Implementing Dice Lose

Question

I would like to implement Dice Lose like this from dice_loss_for_keras.py:

from keras import backend as K

def dice_coef(y_true, y_pred, smooth=1):
    """
    Dice = (2*|X & Y|)/ (|X|+ |Y|)
         =  2*sum(|A*B|)/(sum(A^2)+sum(B^2))
    ref: https://arxiv.org/pdf/1606.04797v1.pdf
    """
    intersection = K.sum(K.abs(y_true * y_pred), axis=-1)
    return (2. * intersection + smooth) / (K.sum(K.square(y_true),-1) + K.sum(K.square(y_pred),-1) + smooth)

The only problem is that, in most UNET implementation, y_true have 1 channel while y_pred have 3 channels (for example for 3 classes) because y_pred is represented by hot vector. Is there away either to convert y_true like y_pred or vice versa or make the output of UNET as 1 channel like y_true

score 1 · Accepted Answer · answered Jan 13 '22 at 08:15

You need to convert y_true to 1-hot representation in order to apply per-class dice loss. It seems like you have tf.one_hot function that does it for you.

Once you have y_true in the same shape as y_pred, you can use your code to compute the dice score for each class separately, and then combine the scores of all classes to get the final scalar loss.

Implementing Dice Lose

1 Answers1