By mistake I forgot to reduce the mean of the output from the cross entropy before I fed it as the loss, but the training ran anyways and produced reasonable results.
Now I'm wondering if what I did:
loss = tf.nn.sparse_softmax_cross_entropy_with_logits(labels=labels, logits=logits, name='cross_entropy_per_example')
op = tf.train.AdamOptimizer(0.01).minimize(loss)
Is the same as:
loss = tf.reduce_mean(tf.nn.sparse_softmax_cross_entropy_with_logits(labels=labels, logits=logits, name='cross_entropy_per_example'))
op = tf.train.AdamOptimizer(0.01).minimize(loss)
I was under the impression that the optimization of the cost function required a single value tensor, but I'm confused why the training ran despite passing a tensor with more than one value.
tf.gradients (and therefore most higher-level interfaces to it, including Optimizers) implicitly sums whatever you're differentiating. tf.gradients will only compute gradients with respect to a scalar. There is some mention of this in the tf.gradients documentation.
So in your case it's just off by whatever reduce_mean was dividing by.
Related
I have a Tensorflow 2.0 tf.keras.Sequential model. Now, my technical specification prescribes using the Levenberg-Marquardt optimizer to fit the model. Tensorflow 2.0 doesn't provide it as an optimizer out of the box, but it is available in the Tensorflow Graphics module.
tfg.math.optimizer.levenberg_marquardt.minimize function accepts residuals ( a residual is a Python callable returning a tensor) and variables (list of tensors corresponding to my model weights) as parameters.
What would be the best way to convert my model into residuals and variables?
If I understand correctly how the minimize function works, I have to provide two residuals. The first residual must call my model for every learning case and aggregate all the results into a tensor. The second residuals must return all labels as a single constant tensor. The problem is that tf.keras.Sequential.predict function returns a numpy array instead of tensor. I believe that if I convert it to a tensor, the minimizer won't be able to calculate jacobians with respect to variables.
The same problem is with variables. It doesn't seem like there's a way to extract all weights from a model into a list of tensors.
There's a major difference between tfg.math.optimizer.levenberg_marquardt.minimize and Keras optimizers from the implementation/API perspective.
Keras optimizers, such as tf.keras.optimizers.Adam consume gradients as input and updates tf.Variables.
In contrast, tfg.math.optimizer.levenberg_marquardt.minimize essentially unrolls the optimization loop in graph mode (using a tf.while_loop construct). It takes initial parameter values and produces updated parameter values, unlike Adam & co, which only apply one iteration and actually change the values of tf.Variables via assign_add.
Stepping back a bit to the theoretical big picture, Levenberg-Marquardt is not a general gradient descent-like solver for any nonlinear optimization problem (such as Adam is). It specifically addresses nonlinear least-squares optimization, so it's not a drop-in replacement for optimizers like Adam. In gradient descent, we compute the gradient of the loss with respect to the parameters. In Levenberg-Marquardt, we compute the Jacobian of the residuals with respect to the parameters. Concretely, it repeatedly solves the linearized problem Jacobian # delta_params = residuals for delta_params using tf.linalg.lstsq (which internally uses Cholesky decomposition on the Gram matrix computed from the Jacobian) and applies delta_params as the update.
Note that this lstsq operation has cubic complexity in the number of parameters, so in case of neural nets it can only be applied for fairly small ones.
Also note that Levenberg-Marquardt is usually applied as a batch algorithm, not a minibatch algorithm like SGD, though there's nothing stopping you from applying the LM iteration on different minibatches in each iteration.
I think you may only be able to get one iteration out of tfg's LM algorithm, through something like
from tensorflow_graphics.math.optimizer.levenberg_marquardt import minimize as lm_minimize
for input_batch, target_batch in dataset:
def residual_fn(trainable_params):
# do not use trainable params, it will still be at its initial value, since we only do one iteration of Levenberg Marquardt each time.
return model(input_batch) - target_batch
new_objective_value, new_params = lm_minimize(residual_fn, model.trainable_variables, max_iter=1)
for var, new_param in zip(model.trainable_variables, new_params):
var.assign(new_param)
In contrast, I believe the following naive method will not work where we assign model parameters before computing the residuals:
from tensorflow_graphics.math.optimizer.levenberg_marquardt import minimize as lm_minimize
dataset_iterator = ...
def residual_fn(params):
input_batch, target_batch = next(dataset_iterator)
for var, param in zip(model.trainable_variables, params):
var.assign(param)
return model(input_batch) - target_batch
final_objective, final_params = lm_minimize(residual_fn, model.trainable_variables, max_iter=10000)
for var, final_param in zip(model.trainable_variables, final_params):
var.assign(final_param)
The main conceptual problem is that residual_fn's output has no gradients wrt its input params, since this dependency goes through a tf.assign. But it might even fail before that due to using constructs that are disallowed in graph mode.
Overall I believe it's best to write your own LM optimizer that works on tf.Variables, since tfg.math.optimizer.levenberg_marquardt.minimize has a very different API that is not really suited for optimizing Keras model parameters since you can't directly compute model(input, parameters) - target_value without a tf.assign.
I am using autoencoders to do anomaly detection. So, I have finished training my model and now I want to calculate the reconstruction loss for each entry in the dataset. so that I can assign anomalies to data points with high reconstruction loss.
This is my current code to calculate the reconstruction loss
But this is really slow. By my estimation, it should take 5 hours to go through the dataset whereas training one epoch occurs in approx 55 mins.
I feel that converting to tensor operation is bottlenecking the code, but I can't find a better way to do it.
I've tried changing the batch sizes but it does not make much of a difference. I have to use the convert to tensor part because K.eval is throwing an error if I do it normally.
python
for i in range(0, encoded_dataset.shape[0], batch_size):
y_true = tf.convert_to_tensor(encoded_dataset[i:i+batch_size].values,
np.float32)
y_pred= tf.convert_to_tensor(ae1.predict(encoded_dataset[i:i+batch_size].values),
np.float32)
# Append the batch losses (numpy array) to the list
reconstruction_loss_transaction.append(K.eval(loss_function( y_true, y_pred)))
I was able to train in 55 mins per epoch. So I feel prediction should not take 5 hours per epoch. encoded_dataset is a variable that has the entire dataset in main memory as a data frame.
I am using Azure VM instance.
K.eval(loss_function(y_true,y_pred) is to find the loss for each row of the batch
So y_true will be of size (batch_size,2000) and so will y_pred
K.eval(loss_function(y_true,y_pred) will give me an output of
(batch_size,1) evaluating binary cross entropy on each row of y
_true and y_pred
Moved from comments:
My suspicion is that ae1.predict and K.eval(loss_function) are behaving in unexpected ways. ae1.predict should normally be used to output the loss function value as well as y_pred. When you create the model, specify that the loss value is another output (you can have a list of multiple outputs), then just call predict here once to get both y_pred the loss value in one call.
But I want the loss for each row . Won't the loss returned by the predict method be the mean loss for the entire batch?
The answer depends on how the loss function is implemented. Both ways produce perfectly valid and identical results in TF under the hood. You could average the loss over the batch before taking the gradient w.r.t. the loss, or take the gradient w.r.t. a vector of losses. The gradient operation in TF will perform the averaging of the losses for you if you use the latter approach (see SO articles on taking the per-sample gradient, it's actually hard to do).
If Keras implements the loss with reduce_mean built into the loss, you could just define your own loss. If you're using square loss, replacing 'mean_squared_error' with lambda y_true, y_pred: tf.square(y_pred - y_true). That would produce square error instead of MSE (no difference to the gradient), but look here for the variant including the mean.
In any case this produces a per sample loss so long as you don't use tf.reduce_mean, which is purely optional in the loss. Another option is to simply compute the loss separately from what you optimize for and make that an output of the model, also perfectly valid.
I would like to get the values of the y_pred and y_true tensors of this keras backend function. I need this to be able to perform some custom calculations and change the loss, these calculations are just possible with the real array values.
def mean_squared_error(y_true, y_pred):
#some code here
return K.mean(K.square(y_pred - y_true), axis=-1)
There is a way to do this in keras? Or in any other ML framework (tf, pytorch, theano)?
No, in general you can't compute the loss that way, because Keras is based on frameworks that do automatic differentiation (like Theano, TensorFlow) and they need to know which operations you are doing in between in order to compute the gradients of the loss.
You need to implement your loss computations using keras.backend functions, else there is no way to compute gradients and optimization won't be possible.
Try including this within the loss function:
y_true = keras.backend.print_tensor(y_true, message='y_true')
Following is an excerpt from the Keras documentation (https://keras.io/backend/):
print_tensor
keras.backend.print_tensor(x, message='')
Prints message and the tensor value when evaluated.
Note that print_tensor returns a new tensor identical to x which should be used in the later parts of the code. Otherwise, the print operation is not taken into account during evaluation.
While Calculating the Loss Function. Can i manually Calculate Loss like
Loss = tf.reduce_mean(tf.square(np.array(Prediction) - np.array(Y)))
and then Optimize this Loss using Adam Optimizer
No.
Tensorflow loss functions typically accept tensors as input and also outputs a tensor. So np.array() wouldn't work.
In case of CNNs, you'd generally come across loss functions like cross-entropy, softmax corss-entropy, sigmoid cross-entropy etc. These are already in-built in tf.losses module. So you can use them directly.
The loss function that you're trying to apply looks like a Mean-squared loss. This is built in tf.losses as well. tf.losses.mean_squared_error.
Having said that, I've also implemented a few loss functions like cross-entropy using hand-coded formula such as: -tf.reduce_mean(tf.reduce_sum(targets * logProb)). This works equally fine, as long as the inputs targets and logProb are computed as tensors and not as numpy arrays.
No, actually you need to use tensor Variable for Loss, not use numpy.array(np.array(Prediction)).
Since tensorflow will eval these tensors in tensorflow engine.
I am trying to implement multi-label classification using TensorFlow (i.e., each output pattern can have many active units). The problem has imbalanced classes (i.e., much more zeros than ones in the labels distribution, which makes label patterns very sparse).
The best way to tackle the problem should be to use the tf.nn.weighted_cross_entropy_with_logits function. However, I get this runtime error:
ValueError: Tensor conversion requested dtype uint8 for Tensor with dtype float32
I can't understand what is wrong here. As input to the loss function, I pass the labels tensor, the logits tensor, and the positive class weight, which is a constant:
positive_class_weight = 10
loss = tf.nn.weighted_cross_entropy_with_logits(targets=labels, logits=logits, pos_weight=positive_class_weight)
Any hints about how to solve this? If I just pass the same labels and logits tensors to the tf.losses.sigmoid_cross_entropy loss function, everything works well (in the sense that Tensorflow runs properly, but of course following training predictions are always zero).
See related problem here.
The error is likely to be thrown after the loss function, because the only significant difference between tf.losses.sigmoid_cross_entropy and tf.nn.weighted_cross_entropy_with_logits is the shape of the returned tensor.
Take a look at this example:
logits = tf.linspace(-3., 5., 10)
labels = tf.fill([10,], 1.)
positive_class_weight = 10
weighted_loss = tf.nn.weighted_cross_entropy_with_logits(targets=labels, logits=logits, pos_weight=positive_class_weight)
print(weighted_loss.shape)
sigmoid_loss = tf.losses.sigmoid_cross_entropy(multi_class_labels=labels, logits=logits)
print(sigmoid_loss.shape)
Tensors logits and labels are kind of artificial and both have shape (10,). But it's important that weighted_loss and sigmoid_loss are different. Here's the output:
(10,)
()
This is because tf.losses.sigmoid_cross_entropy performs reduction (the sum by default). So in order to replicate it, you have to wrap the weighted loss with tf.reduce_sum(...).
If this doesn't help, make sure that labels tensor has type float32. This bug is very easy to make, e.g., the following declaration won't work:
labels = tf.fill([10,], 1) # the type is not float!
You might be also interested to read this question.