Tensorflow: tensor binarization

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 (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 know how to apply the modulo to a tensor, but I don't find how to do the rest of the transformation
thanks

asked Nov 21 '18 at 14:37

taktak004

410524

add a comment |

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 (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 know how to apply the modulo to a tensor, but I don't find how to do the rest of the transformation
thanks

asked Nov 21 '18 at 14:37

taktak004

410524

add a comment |

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 (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 know how to apply the modulo to a tensor, but I don't find how to do the rest of the transformation
thanks

asked Nov 21 '18 at 14:37

taktak004

410524

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 (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 know how to apply the modulo to a tensor, but I don't find how to do the rest of the transformation
thanks

python tensorflow

asked Nov 21 '18 at 14:37

taktak004

410524

asked Nov 21 '18 at 14:37

taktak004

410524

asked Nov 21 '18 at 14:37

taktak004

410524

asked Nov 21 '18 at 14:37

taktak004

410524

asked Nov 21 '18 at 14:37

taktak004

410524

add a comment |

1 Answer
1

active

oldest

votes

That is similar to tf.one_hot, only for multiple values at the same time. Here is a way to do that:

import tensorflow as tf



def binarization(t, n):

    # One-hot encoding of each value

    t_1h = tf.one_hot(t % n, n, dtype=tf.bool, on_value=True, off_value=False)

    # Reduce across last dimension of the original tensor

    return tf.cast(tf.reduce_any(t_1h, axis=-2), t.dtype)



# Test

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

    t = tf.constant([

        [ 0,  3,  4],

        [12,  2,  4]

    ])

    t_m1h = binarization(t, 9)

    print(sess.run(t_m1h))

Output:

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

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

edited Nov 21 '18 at 15:08

answered Nov 21 '18 at 14:46

jdehesa

24.1k43554

Thanks, can you explain the axis=-2. I understand what you are doing, but I don't understand how you come up with -2.

– taktak004
Nov 21 '18 at 15:05

1

@taktak004 Right, so after you do tf.one_hot a new dimension is added at the end. However, you want to reduce across the last dimension of the input tensor (e.g. the "columns" in your example, 0, 3, 4 and 12, 2, 4. This dimension will not be the last one in t_1h (because that is the new one-hot dimension), but the previous-to-last one, which can be referred to with the -2. I hope that makes it clearer?

– jdehesa
Nov 21 '18 at 15:09

ok, and by puting -2 instead of 1, you make it more generic

– taktak004
Nov 21 '18 at 15:13

1

@taktak004 Yes exactly the idea was that it worked either with a vector or a matrix or whatever (well actually it would fail for a scalar I think). You could also add explicit parameters for the axis that is reduced and/or the position of the new axis, but that seemed overcomplicated.

– jdehesa
Nov 21 '18 at 15:18

1

I created a new question as I think the approch will be totally different actually :stackoverflow.com/questions/53450218/…

– taktak004
Nov 23 '18 at 16:35

|
show 1 more comment

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

That is similar to tf.one_hot, only for multiple values at the same time. Here is a way to do that:

import tensorflow as tf



def binarization(t, n):

    # One-hot encoding of each value

    t_1h = tf.one_hot(t % n, n, dtype=tf.bool, on_value=True, off_value=False)

    # Reduce across last dimension of the original tensor

    return tf.cast(tf.reduce_any(t_1h, axis=-2), t.dtype)



# Test

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

    t = tf.constant([

        [ 0,  3,  4],

        [12,  2,  4]

    ])

    t_m1h = binarization(t, 9)

    print(sess.run(t_m1h))

Output:

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

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

edited Nov 21 '18 at 15:08

answered Nov 21 '18 at 14:46

jdehesa

24.1k43554

Thanks, can you explain the axis=-2. I understand what you are doing, but I don't understand how you come up with -2.

– taktak004
Nov 21 '18 at 15:05

1

@taktak004 Right, so after you do tf.one_hot a new dimension is added at the end. However, you want to reduce across the last dimension of the input tensor (e.g. the "columns" in your example, 0, 3, 4 and 12, 2, 4. This dimension will not be the last one in t_1h (because that is the new one-hot dimension), but the previous-to-last one, which can be referred to with the -2. I hope that makes it clearer?

– jdehesa
Nov 21 '18 at 15:09

ok, and by puting -2 instead of 1, you make it more generic

– taktak004
Nov 21 '18 at 15:13

1

@taktak004 Yes exactly the idea was that it worked either with a vector or a matrix or whatever (well actually it would fail for a scalar I think). You could also add explicit parameters for the axis that is reduced and/or the position of the new axis, but that seemed overcomplicated.

– jdehesa
Nov 21 '18 at 15:18

1

I created a new question as I think the approch will be totally different actually :stackoverflow.com/questions/53450218/…

– taktak004
Nov 23 '18 at 16:35

|
show 1 more comment

That is similar to tf.one_hot, only for multiple values at the same time. Here is a way to do that:

import tensorflow as tf



def binarization(t, n):

    # One-hot encoding of each value

    t_1h = tf.one_hot(t % n, n, dtype=tf.bool, on_value=True, off_value=False)

    # Reduce across last dimension of the original tensor

    return tf.cast(tf.reduce_any(t_1h, axis=-2), t.dtype)



# Test

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

    t = tf.constant([

        [ 0,  3,  4],

        [12,  2,  4]

    ])

    t_m1h = binarization(t, 9)

    print(sess.run(t_m1h))

Output:

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

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

edited Nov 21 '18 at 15:08

answered Nov 21 '18 at 14:46

jdehesa

24.1k43554

Thanks, can you explain the axis=-2. I understand what you are doing, but I don't understand how you come up with -2.

– taktak004
Nov 21 '18 at 15:05

1

@taktak004 Right, so after you do tf.one_hot a new dimension is added at the end. However, you want to reduce across the last dimension of the input tensor (e.g. the "columns" in your example, 0, 3, 4 and 12, 2, 4. This dimension will not be the last one in t_1h (because that is the new one-hot dimension), but the previous-to-last one, which can be referred to with the -2. I hope that makes it clearer?

– jdehesa
Nov 21 '18 at 15:09

ok, and by puting -2 instead of 1, you make it more generic

– taktak004
Nov 21 '18 at 15:13

1

@taktak004 Yes exactly the idea was that it worked either with a vector or a matrix or whatever (well actually it would fail for a scalar I think). You could also add explicit parameters for the axis that is reduced and/or the position of the new axis, but that seemed overcomplicated.

– jdehesa
Nov 21 '18 at 15:18

1

I created a new question as I think the approch will be totally different actually :stackoverflow.com/questions/53450218/…

– taktak004
Nov 23 '18 at 16:35

|
show 1 more comment

That is similar to tf.one_hot, only for multiple values at the same time. Here is a way to do that:

import tensorflow as tf



def binarization(t, n):

    # One-hot encoding of each value

    t_1h = tf.one_hot(t % n, n, dtype=tf.bool, on_value=True, off_value=False)

    # Reduce across last dimension of the original tensor

    return tf.cast(tf.reduce_any(t_1h, axis=-2), t.dtype)



# Test

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

    t = tf.constant([

        [ 0,  3,  4],

        [12,  2,  4]

    ])

    t_m1h = binarization(t, 9)

    print(sess.run(t_m1h))

Output:

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

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

edited Nov 21 '18 at 15:08

answered Nov 21 '18 at 14:46

jdehesa

24.1k43554

That is similar to tf.one_hot, only for multiple values at the same time. Here is a way to do that:

import tensorflow as tf



def binarization(t, n):

    # One-hot encoding of each value

    t_1h = tf.one_hot(t % n, n, dtype=tf.bool, on_value=True, off_value=False)

    # Reduce across last dimension of the original tensor

    return tf.cast(tf.reduce_any(t_1h, axis=-2), t.dtype)



# Test

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

    t = tf.constant([

        [ 0,  3,  4],

        [12,  2,  4]

    ])

    t_m1h = binarization(t, 9)

    print(sess.run(t_m1h))

Output:

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

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

edited Nov 21 '18 at 15:08

answered Nov 21 '18 at 14:46

jdehesa

24.1k43554

edited Nov 21 '18 at 15:08

answered Nov 21 '18 at 14:46

jdehesa

24.1k43554

answered Nov 21 '18 at 14:46

jdehesa

24.1k43554

answered Nov 21 '18 at 14:46

jdehesa

24.1k43554

Thanks, can you explain the axis=-2. I understand what you are doing, but I don't understand how you come up with -2.

– taktak004
Nov 21 '18 at 15:05

1

@taktak004 Right, so after you do tf.one_hot a new dimension is added at the end. However, you want to reduce across the last dimension of the input tensor (e.g. the "columns" in your example, 0, 3, 4 and 12, 2, 4. This dimension will not be the last one in t_1h (because that is the new one-hot dimension), but the previous-to-last one, which can be referred to with the -2. I hope that makes it clearer?

– jdehesa
Nov 21 '18 at 15:09

ok, and by puting -2 instead of 1, you make it more generic

– taktak004
Nov 21 '18 at 15:13

1

@taktak004 Yes exactly the idea was that it worked either with a vector or a matrix or whatever (well actually it would fail for a scalar I think). You could also add explicit parameters for the axis that is reduced and/or the position of the new axis, but that seemed overcomplicated.

– jdehesa
Nov 21 '18 at 15:18

1

I created a new question as I think the approch will be totally different actually :stackoverflow.com/questions/53450218/…

– taktak004
Nov 23 '18 at 16:35

|
show 1 more comment

Thanks, can you explain the axis=-2. I understand what you are doing, but I don't understand how you come up with -2.

– taktak004
Nov 21 '18 at 15:05

1

@taktak004 Right, so after you do tf.one_hot a new dimension is added at the end. However, you want to reduce across the last dimension of the input tensor (e.g. the "columns" in your example, 0, 3, 4 and 12, 2, 4. This dimension will not be the last one in t_1h (because that is the new one-hot dimension), but the previous-to-last one, which can be referred to with the -2. I hope that makes it clearer?

– jdehesa
Nov 21 '18 at 15:09

ok, and by puting -2 instead of 1, you make it more generic

– taktak004
Nov 21 '18 at 15:13

1

@taktak004 Yes exactly the idea was that it worked either with a vector or a matrix or whatever (well actually it would fail for a scalar I think). You could also add explicit parameters for the axis that is reduced and/or the position of the new axis, but that seemed overcomplicated.

– jdehesa
Nov 21 '18 at 15:18

1

I created a new question as I think the approch will be totally different actually :stackoverflow.com/questions/53450218/…

– taktak004
Nov 23 '18 at 16:35

Thanks, can you explain the axis=-2. I understand what you are doing, but I don't understand how you come up with -2.

– taktak004
Nov 21 '18 at 15:05

@taktak004 Right, so after you do tf.one_hot a new dimension is added at the end. However, you want to reduce across the last dimension of the input tensor (e.g. the "columns" in your example, 0, 3, 4 and 12, 2, 4. This dimension will not be the last one in t_1h (because that is the new one-hot dimension), but the previous-to-last one, which can be referred to with the -2. I hope that makes it clearer?

– jdehesa
Nov 21 '18 at 15:09

ok, and by puting -2 instead of 1, you make it more generic

– taktak004
Nov 21 '18 at 15:13

@taktak004 Yes exactly the idea was that it worked either with a vector or a matrix or whatever (well actually it would fail for a scalar I think). You could also add explicit parameters for the axis that is reduced and/or the position of the new axis, but that seemed overcomplicated.

– jdehesa
Nov 21 '18 at 15:18

I created a new question as I think the approch will be totally different actually :stackoverflow.com/questions/53450218/…

– taktak004
Nov 23 '18 at 16:35

|
show 1 more comment

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