This would be very helpful if, say, one were using a curvature matrix that doesn't need to be evaluated every iteration.

Related

Tensorflow error:The graph couldn't be sorted in topological order

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

Why AdamOptimizer fails to find optimal value to minimize x*x?

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.

What makes dynamic_rnn faster to compile?

I have a complicated network with many repeated RNN steps. Compiling takes a long time (30+ minutes, mostly stuck at the gradient step) and I found that this issue might be related, which mentions dynamic_rnn as a much faster way to compile:
Looking over dynamic_rnn, I then reformatted my network to include a while_loop, like so:
#input: tensor with 1000 time steps
def body(i, prev_state):
inp = tf.slice(input, i, 1)
new_state = cell(tf.squeeze(int), prev_state) # Includes scope reuse
return [tf.add(i, tf.constant(1)), new_state]
def cond(i):
return some_cond(i)
tf.while_loop(cond, body, [tf.constant(0), initial_state])
But this didn't seem to help. What besides simply putting the cell call in a loop makes dynamic_rnn so much faster to compile?

Tensorflow: opt.compute_gradients() returns values different from the weight difference of opt.apply_gradients()

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.

What is the best way to implement weight constraints in TensorFlow?

Suppose we have weights
x = tf.Variable(np.random.random((5,10)))
cost = ...
And we use the GD optimizer:
upds = tf.train.GradientDescentOptimizer(lr).minimize(cost)
session.run(upds)
How can we implement for example non-negativity on weights?
I tried clipping them:
upds = tf.train.GradientDescentOptimizer(lr).minimize(cost)
session.run(upds)
session.run(tf.assign(x, tf.clip_by_value(x, 0, np.infty)))
But this slows down my training by a factor of 50.
Does anybody know a good way to implement such constraints on the weights in TensorFlow?
P.S.: in the equivalent Theano algorithm, I had
T.clip(x, 0, np.infty)
and it ran smoothly.
You can take the Lagrangian approach and simply add a penalty for features of the variable you don't want.
e.g. To encourage theta to be non-negative, you could add the following to the optimizer's objective function.
added_loss = -tf.minimum( tf.reduce_min(theta),0)
If any theta are negative, then add2loss will be positive, otherwise zero. Scaling that to a meaningful value is left as an exercise to the reader. Scaling too little will not exert enough pressure. Too much may make things unstable.
As of TensorFlow 1.4, there is a new argument to tf.get_variable that allows to pass a constraint function that is applied after the update of the optimizer. Here is an example that enforces a non-negativity constraint:
with tf.variable_scope("MyScope"):
v1 = tf.get_variable("v1", …, constraint=lambda x: tf.clip_by_value(x, 0, np.infty))
constraint: An optional projection function to be applied to the
variable
after being updated by an Optimizer (e.g. used to implement norm
constraints or value constraints for layer weights). The function must
take as input the unprojected Tensor representing the value of the
variable and return the Tensor for the projected value
(which must have the same shape). Constraints are not safe to
use when doing asynchronous distributed training.
By running
sess.run(tf.assign(x, tf.clip_by_value(x, 0, np.infty)))
you are consistently adding nodes to the graph and making it slower and slower.
Actually you may just define a clip_op when building the graph and run it each time after updating the weights:
# build the graph
x = tf.Variable(np.random.random((5,10)))
loss = ...
train_op = tf.train.GradientDescentOptimizer(lr).minimize(loss)
clip_op = tf.assign(x, tf.clip(x, 0, np.infty))
# train
sess.run(train_op)
sess.run(clip_op)
I recently had this problem as well. I discovered that you can import keras which has nice weight constraint functions as use them directly in the kernen constraint in tensorflow. Here is an example of my code. You can do similar things with kernel regularizer
from keras.constraints import non_neg
conv1 = tf.layers.conv2d(
inputs=features['x'],
filters=32,
kernel_size=[5,5],
strides = 2,
padding='valid',
activation=tf.nn.relu,
kernel_regularizer=None,
kernel_constraint=non_neg(),
use_bias=False)
There is a practical solution: Your cost function can be written by you, to put high cost onto negative weights. I did this in a matrix factorization model in TensorFlow with python, and it worked well enough. Right? I mean it's obvious. But nobody else mentioned it so here you go. EDIT: I just saw that Mark Borderding also gave another loss and cost-based solution implementation before I did.
And if "the best way" is wanted, as the OP asked, what then? Well "best" might actually be application-specific, in which case you'd need to try a few different ways with your dataset and consider your application requirements.
Here is working code for increasing the cost for unwanted negative solution variables:
cost = tf.reduce_sum(keep_loss) + Lambda * reg # Cost = sum of losses for training set, except missing data.
if prefer_nonneg: # Optionally increase cost for negative values in rhat, if you want that.
negs_indices = tf.where(rhat < tf.constant(0.0))
neg_vals = tf.gather_nd(rhat, negs_indices)
cost += 2. * tf.reduce_sum(tf.abs(neg_vals)) # 2 is a magic number (empirical parameter)
You are free to use my code but please give me some credit if you choose to use it. Give a link to this answer on stackoverflow.com please.
This design would be considered a soft constraint, because you can still get negative weights, if you let it, depending on your cost definition.
It seems that constraint= is also available in TF v1.4+ as a parameter to tf.get_variable(), where you can pass a function like tf.clip_by_value. This seems like another soft constraint, not hard constraint, in my opinion, because it depends on your function to work well or not. It also might be slow, as the other answerer tried the same function and reported it was slow to converge, although they didn't use the constraint= parameter to do this. I don't see any reason why one would be any faster than the other since they both use the same clipping approach. So if you use the constraint= parameter then you should expect slow convergence in the context of the original poster's application.
It would be nicer if also TF provided true hard constraints to the API, and let TF figure out how to both implement that as well as make it efficient on the back end. I mean, I have seen this done in linear programming solvers already for a long time. The application declares a constraint, and the back end makes it happen.