fix: add cuda backend support for to_raggedtensor and from_raggedtensor functions

[maxymnaumchyk]

No description provided.

[codecov]

Codecov Report

Attention: Patch coverage is 10.71429% with 25 lines in your changes missing coverage. Please review.

Project coverage is 81.98%. Comparing base (b749e49) to head (7f9bb74).
Report is 165 commits behind head on main.

Files with missing lines	Patch %	Lines
src/awkward/operations/ak_to_raggedtensor.py	10.71%	25 Missing ⚠️

Additional details and impacted files

Files with missing lines	Coverage Δ
src/awkward/operations/ak_to_raggedtensor.py	21.15% <10.71%> (ø)

... and 105 files with indirect coverage changes

[maxymnaumchyk]

@jpivarski while trying to make the to_raggedtensor function keep the device of the original awkward array I stumbled upon an issue. The thing is, tensorflow automatically selects gpu for computation, if it's available. And if I try to run the following code on gpu, it does return a tensor on cpu:

import tensorflow as tf

def function():
    with tf.device('CPU:0'):
        a = tf.constant([[1.0, 2.0, 3.0], [4.0, 5.0, 6.0]])
        return a

a = function()
a.device

>>/job:localhost/replica:0/task:0/device:CPU:0

However if try to do the same with the to_raggedtensor function, the intermediate ragged tensor is allocated on cpu (the line 78 print says that it's on cpu) but the resulting tensor is allocated on gpu:

to_raggedtensor(ak.Array([[[1.1, 2.2], [3.3]], [], [[4.4, 5.5]]]))[0][0].device
>>/job:localhost/replica:0/task:0/device:GPU:0

Should I make the function use a TensorFlow policy and automatically select a device or create some kind of workaround?

[jpivarski]

ak.to_raggedtensor should return a RaggedTensor on the same device as the Awkward Array, as a view (no copy) if possible. That may mean that the implementation needs to specify non-default arguments of the RaggedTensor constructor (or use the with block) in order to control it.

If this is not possible and TensorFlow returns an object whose backend depends on what hardware is available (a terrible practice! shame on TensorFlow!), then we'll have to explain that (apologetically) in our documentation.