0
$\begingroup$

I have a general question about supervised ANNs that map inputs to outputs. It is possible to vary the length of the input and output vectors by inserting some dummy variables that will not be considered in the mapping (or will be mapped to other dummy variables). So basically the mapping should look like this (v: value, d: dummy)

Input vector 1 $[v,v,v,v,v] \rightarrow$ Output vector 1 $[v,v,v,v,v]$

Input vector 2 $[v,v,v,v,v]\rightarrow$ Output vector 2 $[v,v,v,v,v]$

Input vector 3 $[v,v,v,d,d] \rightarrow$ Output vector 3 $[v,v,v,d,d]$

Input vector 4 $[v,v,d,d,d] \rightarrow$ Output vector 4 $[v,v,d,d,d]$

Input vector 5 $[v,d,d,d,d] \rightarrow$ Output vector 5 $[v,d,d,d,d]$

The input and output vectors have a length of 5 with 5 values. However, sometimes only a vector of size e.g. 3 (which is basically a vector of length 5 with 2 dummy variables) should be mapped to an output vector of length 3. So after training the ANN should know that if it for example gets an input vector of length 3 it should produce an output vector of length 3.

Is something like this generally possible with ANNs or other machine learning approaches? If so, what type of ANN or machine learning approach can be used for this? I'll appreciate every comment.

Reminder: Can anybody give me more insights into this?

$\endgroup$
15
  • $\begingroup$ What is the difference between first two pairs of vectors? $\endgroup$
    – hanugm
    Jul 29, 2021 at 10:58
  • $\begingroup$ Thanks hanugm for your answer. Basically there is no difference between the first two pair of vectors. I just wanted to show that we have multiple training data with varying length. The first two pair of vectors turn out to have the same size (same number of v) in this example. $\endgroup$
    – PeterBe
    Jul 29, 2021 at 12:17
  • $\begingroup$ Im saying that first I/O vectors and second I/O vectors are same. $\endgroup$
    – hanugm
    Jul 29, 2021 at 12:18
  • $\begingroup$ Well this is just an example and the v can stand for any number. The first two pair of vectors can but do not have to be identical (most probably they are not). I just have let's say 1000 input and output vectors and an ANN should be trained to map them correctly. The v stand for any number. So most of the input vectors (even if they have the same size) are not identical. But my question is trageting at having input and output vectors with different lengths $\endgroup$
    – PeterBe
    Jul 29, 2021 at 12:22
  • $\begingroup$ Is the output always going to have the same length as the input? In that case you can just simply ignore the part of the output that is past the length of the input. $\endgroup$
    – Taw
    Aug 17, 2021 at 19:03

1 Answer 1

1
$\begingroup$

This should be possible but I've never seen it done in practice. Whether or not this will even actually work is unclear to me and will be highly dependent both on your training data and choice of loss. I'd take a step back and look into the literature to see if you can't find a more established approach to your problem, perhaps with RNNs. That being said, I believe the following should do what you're asking.

Consider network $N$ to be a dense neural net with $k$ layers, $N_i$ to be the $i$th layer of $N$, $L$ to be the max length of the input, and $V$ to be the number of terms in the input (the number of $v$s). To accomplish what you want in the above scenario, you can add three additional layers to $N$, $N_{k+1}$ $N_{k+2}$, and $N_{k+3}$:

$N_{k+1}$ is a simple dense layer that has $L$ neurons and takes as input the output of $N_k$. This layer can be skipped if layer $N_k$ already has $L$ neurons.

$N_{k+2}$ takes as input the output $N_{k+1}$ and takes the Hadamard product (elementwise multiplication) of it with a second input, a binary vector of length $L$ with a prefix of $V$ $1$s and suffix of $L-V$ $0$s. For example if $L=5$ and $V=2$, you would supply as the second input the vector $[1, 1, 0, 0, 0]$, which effectively "zeroes out" the third, fourth, and fifth positions.

$N_{k+3}$ is your new output layer, which also has $L$ neurons. $N$ can now be trained and, given the target data is in the proper format to achieve this, it should output results like in your question.

$\endgroup$
1
  • $\begingroup$ Thanks Oso for your answer. I understand the concept behind it but I would assume that putting the target data in the proper format might be quite challenging. Nontheless I will consider using your approach. Thanks for your help. I upvoted and accepted your answer. $\endgroup$
    – PeterBe
    Sep 21, 2021 at 11:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .