CODEWITHSUNDEEP
python-2.7machine-learningtensorflowkeras
beewere
beewere

Reputation: 11

keras custom loss function

I'm new to Keras framework and I want to implement the following loss function of Root Mean Squared Logarithmic Error

Here is my code for Keras with tensorflow backend

def loss_function(y_true, y_pred):
    ones = K.ones(shape=K.shape(y_pred).shape)
    y_pred = tf.add(y_pred,ones)
    y_true = tf.add(y_true,ones)
    val = K.sqrt(K.mean(K.sum(K.log(y_pred)-K.log(y_true))))
    return val

But I end up getting the following error: 

ValueError: Error when checking input: expected dense_1_input to have shape (None, 16) but got array with shape (1312779, 11)

with the val returned to be 0.

Upvotes: 1

Views: 936

Answers (1)

Daniel Möller
Daniel Möller

Reputation: 86640

The order of your operations is inverted.

Since "log(true) - log(pred)" can be either negative or positive (the result may be a little higher or a little lower than the expected), the square is the first thing that must happen. (The square is responsible for eliminating the negative signs).

And the mean is the last one (the most external), because you want first to compute the error for each element, and only after that you get the mean of the error. (The mean function already carries the sum function in it).

So:

def loss_function(y_true, y_pred):

    y_pred = y_pred + 1
    y_true = y_true + 1
    return K.mean(K.square(K.log(y_pred)-K.log(y_true)))

Please note that this does not carry the "root" part. If you want to add it, I'd say that the root should go before the mean (different from the formula in the picture)

I'd use this instead:

return K.mean(K.sqrt(K.square(K.log(y_pred)-K.log(y_true))))

Make sure that your model ends with an activation that outputs numbers greater or equal to zero:

  • Relu is ok
  • Sigmoid is ok
  • Softmax is ok

Other activations may have negative values and will bring errors with log:

  • linear is not ok
  • tanh is not ok

Upvotes: 1

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