raises error in continuous entropy

[un-lock-me]

Hi,

Thanks for sharing you work. I want to use the continuous entropy of your project in mine.

I have a matrice like this:

x =  tf.Variable(   [   [0.96,    -0.65,    0.99,    -0.1   ],
                        [0.97,    0.33,    0.25  ,    0.05  ],
                        [0.9,     0.001,    0.009,    0.33  ],
                        [-0.60,   -0.1,    -0.3,     -0.5   ],
                        [0.49,    -0.8,     -0.05,   -0.0036],
                        [0.0  ,   -0.45,    0.087,    0.023 ],
                        [0.3,     -0.23,    0.82,    -0.28  ]])

When I apply the ee.entropy, I receive this error:

    rev = 1/ee.entropy(row)
  File "/home/sgnbx/Downloads/NPEET/npeet/entropy_estimators.py", line 21, in entropy
    assert k <= len(x) - 1, "Set k smaller than num. samples - 1"
TypeError: object of type 'Tensor' has no len() 

This is my code:

def rev_entropy(x):
    def row_entropy(row):
        rev = 1/ee.entropy(row)
        return rev
    rev= tf.map_fn(row_entropy, x, dtype=tf.float32)
    return rev

x =  tf.Variable(   [   [0.96,    -0.65,    0.99,    -0.1   ],
                        [0.97,    0.33,    0.25  ,    0.05  ],
                        [0.9,     0.001,    0.009,    0.33  ],
                        [-0.60,   -0.1,    -0.3,     -0.5   ],
                        [0.49,    -0.8,     -0.05,   -0.0036],
                        [0.0  ,   -0.45,    0.087,    0.023 ],
                        [0.3,     -0.23,    0.82,    -0.28  ]])

p = (x + tf.abs(x)) / 2
ent_p = rev_entropy(p)

Can you please explain how can I know the `k` here?
print(ent_p)

[gregversteeg]

Sorry to take so long responding. "k" is one of the arguments for ee.entropy. def entropy(x, k=3, base=2):

The default is 3, but you clearly have more than 3 samples. I'm guessing the issue has to do with using a tensorflow tensor. I don't think it will work, for a variety of reasons. The most fundamental one is that the k nearest neighbor library that I use in numpy probably won't work on tensors.

I'd be interested if you try x = np.array( same matrix ) whether it works.

[un-lock-me]

Thank you so much for getting back to me with this issue.

Actually I will be able to convert the code to tensorflow so that it can work on tensors.

The only issue I have is that, I need to calculate the entropy over each row, say I need to know what is the entropy of a single row in my tensor.

Taking this to account, and looking at most of the implementations for the KNN on continuous entropy, they consider the whole matrix.

Now I got stuck in as I dont know how to change the code to apply knn only one row by one row. In terms of the coding, I did it but it raises error as it compare each row with other rows(please correct me if Im wrong).

I appreciate it if you could share your idea regarding this with me.

Thanks~

[gregversteeg]

I'm not sure if it's relevant, but I have seen entropy estimators based on pairwise distances that can be implemented as differentiable expressions in tensorflow. For instance, Eq. 10 of this paper: https://arxiv.org/pdf/1705.02436.pdf. I think Artemy has a few papers discussing that "mixture of Gaussians" estimator for entropy and mutual information.