# Differentiable version of numpy.diff()

Hi everyone,
I’d need to count the number of times that a tensor changes its values(from the previous one to the next one), in order to include this information as a term in my loss function.
Example:

``````>>> import numpy as np
>>> a = [1, 1, 1, 0, 0, 1, 0]
>>> diffs = np.diff(a)
>>> diffs
array([ 0,  0, -1,  0,  1, -1])
>>> (diffs != 0).sum()
3
``````

In PyTorch I did it in the following way:

``````def count_diffs(x):
diff_x = x[1:] - x[:-1]
return (diff_x != 0).sum()

def loss_diffs(x, batch_size, num_classes):
# x.shape == (batch_size, steps, num_classes)
loss = 0.0
for idx_batch in range(batch_size):
for idx_class in range(num_classes):
loss += count_diffs(x[idx_batch, :, idx_class])
loss /= (batch_size * num_classes)

return loss
``````

In other words, choosing a sample and a class I have a sequence, and I would like to explicitly tell to the loss to prefer this sequence

``[1, 1, 1, 0, 0, 0, 0]``
over this one
``[1, 1, 1, 0, 1, 0, 0]``

Any suggestions on how to do it? My PyTorch implementation does not work, I think because it is not differentiable.
Thank you Hi,

The problem is that your `count_diffs()` method actually returns a LongTensor. And we can’t really have gradients for integer types.
You will need to find a version of your loss that only uses continuous values for it to be differentiable.

Sorry but I never faced this problem before, could you kindly show me an example where this problem occurs?

This function takes a `x` that is a float Tensor (potentially with autograd info attached) and returns an integer (inside a Tensor) that represent the value you’re looking for.
But integers are not continuous, so you cannot compute gradients for them.
So you won’t be able to get gradients flowing back if you use this function.

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