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I want to transform this dataset in such a way that each tensor has a given size n and that a feature at index i of this new tensor is set to 1 if and only if there is a i in the original feature (modulo n).

I hope the following example will make things clearer

Let's suppose I have a dataset like:

t = tf.constant([

  [0, 3, 4],

  [12, 2 ,4]])

ds = tf.data.Dataset.from_tensors(t)

I want to get the sparse equivalent of (if n = 9)

t = tf.constant([

  [1, 0, 0, 1, 1, 0, 0, 0, 0], # index set to 1 are 0, 3 and 4

  [0, 0, 1, 1, 1, 0, 0, 0, 0]]) # index set to 1 are 2, 4, and 12%9 = 3

I already know how to obtain a not sparse representation (Tensorflow: tensor binarization) and as I will end up with n > 1 million, I do not want to pass by the dense tensor to get the sparse one

thanks

Here is a possible implementation for that:

import tensorflow as tf



def binarization_sparse(t, n):

    # Input size

    t_shape = tf.shape(t)

    t_rows = t_shape[0]

    t_cols = t_shape[1]

    # Make sparse row indices for each value

    row_idx = tf.tile(tf.range(t_rows)[: ,tf.newaxis], [1, t_cols])

    # Sparse column indices

    col_idx = t % n

    # "Flat" indices - needed to discard repetitions

    total_idx = row_idx * n + col_idx

    # Remove repeated elements

    out_idx, _ = tf.unique(tf.reshape(total_idx, [-1]))

    # Back to row and column indices

    sparse_idx = tf.stack([out_idx // n, out_idx % n], axis=-1)

    # Sparse values

    sparse_values = tf.ones([tf.shape(sparse_idx)[0]], dtype=t.dtype)

    # Make sparse tensor

    out = tf.sparse.SparseTensor(tf.cast(sparse_idx, tf.int64),

                                 sparse_values,

                                 [t_rows, n])

    # Reorder indices

    out = tf.sparse.reorder(out)

    return out



# Test

with tf.Graph().as_default(), tf.Session() as sess:

    t = tf.constant([

        [ 0,  3,  4],

        [12,  2,  4]

    ])

    # Sparse result

    t_m1h_sp = binarization_sparse(t, 9)

    # Convert to dense to check output

    t_m1h = tf.sparse.to_dense(t_m1h_sp)

    print(sess.run(t_m1h))

Output:

[[1 0 0 1 1 0 0 0 0]

 [0 0 1 1 1 0 0 0 0]]

I added the logic to remove repeated elements because in principle it could happen, but if you have a guarantee that there are no repetitions (including modulo), you may skip that step. Also, I reorder the sparse tensor at the end. That is not strictly necessary here, but (I think) sparse operations sometimes expect the indices to be ordered (and sparse_idx may not be ordered).

Also, this solution is specific to 2D inputs. For 1D inputs would be simpler, and it can be written for higher-dimensional inputs as well if necessary. I think a completely general solution is possible but it would be more complicated (specially if you want to consider tensors with unknown number of dimensions).