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I'm computing and optimization of some Variables that are used on an external process, but I get the error "No Gradient".

A heavily simplified (not tested) version of the code, but you can get the idea:

def external_process (myvar):
    subprocess.call("process.sh", myvar)
    with open('result.json', 'r') as f:
        result = json.load(data, f)
    return np.array(result["result"])

myvar = tf.Variable(1.0, dtype = 'float32', trainable = True)

loss = tf.reduce_sum( tf.py_func(external_process, [myvar], [tf.float32])[0] )

optimizer = tf.train.AdamOptimizer(0.05)
train_step = optimizer.minimize(loss)
sess.run(train_step)

I saw this discussion but I don't fully understand it: https://github.com/tensorflow/tensorflow/issues/1095

Thanks!

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Here https://www.tensorflow.org/versions/r0.9/api_docs/python/framework.html (search gradient_override_map) is an example on gradient_override_map:

@tf.RegisterGradient("CustomSquare")
def _custom_square_grad(op, grad):
  # ...

with tf.Graph().as_default() as g:
  c = tf.constant(5.0)
  s_1 = tf.square(c)  # Uses the default gradient for tf.square.
  with g.gradient_override_map({"Square": "CustomSquare"}):
    s_2 = tf.square(s_2)  # Uses _custom_square_grad to compute the
                          # gradient of s_2.

So, a possible solution could be:

@tf.RegisterGradient("ExternalGradient")
def _custom_external_grad(unused_op, grad):
    # I don't know yet how to compute a gradient
    # From Tensorflow documentation:
    return grad, tf.neg(grad)

def external_process (myvar):
    subprocess.call("process.sh", myvar)
    with open('result.json', 'r') as f:
        result = json.load(data, f)
    return np.array(result["result"])

myvar = tf.Variable(1.0, dtype = 'float32', trainable = True)

g = tf.get_default_graph()
with g.gradient_override_map({"PyFunc": "ExternalGradient"}):
    external_data =  tf.py_func(external_process, [myvar], [tf.float32])[0]

loss =  tf.reduce_sum(external_data)

optimizer = tf.train.AdamOptimizer(0.05)
train_step = optimizer.minimize(loss)
sess.run(train_step)
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  • $\begingroup$ Note that an alternative approach would be to wrap it using function.Defun instead of using gradient_override_map. $\endgroup$ – Albert Jun 8 '17 at 8:03
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The correct answer is: "An external process is not differentiable (unless you know each detail, what is impossible in this case), so this problem should be faced as a Reinforcement Learning problem"

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