tensorflow about pycall.rb HOT 10 CLOSED

I'm planning to redefine PyObjectWrapper#method method to return a wrapper object of Python's function object so that we can write tf.keras.layers.Dense(512, activation=tf.nn.method(:relu)).

On ruby-core, people are discussing a syntax sugar for method reference, which is maybe introduced in the future Ruby. I think it's very good that this syntax sugar can be used for this case.

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This is an awesome project. Please continue to innovate here as I think it opens a world of opportunity to Ruby developers.

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@mightystrong I've implemented easy ways to obtain functional attributes in this pull-request #72

After merging it, you can write it like tf.keras.layers.Dense(512, activation: tf.nn[:relu]) because tf.nn is a module.
And if you want to get functional attributes from non-module objects, you can use PyCall.getattr method.

Do these resolve your issue?

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@sp00ck there is a dedicated project for that:
https://github.com/somaticio/tensorflow.rb

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@sp00ck I heared that some people tried to use Python's tensorflow via pycall.rb, and they've succeeded. As tensorflow is very changed day by day, it is unclear that Python's tensorflow can work on pycall.rb now. Please check it by yourself. And if you encounter any troubles, please report it as an issue.

As @arbox mentioned, you can use tensorflow.rb, too. It is just a binding of Tensorflow's C API. But you cannot use such features in tf.contrib.*, which are provided only in Python API, with tensorflow.rb. If you want to use them from Ruby, you need to use pycall.rb.

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I've been using tensorflow via PyCall but it has some issues. For example, if you simply try running the example on the tensorflow tutorials page (below) you first run into a lot of conversion challenges dealing with tuples. You also need to convert other lines such as tf.keras.layers.Dense(512, activation=tf.nn.relu) to tf.keras.layers.Dense(512, activation: tf.nn.relu).

While setting the model variable, PyCall throws an error saying <type 'exceptions.TypeError'>: relu() takes at least 1 argument (0 given). Running this code directly in Python, doesn't require the argument to be given, so it seems there is some work to do here. Maybe there's an answer of which I'm not aware.

import tensorflow as tf
mnist = tf.keras.datasets.mnist

(x_train, y_train),(x_test, y_test) = mnist.load_data()
x_train, x_test = x_train / 255.0, x_test / 255.0

model = tf.keras.models.Sequential([
  tf.keras.layers.Flatten(),
  tf.keras.layers.Dense(512, activation=tf.nn.relu),
  tf.keras.layers.Dropout(0.2),
  tf.keras.layers.Dense(10, activation=tf.nn.softmax)
])
model.compile(optimizer='adam',
              loss='sparse_categorical_crossentropy',
              metrics=['accuracy'])

model.fit(x_train, y_train, epochs=5)
model.evaluate(x_test, y_test)

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@mightystrong Thank you for using PyCall, and reporting back your feeling of use.

This issue comes from the difference that Ruby is lisp-2 but Python is lisp-1.

Now you can get function object by using getattr method defined in LibPython::Helpers module, but I understand this is too much uncomfortable.

I’ve decided to introduce other easy way to get function attributes.

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There is another idea I got from @znz, that is letting obj.name! returns a function reference.
It is possible because Pytohn doesn't permit ! in a method name.

But, I currently don't prefer this idea...

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@mrkn How would you call tf.keras.layers.Dense(512, activation: tf.nn.relu) instead in the current version of PyCall so that it doesn't throw the argument required error?

Now you can get function object by using getattr method defined in LibPython::Helpers module, but I understand this is too much uncomfortable.

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