CODEWITHSUNDEEP
pythontensorflowkerasnanloss
AljoSt
AljoSt

Reputation: 439

Keras loss function understanding

In order to understand some callbacks of Keras better, I want to artificially create a nan loss.

This is the function

def soft_dice_loss(y_true, y_pred):

  from keras import backend as K
  if K.eval(K.random_normal((1, 1), mean=2, stddev=2))[0][0] // 1 == 2.0:
    # return nan
    return K.exp(1.0) / K.exp(-10000000000.0) - K.exp(1.0) / K.exp(-10000000000.0)

  epsilon = 1e-6

  axes = tuple(range(1, len(y_pred.shape) - 1))
  numerator = 2. * K.sum(y_pred * y_true, axes)
  denominator = K.sum(K.square(y_pred) + K.square(y_true), axes)

 return 1 - K.mean(numerator / (denominator + epsilon))

So normally, it calculates the dice loss, but from time to time it should randomly return a nan. However, this does not seem to happen:

keras outputs

From time to time though, when I try to run the code, it stops right at the start (before the first epoch) with an error, saying that An operation has None for gradient. Please make sure that all of your ops have a gradient defined

Does that mean, that the the random function of Keras is just evaluated once and then always returns the same value? If so, why is that and how can I create a loss function that returns nan from time to time?

Upvotes: 0

Views: 307

Answers (1)

rvinas
rvinas

Reputation: 11895

Your first conditional statement is only evaluated once the loss function is defined (i.e. called; that is why Keras stops right at the start). Instead, you could use keras.backend.switch to integrate your conditional into the graph's logic. Your loss function could be something along the lines of:

import keras.backend as K
import numpy as np


def soft_dice_loss(y_true, y_pred):
    epsilon = 1e-6
    axes = tuple(range(1, len(y_pred.shape) - 1))
    numerator = 2. * K.sum(y_pred * y_true, axes)
    denominator = K.sum(K.square(y_pred) + K.square(y_true), axes)
    loss = 1 - K.mean(numerator / (denominator + epsilon))

    return K.switch(condition=K.random_normal((), mean=0, stddev=1) > 3,
                    then_expression=K.variable(np.nan),
                    else_expression=loss)

Upvotes: 1

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