In keras you can define a custom loss with arguments (y_true, y_pred).
How do I know to which patterns are they correlated?
I mean, y_true is a tensor with batchSize elements. How can I relate those element to the original X?
I would like to know the correspondence between y_true[0] and the relative X[i].
So what you would like to have is a loss function like this
def custom_loss(y_true, y_pred, X):
because you need the input for your loss calculation.
That's not directly possible in Keras, as far as I know.
One possible workaround could be to have a running index:
X = ...
Y = ...
batch_size = ...
i = 0
def custom_loss(y_true, y_pred):
x = X[i*batch_size:(i+1)*batch_size]
loss = ...
i += 1
return loss
Make sure to reset i after each epoch. You can do this in a LambdaCallback that you pass to model.fit(). Also make sure to pass shuffle=False to model.fit().
Related
I am having trouble with Keras Custom loss function. I want to be able to access truth as a numpy array.
Because it is a callback function, I think I am not in eager execution, which means I can't access it using the backend.get_value() function. i also tried different methods, but it always comes back to the fact that this 'Tensor' object doesn't exist.
Do I need to create a session inside the custom loss function ?
I am using Tensorflow 2.2, which is up to date.
def custom_loss(y_true, y_pred):
# 4D array that has the label (0) and a multiplier input dependant
truth = backend.get_value(y_true)
loss = backend.square((y_pred - truth[:,:,0]) * truth[:,:,1])
loss = backend.mean(loss, axis=-1)
return loss
model.compile(loss=custom_loss, optimizer='Adam')
model.fit(X, np.stack(labels, X[:, 0], axis=3), batch_size = 16)
I want to be able to access truth. It has two components (Label, Multiplier that his different for each item. I saw a solution that is input dependant, but I am not sure how to access the value. Custom loss function in Keras based on the input data
I think you can do this by enabling run_eagerly=True in model.compile as shown below.
model.compile(loss=custom_loss(weight_building, weight_space),optimizer=keras.optimizers.Adam(), metrics=['accuracy'],run_eagerly=True)
I think you also need to update custom_loss as shown below.
def custom_loss(weight_building, weight_space):
def loss(y_true, y_pred):
truth = backend.get_value(y_true)
error = backend.square((y_pred - y_true))
mse_error = backend.mean(error, axis=-1)
return mse_error
return loss
I am demonstrating the idea with a simple mnist data. Please take a look at the code here.
I would like to replace to gradient function for a loss function in tensorflow 2.0.
Say for example I have a loss function which looks like:
def loss_function(prediction):
# do some standard tensorflow things here
return loss
I then apply the gradients using the tf.GradientTape method i.e.
with tf.GradientTape() as tape:
prediction = model(input)
loss = loss_function(prediction)
gradients = tf.gradient(loss, model.trainable_variables)
optimizer.apply_gradients(zip(gradients, model.trainable_variables))
My issue is that I want to change the gradient computation and explicitly calculate it myself for just the loss_function() that's currently automatically computed.
I would guess this has something to do with the #tf.custom_gradient decorator, but unsure how I can make it work for the loss.
I am using a custom training loop as apposed to sequential/functional api.
The answer was actually quite straight forward. First define a #tf.custom_gradient() function which defined the gradient and pass the loss through it i.e.
#tf.custom_gradient
def custom_grad(x):
def grad(dy, **kwargs):
# do something with dy here
return dy
return tf.identity(x), grad
and pass the loss variable from the original function through this:
def loss_function(prediction):
# calculate the loss
return custom_grad(loss)
I am trying to optimize a model with the following two loss functions
def loss_1(pred, weights, logits):
weighted_sparse_ce = kls.SparseCategoricalCrossentropy(from_logits=True)
policy_loss = weighted_sparse_ce(pred, logits, sample_weight=advantages)
and
def loss_2(y_pred, y):
return kls.mean_squared_error(y_pred, y)
however, because TensorFlow 2 expects loss function to be of the form
def fn(y_pred, y_true):
...
I am using a work-around for loss_1 where I pack pred and weights into a single tensor before passing to loss_1 in the call to model.fit and then unpack them in loss_1. This is inelegant and nasty because pred and weights are of different data types and so this requires an additional cast, pack, un-pack and un-cast each time I call model.fit.
Furthermore, I am aware of the sample_weight argument to fit, which is kind of like the solution to this question. This might be a workable solution were it not for the fact that I am using two loss functions and I only want the sample_weight applied to one of them. Also, even if this were a solution, would it not be generalizable to other types of custom loss functions.
All that being said, my question, said concisely, is:
What is the best way to create a loss function with an arbitrary number of
arguments in TensorFlow 2?
Another thing I have tried is passing a tf.tuple but that also seems to violate TensorFlow's desires for a loss function input.
This problem can be easily solved using custom training in TF2. You need only compute your two-component loss function within a GradientTape context and then call an optimizer with the produced gradients. For example, you could create a function custom_loss which computes both losses given the arguments to each:
def custom_loss(model, loss1_args, loss2_args):
# model: tf.model.Keras
# loss1_args: arguments to loss_1, as tuple.
# loss2_args: arguments to loss_2, as tuple.
with tf.GradientTape() as tape:
l1_value = loss_1(*loss1_args)
l2_value = loss_2(*loss2_args)
loss_value = [l1_value, l2_value]
return loss_value, tape.gradient(loss_value, model.trainable_variables)
# In training loop:
loss_values, grads = custom_loss(model, loss1_args, loss2_args)
optimizer.apply_gradients(zip(grads, model.trainable_variables))
In this way, each loss function can take an arbitrary number of eager tensors, regardless of whether they are inputs or outputs to the model. The sets of arguments to each loss function need not be disjoint as shown in this example.
To expand on Jon's answer. In case you want to still have the benefits of a Keras Model you can expand the model class and write your own custom train_step:
from tensorflow.python.keras.engine import data_adapter
# custom loss function that takes two outputs of the model
# as input parameters which would otherwise not be possible
def custom_loss(gt, x, y):
return tf.reduce_mean(x) + tf.reduce_mean(y)
class CustomModel(keras.Model):
def compile(self, optimizer, my_loss):
super().compile(optimizer)
self.my_loss = my_loss
def train_step(self, data):
data = data_adapter.expand_1d(data)
input_data, gt, sample_weight = data_adapter.unpack_x_y_sample_weight(data)
with tf.GradientTape() as tape:
y_pred = self(input_data, training=True)
loss_value = self.my_loss(gt, y_pred[0], y_pred[1])
grads = tape.gradient(loss_value, self.trainable_variables)
self.optimizer.apply_gradients(zip(grads, self.trainable_variables))
return {"loss_value": loss_value}
...
model = CustomModel(inputs=input_tensor0, outputs=[x, y])
model.compile(optimizer=tf.keras.optimizers.Adam(), my_loss=custom_loss)
In tf 1.x we have tf.nn.weighted_cross_entropy_with_logits function which allows us trade off recall and precision by adding extra positive weights for each class. In multi-label classification, it should be a (N,) tensor or numpy array. However, in tf 2.0, I haven't found similar loss functions yet, so I wrote my own loss function with extra arguments pos_w_arr.
from tensorflow.keras.backend import epsilon
def pos_w_loss(pos_w_arr):
"""
Define positive weighted loss function
"""
def fn(y_true, y_pred):
_epsilon = tf.convert_to_tensor(epsilon(), dtype=y_pred.dtype.base_dtype)
_y_pred = tf.clip_by_value(y_pred, _epsilon, 1. - _epsilon)
cost = tf.multiply(tf.multiply(y_true, tf.math.log(
_y_pred)), pos_w_arr)+tf.multiply((1-y_true), tf.math.log(1-_y_pred))
return -tf.reduce_mean(cost)
return fn
Not sure what do you mean it wouldn't work when using eager tensors or numpy array as inputs though. Please correct me if I'm wrong.
I'm trying to train DNN that outputs 3 values (x,y,z) where x and y are coordinates of the object I'm looking for and z is the probability that object is present
I need custom loss function:
If z_true<0.5 I don't care of x and y values, so error should be equal to (0, 0, sqr(z_true - z_pred))
otherwise error should be like (sqr(x_true - x_pred), sqr(y_true - y_pred), sqr(z_true - z_pred))
I'm in a struggle with mixing tensors and if statements together.
Maybe this example of a custom loss function will get you up and running. It shows how you can mix tensors with if statements.
def conditional_loss_function(l):
def loss(y_true, y_pred):
if l == 0:
return loss_funtion1(y_true, y_pred)
else:
return loss_funtion2(y_true, y_pred)
return loss
model.compile(loss=conditional_loss_function(l), optimizer=...)
Use switch from Keras backend: https://keras.io/backend/#switch
It is similar to tf.cond
How to create a custom loss in Keras described here: Make a custom loss function in keras
I am trying to write a cusom Keras loss function in which I process the tensors in sub-vector chunks. For example, if an output tensor represented a concatenation of quaternion coefficients (i.e. w,x,y,z,w,x,y,z...) I might wish to normalize each quaternion before calculating the mean squared error in a loss function like:
def norm_quat_mse(y_true, y_pred):
diff = y_pred - y_true
dist = 0
for i in range(0,16,4):
dist += K.sum( K.square(diff[i:i+4] / K.sqrt(K.sum(K.square(diff[i:i+4])))))
return dist/4
While Keras will accept this function without error and use in training, it outputs a different loss value from when applied as an independent function and when using model.predict(), so I suspect it is not working properly. None of the built-in Keras loss functions use this per-chunk processing approach, is it possible to do this within Keras' auto-differentiation framework?
Try:
def norm_quat_mse(y_true, y_pred):
diff = y_pred - y_true
dist = 0
for i in range(0,16,4):
dist += K.sum( K.square(diff[:,i:i+4] / K.sqrt(K.sum(K.square(diff[:,i:i+4])))))
return dist/4
You need to know that shape of y_true and y_pred is (batch_size, output_size) so you need to skip first dimension during computations.