tf.custom_gradient with multiple inputs

Question

tf.custom_gradient accepts only one Tensor x, what if this op needs more than one inputs?

For example, to define the gradient of Softmax which needs input x and label?

Update

Thanks for the suggestion from @AllenLavoie, I use a Python list as input.

def self_define_op_multiple_inputs():
    @tf.custom_gradient
    def loss_func(input_):
        x = input_[0]
        label = input_[2]

        def grad(dy):
            return [dy, dy]

        return x - label, grad

    x = tf.range(10, dtype=tf.float32)
    y = tf.range(10, dtype=tf.int32)

    loss = loss_func([x, y])


if __name__ == '__main__':
    self_define_op_multiple_inputs()

It seems that it will convert the Python list to a Tensor. The snippet above will raise a TypeError: TypeError: Cannot convert a list containing a tensor of dtype <dtype: 'int32'> to <dtype: 'float32'> (Tensor is: <tf.Tensor 'range_1:0' shape=(10,) dtype=int32>)

How to fix it?

The documentation says x and y can both either be Tensors or sequences of Tensors. Did this not work for you? — Allen Lavoie
– Allen Lavoie, Commented Aug 14, 2018 at 16:32
@AllenLavoie Actually this is exactly what confused me. I don't understand what's sequences of Tensors, does it mean a Python list of Tensor? — huangbiubiu
– huangbiubiu, Commented Aug 15, 2018 at 2:37
My interpretation is Python list (or tuple, etc.). So len(x) is the number of inputs to the operation, len(y) is the number of outputs. Then the gradient function takes len(y) Tensor argument and returns len(x) Tensors. — Allen Lavoie
– Allen Lavoie, Commented Aug 15, 2018 at 18:33
@AllenLavoie I tried to use list but it seems like a list will be converted as a Tensor, which will cause an error if there are multiple inputs with different type and matched shape. The question has been updated. — huangbiubiu
– huangbiubiu, Commented Aug 16, 2018 at 2:23

StoneFree · Accepted Answer · 2019-03-13 02:10:22Z

2

I ran into a similar problem yesterday and found this post, and I believe I know what you are running into. Problem is that while using @tf.custom_gradient, the function that it decorates can have multiple inputs (instead of a list of tensors). Look at the following code(note that it's just a test code with no actual meaning):

@tf.custom_gradient
def loop1(x,a):
    def grad(dy):
        return dy*3,dy*2
    n = tf.multiply(x,a)
    return n,grad

By using two inputs x and a, you have to return two gradients respectively in the grad function. dy*3 corresponds to the gradient of x and dy*2 corresponds to the gradient of a.

I think in this function the documents make people very confusing, but you can still use multiple inputs, just make sure that you also have the same number of gradients, or else you will run into errors.

answered Mar 13, 2019 at 2:10

StoneFree

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1 Comment

pateheo Over a year ago

Can we return None as gradients for unused terms ?

Sim · Accepted Answer · 2018-08-14 07:52:00Z

0

I believe you need something like this a tf Graph input:+ n_input is the input number

x = tf.placeholder("float", [None, n_input])
y = tf.placeholder("float", [None])

Does this answer your question ?

answered Aug 14, 2018 at 7:52

Sim

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1 Comment

huangbiubiu Over a year ago

Thanks for your help. But it seems that you didn't understand what I am talking about. tf.custom_gradient is not a computational graph. You can read docs for more details.

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