How to create an LSTM with no input in Tensorflow?










0















I am implementing an attention mechanism which uses an LSTM cell as encoder where the LSTM does not take/have any input (only a hidden state). How would one specify an LSTM with no input in TensorFlow?



Essentially, the hidden state would get concatenated with an empty input at the "entrypoint" of the LSTM cell.



Thanks!



Edit: related question I have asked at Attention mechanism with an LSTM which takes no input










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  • 1





    If you feed a tf.constant(0) as an input - would it be the same as "no input"?

    – Vlad-HC
    Nov 13 '18 at 14:59











  • Thanks for the reply. I thought of feeding the LSTM a tensor of zeros at each time step. However, that would create extra parameters within the LSTM? Although I guess they would not really count as such since they would not get updated during backprop? I am making an edit to my original question and link another question I have asked, for more details.

    – Tom
    Nov 13 '18 at 16:32












  • This input won't create extra parameters because the architecture of the LSTM cell will stay the same. The parameters just won't be updated during backpropagation because the dLoss/dLstmInput for these weights would be 0.

    – Vlad-HC
    Nov 13 '18 at 17:15











  • Right, so it would create parameters but they would not take any degree of freedom? Or can you just specify that the LSTM weights corresponding to the input are not trainable. Thanks!

    – Tom
    Nov 13 '18 at 17:38















0















I am implementing an attention mechanism which uses an LSTM cell as encoder where the LSTM does not take/have any input (only a hidden state). How would one specify an LSTM with no input in TensorFlow?



Essentially, the hidden state would get concatenated with an empty input at the "entrypoint" of the LSTM cell.



Thanks!



Edit: related question I have asked at Attention mechanism with an LSTM which takes no input










share|improve this question



















  • 1





    If you feed a tf.constant(0) as an input - would it be the same as "no input"?

    – Vlad-HC
    Nov 13 '18 at 14:59











  • Thanks for the reply. I thought of feeding the LSTM a tensor of zeros at each time step. However, that would create extra parameters within the LSTM? Although I guess they would not really count as such since they would not get updated during backprop? I am making an edit to my original question and link another question I have asked, for more details.

    – Tom
    Nov 13 '18 at 16:32












  • This input won't create extra parameters because the architecture of the LSTM cell will stay the same. The parameters just won't be updated during backpropagation because the dLoss/dLstmInput for these weights would be 0.

    – Vlad-HC
    Nov 13 '18 at 17:15











  • Right, so it would create parameters but they would not take any degree of freedom? Or can you just specify that the LSTM weights corresponding to the input are not trainable. Thanks!

    – Tom
    Nov 13 '18 at 17:38













0












0








0








I am implementing an attention mechanism which uses an LSTM cell as encoder where the LSTM does not take/have any input (only a hidden state). How would one specify an LSTM with no input in TensorFlow?



Essentially, the hidden state would get concatenated with an empty input at the "entrypoint" of the LSTM cell.



Thanks!



Edit: related question I have asked at Attention mechanism with an LSTM which takes no input










share|improve this question
















I am implementing an attention mechanism which uses an LSTM cell as encoder where the LSTM does not take/have any input (only a hidden state). How would one specify an LSTM with no input in TensorFlow?



Essentially, the hidden state would get concatenated with an empty input at the "entrypoint" of the LSTM cell.



Thanks!



Edit: related question I have asked at Attention mechanism with an LSTM which takes no input







tensorflow lstm attention-model






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 13 '18 at 16:34







Tom

















asked Nov 13 '18 at 13:16









TomTom

1517




1517







  • 1





    If you feed a tf.constant(0) as an input - would it be the same as "no input"?

    – Vlad-HC
    Nov 13 '18 at 14:59











  • Thanks for the reply. I thought of feeding the LSTM a tensor of zeros at each time step. However, that would create extra parameters within the LSTM? Although I guess they would not really count as such since they would not get updated during backprop? I am making an edit to my original question and link another question I have asked, for more details.

    – Tom
    Nov 13 '18 at 16:32












  • This input won't create extra parameters because the architecture of the LSTM cell will stay the same. The parameters just won't be updated during backpropagation because the dLoss/dLstmInput for these weights would be 0.

    – Vlad-HC
    Nov 13 '18 at 17:15











  • Right, so it would create parameters but they would not take any degree of freedom? Or can you just specify that the LSTM weights corresponding to the input are not trainable. Thanks!

    – Tom
    Nov 13 '18 at 17:38












  • 1





    If you feed a tf.constant(0) as an input - would it be the same as "no input"?

    – Vlad-HC
    Nov 13 '18 at 14:59











  • Thanks for the reply. I thought of feeding the LSTM a tensor of zeros at each time step. However, that would create extra parameters within the LSTM? Although I guess they would not really count as such since they would not get updated during backprop? I am making an edit to my original question and link another question I have asked, for more details.

    – Tom
    Nov 13 '18 at 16:32












  • This input won't create extra parameters because the architecture of the LSTM cell will stay the same. The parameters just won't be updated during backpropagation because the dLoss/dLstmInput for these weights would be 0.

    – Vlad-HC
    Nov 13 '18 at 17:15











  • Right, so it would create parameters but they would not take any degree of freedom? Or can you just specify that the LSTM weights corresponding to the input are not trainable. Thanks!

    – Tom
    Nov 13 '18 at 17:38







1




1





If you feed a tf.constant(0) as an input - would it be the same as "no input"?

– Vlad-HC
Nov 13 '18 at 14:59





If you feed a tf.constant(0) as an input - would it be the same as "no input"?

– Vlad-HC
Nov 13 '18 at 14:59













Thanks for the reply. I thought of feeding the LSTM a tensor of zeros at each time step. However, that would create extra parameters within the LSTM? Although I guess they would not really count as such since they would not get updated during backprop? I am making an edit to my original question and link another question I have asked, for more details.

– Tom
Nov 13 '18 at 16:32






Thanks for the reply. I thought of feeding the LSTM a tensor of zeros at each time step. However, that would create extra parameters within the LSTM? Although I guess they would not really count as such since they would not get updated during backprop? I am making an edit to my original question and link another question I have asked, for more details.

– Tom
Nov 13 '18 at 16:32














This input won't create extra parameters because the architecture of the LSTM cell will stay the same. The parameters just won't be updated during backpropagation because the dLoss/dLstmInput for these weights would be 0.

– Vlad-HC
Nov 13 '18 at 17:15





This input won't create extra parameters because the architecture of the LSTM cell will stay the same. The parameters just won't be updated during backpropagation because the dLoss/dLstmInput for these weights would be 0.

– Vlad-HC
Nov 13 '18 at 17:15













Right, so it would create parameters but they would not take any degree of freedom? Or can you just specify that the LSTM weights corresponding to the input are not trainable. Thanks!

– Tom
Nov 13 '18 at 17:38





Right, so it would create parameters but they would not take any degree of freedom? Or can you just specify that the LSTM weights corresponding to the input are not trainable. Thanks!

– Tom
Nov 13 '18 at 17:38












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