变量 tf.Variable 在 TensorFlow Probability 中的梯度为“无”

Question

I'm having trouble constructing a basic BNN in TFP. I'm new to TFP and BNNs in general, so I apologize if I've missed something simple.

I can train a basic NN in Tensorflow by doing the following:

model = keras.Sequential([
    keras.layers.Dense(units=100, activation='relu'),
    keras.layers.Dense(units=50, activation='relu'),
    keras.layers.Dense(units=5, activation='softmax')
])

model.compile(optimizer=optimizer, 
              loss=tf.losses.CategoricalCrossentropy(from_logits=True),
              metrics=['accuracy'])

history = model.fit(
    training_data.repeat(), 
    epochs=100, 
    steps_per_epoch=(X_train.shape[0]//1024),
    validation_data=test_data.repeat(), 
    validation_steps=2
)

However, I have trouble when trying to implement a similar architecture with tfp DenseFlipout layers:

model = keras.Sequential([
    tfp.layers.DenseFlipout(units=100, activation='relu'),
    tfp.layers.DenseFlipout(units=10, activation='relu'),
    tfp.layers.DenseFlipout(units=5, activation='softmax')
])

model.compile(optimizer=optimizer, 
              loss=tf.losses.CategoricalCrossentropy(from_logits=True),
              metrics=['accuracy'])

history = model.fit(
    training_data.repeat(), 
    epochs=100, 
    steps_per_epoch=(X_train.shape[0]//1024),
    validation_data=test_data.repeat(), 
    validation_steps=2
)

I get the following Value error:

ValueError: 
Variable <tf.Variable 'sequential_11/dense_flipout_15/kernel_posterior_loc:0' 
shape=(175, 100) dtype=float32> has `None` for gradient. 
Please make sure that all of your ops have a gradient defined (i.e. are differentiable). 
Common ops without gradient: K.argmax, K.round, K.eval.

I've done some googling, and have looked around the TFP docs, but am at a loss so thought I would share the issue. Have I missed something obvious?

Thanks in advance.

Answer 1

I expect it's because you're using TensorFlow 2, are you? It isn't fully supported yet. If so, downgrading to 1.14 should get it working.