Where in documentation does it say that fitnet([]) gives a linear model?

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KAE 2018-10-4

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https://ww2.mathworks.cn/matlabcentral/answers/422199-where-in-documentation-does-it-say-that-fitnet-gives-a-linear-model

编辑： KAE 2018-10-9

I have seen Greg Heath's posts such as this one in which fitnet([]) is used to fit a linear model, i.e. with no hidden layers. But I can't find a reference in the fitnet documentation that this is the expected result of [] as an input, and I find this hard to spot in the fitnet.m code. Can someone please point me to a description in the documentation of the expected behavior of fitnet([])? I just want to understand this better.

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Answer 1

Greg Heath 2018-10-4

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fitnet(H) yields H nodes in the hidden layer

A linear function needs no hidden layer

I could only think of 2 possibilities

 a. net1 = fitnet(0), view(net1)
 b. net2 = fitnet([]), view(net2)

So, as I did, try them out and see what happens.

Not sure if I checked the documentation.

Since I don't have MATLAB on this computer I can't check.

Also can't remember what happens when you try fitnet(0).

Hope this helps.

Greg

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KAE 2018-10-8

编辑：KAE 2018-10-9

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Here is some code to show that the output fitnet([]) looks like a linear model, for other learners.

slopeList = [+2 +7 -3 -5]; % Slope values that we will use in different cases
offsetList = [2 -1 5 -3]; % Offset values
x = rand(1, 1000); % Independant variable
yNoise = rand(1, 1000) - 0.5; % Add some zero-mean noise
net = fitnet([]); % Fits a linear model
for iCase = 1:length(slopeList)
  % Target is the line plus some noise
  t = x*slopeList(iCase) + offsetList(iCase) + yNoise; 
  net = configure(net, x, t);
  % Train the network on x and t, and return the prediction y
  [net, tr, y] = train(net,x,t); 
    % Try to undo the automatic normalization to get the 
    % unscaled weight and bias (doesn't work right)
    weightBiasUnscaled = ...
        mapminmax('reverse', [net.IW{1} net.b{1}],...
        net.outputs{1}.processSettings{1});
    weightUnscaled = weightBiasUnscaled(1);
    biasUnscaled = weightBiasUnscaled(2); 
    % Show results
    figure;
    h1 = plot(x, t, 'g.'); % Data
    hold on;
    % Prediction from net
    h2 = plot(x, y, '.', 'color', [1 1 1]*0.5, 'markersize', 10); 
    x0 = get(gca, 'xlim');
    h3 = plot(x0, x0*slopeList(iCase) + offsetList(iCase), 'r'); % Real relation
    box off
    % Explain what was calculated
    modelStr = ['Model: y'' = ' sprintf('%.2f', weightUnscaled) ...
      '''*x + ' sprintf('%.1f', biasUnscaled)];
    realStr = ['Real Relation: y = ' num2str(slopeList(iCase)) ...
      '*x + ' num2str(offsetList(iCase))];
    legend('Noisy data', modelStr, realStr);
    title('Linear Model Check');
end

KAE 2018-10-9

编辑：KAE 2018-10-9

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As a side note, I am not sure how to reverse the scaling to regain the unscaled weight and bias values for the linear model y=mx+b. Here is what I tried,

    weightBiasUnscaled = ...
        mapminmax('reverse', [net.IW{1} net.b{1}],...
        net.outputs{1}.processSettings{1});
    % Unscaled weight
    weightUnscaled = weightBiasUnscaled(1);
    % Unscaled bias
    biasUnscaled = weightBiasUnscaled(2);

However this gives a slope (weight) of -8.07 instead of the expected value of m=-5, and a bias (offset) of -5.5 not the expected value of b=-3 in y=mx+b. What am I doing wrong?

KAE 2018-10-9

编辑：KAE 2018-10-9

MathWorks has confirmed that fitnet([]) produces a linear model like linear regression. They pointed me to this helpful post for background. I have asked them to add info on fitnet([]) to the documentation. Thanks all for your help!

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Answer 2

the cyclist 2018-10-4

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https://ww2.mathworks.cn/matlabcentral/answers/422199-where-in-documentation-does-it-say-that-fitnet-gives-a-linear-model#answer_339760

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In the Input Arguments section of the documentation, where it discusses the hiddenSizes input, there is the sentence, "The length of the vector determines the number of hidden layers in the network."

Since the length of the empty vector is zero:

>> length([])
ans =
     0

one can infer, I suppose, that there will be zero hidden layers. This could certainly be made more explicit, though. (I don't think I would have thought to try, had it not been for your question.)

I don't have access to the code, but I'm guessing that armed with the above knowledge, you can find where MATLAB determines the number of hidden layers from the length of that input.