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How do we decide the number of hiddenlayers in a PatternNet?

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Well, since I'm working on PatternNets (Classification), my doubt is, how do we decide then number of hiddenlayers the network requires.
Before that, I have 8 predictors and 1 output response containing 4 classes.
I know that in order to define a hidden layer, we use :
MyNet = patternnet(10);
But how do I decide I need 10 and not 8 or any other number?
Any suggestion would be helpful. Thank you.

Accepted Answer

Greg Heath
Greg Heath on 7 Aug 2019
Edited: Greg Heath on 7 Aug 2019
  1. patternnet(10) indicates ONE HIDDEN LAYER WITH TEN NODES
  2. It is important to be mindful of the number of layers and nodes.
  3. The set of input nodes IS NOT CONSIDERED A LAYER ( Probably because no mathematical operations are performed there)
  4. Three sets of nodes (INPUT/HIDDEN/OUTPUT) is sufficient. Let
size(inputmatrix) = [ I N ]
size(targetmatrix) = [ O N ]
Then with a single middle "HIDDEN" layer of size H,
The size of the typical two layer net is
size(net) = I - H - O
a. I input nodes
b. H hidden layer nodes
c. O output layer nodes
4. The total number of weights is
Nw = ( I + 1 ) * H + ( H + 1 ) * O
= O + ( I + O + 1 ) * H
where the "1s" represent constant bias connections
5. The data is typically divided into a training subset of size Ntrn ~ 0.7*N, a validation subset of size Nval ~ 0.15*N and a test subset of size Ntst ~ 0.15*N.
6. For good approximate solutions, the number of training equations Ntrneq = Ntrn*O should be sufficiently larger than the number of unknown weights Nw
7. The inequality Nw << Ntrneq results in
H << ( 0.7 * N * O - O )/(I+O+1)
8. A factor of 10 is usually sufficient
Hope this helps.
  1 Comment
abhisrisai on 8 Aug 2019
Hello Greg,
Thank you very much for the answer, it was helpful. Just another doubt, what happens to the prediction performance when we have abundant H nodes. In my case, I have 8 predictors and I feel 10 H nodes are sufficient. What if I up the number to 40 H nodes? Does it affect my prediction and how?
Thank you.

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