Implementing custom loss function in keras with different sizes for y_true and y_pred

Question

I am new to Keras. I need some help in writing a custom loss function in keras with TensorFlow backend for the following loss equation.

Answer 1

Recent versions of Keras actually support losses with different shapes of y_pred and y_true. The build in loss sparse_categorical_crossentropy is an example of this. The TensorFlow implementation of this loss is here: https://github.com/keras-team/keras/blob/0fc33feb5f4efe3bb823c57a8390f52932a966ab/keras/backend/tensorflow_backend.py#L3570

Notice how it says target: An integer tensor. and not target: A tensor of the same shape as `output`. like the others. I tried with a custom loss I made myself and it seems to work fine.

I'm using Keras 2.2.4.

Answer 2

You can pretty much just translate the numpy functions into Keras backend functions. The only thing to notice is to set up the right broadcast shape.

def l2_loss_keras(y_true, y_pred):
    # set up meshgrid: (height, width, 2)
    meshgrid = K.tf.meshgrid(K.arange(im_height), K.arange(im_width))
    meshgrid = K.cast(K.transpose(K.stack(meshgrid)), K.floatx())

    # set up broadcast shape: (batch_size, height, width, num_joints, 2)
    meshgrid_broadcast = K.expand_dims(K.expand_dims(meshgrid, 0), -2)
    y_true_broadcast = K.expand_dims(K.expand_dims(y_true, 1), 2)
    diff = meshgrid_broadcast - y_true_broadcast

    # compute loss: first sum over (height, width), then take average over num_joints
    ground = K.exp(-0.5 * K.sum(K.square(diff), axis=-1) / sigma ** 2)
    loss = K.sum(K.square(ground - y_pred), axis=[1, 2])
    return K.mean(loss, axis=-1)

To verify it:

def l2_loss_numpy(y_true, y_pred):
     loss = 0
     n = y_true.shape[0]
     for j in range(n):
        for i in range(num_joints):
            yv, xv = np.meshgrid(np.arange(0, im_height), np.arange(0, im_width))
            z = np.stack([xv, yv]).transpose(1, 2, 0)
            ground = np.exp(-0.5*(((z - y_true[j, i, :])**2).sum(axis=2))/(sigma**2))
            loss = loss + np.sum((ground - y_pred[j,:, :, i])**2)
     return loss/num_joints

batch_size = 32
num_joints = 10
sigma = 5
im_width = 256
im_height = 256

y_true = 255 * np.random.rand(batch_size, num_joints, 2)
y_pred = 255 * np.random.rand(batch_size, im_height, im_width, num_joints)

print(l2_loss_numpy(y_true, y_pred))
45448272129.0

print(K.eval(l2_loss_keras(K.variable(y_true), K.variable(y_pred))).sum())
4.5448e+10

The number is truncated under the default dtype float32. If you run it with dtype set to float64:

y_true = 255 * np.random.rand(batch_size, num_joints, 2)
y_pred = 255 * np.random.rand(batch_size, im_height, im_width, num_joints)

print(l2_loss_numpy(y_true, y_pred))
45460126940.6

print(K.eval(l2_loss_keras(K.variable(y_true), K.variable(y_pred))).sum())
45460126940.6

---

EDIT:

It seems that Keras requires y_true and y_pred to have the same number of dimensions. For example, on the following testing model:

X = np.random.rand(batch_size, 256, 256, 3)
model = Sequential([Dense(10, input_shape=(256, 256, 3))])
model.compile(loss=l2_loss_keras, optimizer='adam')
model.fit(X, y_true, batch_size=8)

ValueError: Cannot feed value of shape (8, 10, 2) for Tensor 'dense_2_target:0', which has shape '(?, ?, ?, ?)'

To deal with this problem, you can add a dummy dimension with expand_dims before feeding y_true into the model:

def l2_loss_keras(y_true, y_pred):
    ...

    y_true_broadcast = K.expand_dims(y_true, 1)  # change this line

    ...

model.fit(X, np.expand_dims(y_true, axis=1), batch_size=8)

Answer 3

Yu's answer is correct still I want to share my experience. Whenever you want to write custom loss function, beware of certain things:

1. At compile time, Keras will not complain about size mismatch. For example, if y_pred which comes from the output layer of your model has 3-D shape, y_true will default to 3-D shape. BUT, at run time i.e. during fit, if you pass target data as 2-D which many people do, you might end up getting the error in your loss function. E.g., if you are calculating sigmoid_crossentropy_with_logits, it will complain. Hence, do pass targets as 3-D via np.expand_dims.
2. Also, in custom loss make sure you use y_true and y_pred as arguments (if somebody knows that we can use any other names as arguments, pl shout).