Keras: passing the model object as an argument to the loss function

Question

This post almost does what I want. In a nutshell, the suggested solution is:

def custom_loss(y_true, y_pred):
  # Your model exists in global scope global e

  # Get the layers of your model
  layers = [l for l in e.layers]

  # Construct a graph to evaluate your other model on y_pred
  eval_pred = y_pred
  for i in range(len(layers)):
      eval_pred = layers[i](eval_pred)

  # Construct a graph to evaluate your other model on y_true
  eval_true = y_true
  for i in range(len(layers)):
      eval_true = layers[i](eval_true)

  # Now do what you wanted to do with outputs.
  # Note that we are not returning the values, but a tensor.
  return K.mean(K.square(eval_pred - eval_true), axis=-1)

In the function above, e is a global argument, which is the model itself, and the custom loss function uses the model (which is global) without requiring the user to pass in the model. I'm not a big fan of global arguments. Is there a way to construct a custom_loss function such that it takes in the model object itself without using a global argument. For example, can I create a function custom_loss(y_true, y_pred, e) and delete the line global e , such that I can pass my custom_loss as a loss function of a model?

Answer 1

Keras API does not support that. As the documentation states, loss functions take exactly two arguments: y_true and y_pred .

If you what such a feature, you have to modify Keras itself. Take a look at:

• The compile function in keras/engine/training.py
• The weighted_masked_objective function in keras/engine/training_utils.py