Using R^2 results to optimize number of neurons in hidden layer

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KAE 2018-9-4

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评论： KAE 2018-9-26

I am trying to find the optimal number of neurons in a hidden layer following Greg Heath's method of looping over the candidate number of neurons, with an several trials per number of neurons. The resulting R-squared statistic is plotted below. I would like some help in using the R-squared results to pick the optimal number of neurons. Is this right:

23 neurons is a good choice, since all the trials exceed the desired threshold of R-squared > 0.995. To make a prediction, I could pick any of the 10 trial nets that were generated with 23 neurons.
15 neurons is a bad choice because sometimes the threshold is not met
More than 23 neurons is a bad choice because the network will be slower to run
More than 33 neurons is a very bad choice due to overfitting (the code below limits the number of neurons to avoid overfitting).
To save time, I could stop my loop after say 28 neurons, since by then I would have had 5 instances (23-28 neurons) in which all 10 trials resulted in R-squared values above my threshold.

Here is the code, based on Greg's examples with comments for learning.

%%Get data
[x, t] = engine_dataset; % Existing dataset
% Described here, https://www.mathworks.com/help/deeplearning/ug/choose-a-multilayer-neural-network-training-function.html#bss4gz0-26
[ I N ] = size(x); % Size of network inputs, nVar x nExamples
[ O N ] = size(t); % Size of network targets, nVar x nExamples
fractionTrain = 75/100; % How much of data will go to training
fractionValid = 15/100; % How much of data will go to validation
fractionTest = 15/100; % How much of data will go to testing
Ntrn = N - round(fractionTest*N + fractionValid*N); % Number of training examples
Ntrneq = Ntrn*O; % Number of training equations
% MSE for a naïve constant output model that always outputs average of target data
MSE00 = mean(var(t',1));
%%Find a good value for number of neurons H in hidden layer
Hmin = 0; % Lowest number of neurons H to consider (a linear model)
% To avoid overfitting with too many neurons, require that Ntrneq > Nw==> H <= Hub (upper bound)
Hub = -1+ceil( (Ntrneq-O) / ( I+O+1) ); % Upper bound for Ntrneq >= Nw
Hmax = floor(Hub/10); % Stay well below the upper bound by dividing by 10
dH = 1;
% Number of random initial weight trials for each candidate h
Ntrials = 10;
% Randomize data division and initial weights
% Choose a repeatable initial random number seed 
% so that you can reproduce any individual design
rng(0);
j=0; % Loop counter for number of neurons in hidden layer
for h=Hmin:dH:Hmax % Number of neurons in hidden layer
    j = j+1;
    if h == 0 % Linear model
        net = fitnet([]); % Fits a linear model
        Nw = (I+1)*O; % Number of unknown weights
    else % Nonlinear neural network model
        net = fitnet(h); % Fits a network with h nodes
        Nw = (I+1)*h + (h+1)*O; % Number of unknown weights
    end
      %%Divide up data into training, validation, testing sets
      % Data presented to network during training. Network is adjusted according to its error.
      net.divideParam.trainRatio = fractionTrain;
      % Data used to measure network generalization, and to halt training when generalization stops improving
      net.divideParam.valRatio = fractionValid;
      % Data used as independent measure of network performance during and after training
      net.divideParam.testRatio = fractionTest;
      %%Loop through several trials for each candidate number of neurons
      % to increase statistical reliability of results
      for i=1:Ntrials
          % Configure network inputs and outputs to best match input and target data
          net = configure(net, x, t);
          % Train network
          [net, tr, y, e] = train(net,x,t); % Error e is target - output
          % R-squared for normalized MSE. Fraction of target variance modeled by net
          R2(i, j) = 1 - mse(e)/MSE00;
      end
  end
%%Plot R-squared
figure;
% Grid data in preparation for plotting
[nNeuronMat, trialMat] = meshgrid(Hmin:dH:Hmax, 1:Ntrials);
% Color of dot indicates the R-squared value
h1 = scatter(nNeuronMat(:), trialMat(:), 20, R2(:), 'filled'); 
hold on;
% Design is successful if it can account for 99.5% of mean target variance
iGood = R2 > 0.995; 
% Circle the successful values
h2 = plot(nNeuronMat(iGood), trialMat(iGood), 'ko', 'markersize', 10);
xlabel('# Neurons'); ylabel('Trial #'); 
h3 = colorbar;
set(get(h3,'Title'),'String','R^2')
title({'Fraction of target variance that is modeled by net', ...
    'Circled if it exceeds 99.5%'});

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Greg Heath 2018-9-25

在 MATLAB Online 中打开

Interesting post. I usually look for the minimum acceptable H; which in this case is H=12.

What I would like to add is that before I search over number of hidden nodes to minimize H, I usually run through with a single reasonable value. Typically, it would be the MATLAB default H = 10.

However, sometimes when Hub exceeds 100 because N is huge, I use a larger value. For example:

 close all, clear all, clc
[x,t] = engine_dataset;
[I N ] = size(x)     % [2 1199]
[O N ] = size(t)     % [2 1199]
vart1  = var(t',1)   % 1e5*[3.0079 2.1709]
MSEref = mean(vart1) % 2.5894e+05
 x1 = x(1,:); x2 = x(2,:);
 t1 = t(1,:); t2 = t(2,:);
figure(1)
subplot(2,2,1), plot(x1,t1,'.')
subplot(2,2,2), plot(x1,t2,'.')
subplot(2,2,3), plot(x2,t1,'.')
subplot(2,2,4), plot(x2,t2,'.')
 Ntrn   = N-round(0.3*N) % 839
 Ntrneq = Ntrn*O         % 1678
% Nw = (I+1)*H+(H+1)*O
% Ntrneq >= Nw ==> H <= Hmax <= Hub
Hub = floor((Ntrneq-O)/(I+O+1)) % 335
H   = round(Hub/10)             % 34
net = fitnet(H);
[net tr y e ] = train(net,x,t);
NMSE = mse(e)/MSEref  % 0.0025
Rsq  = 1-NMSE         % 0.9975

Hope this helps

Greg

KAE 2018-9-25

编辑：KAE 2018-9-25

在 MATLAB Online 中打开

You say you typically run through a reasonable value, here H=34, before seeking to minimize H. What do the 'reasonable value' results tell you? Maybe a good choice of a R-square threshold to use later in the H minimization, so for your example we might require the smallest H with Rsq>= 0.9975?

Below for other learners is Greg's code with comments added, as I figured out what it was doing.

%%Load data
[x,t] = engine_dataset; 
% Data description, https://www.mathworks.com/help/deeplearning/ug/choose-a-multilayer-neural-network-training-function.html#bss4gz0-26
help engine_dataset % Learn about dataset
[I N ] = size(x)     % [2 1199] Size of network inputs, nVar x nExamples
% Inputs are engine speed and fuel rate
[O N ] = size(t)     % [2 1199] Size of network targets, nVar x nExamples
% Ouputs are torque and emission levels
% Separate out inputs and outputs for later plotting
x1 = x(1,:); x2 = x(2,:);
t1 = t(1,:); t2 = t(2,:);
%%Mean square error for later normalization
vart1  = var(t',1)   % 1e5*[3.0079 2.1709]
% MSE for a naïve constant output model that always outputs average of target data
MSEref = mean(vart1) % 2.5894e+05
%%Look for obvious relations between inputs and outputs
figure(1)
subplot(2,2,1), plot(x1,t1,'.') % Looks linear
title('Torque vs. Engine Speed'); xlabel('Input 1'); ylabel('Target 1'); 
subplot(2,2,2), plot(x1,t2,'.') % Looks like a cubic 
title('Nitrous Oxide Emission vs. Engine Speed'); xlabel('Input 1'); ylabel('Target 2'); 
subplot(2,2,3), plot(x2,t1,'.') % On/off behavior, maybe cubic 
title('Torque vs. Fuel Rate'); xlabel('Input 2'); ylabel('Target 1'); 
subplot(2,2,4), plot(x2,t2,'.') % On/off behavior
title('Nitrous Oxide Emission vs. Fuel Rate'); xlabel('Input 2'); ylabel('Target 2');
Ntrn   = N-round(0.3*N) % 839 Number of training examples
Ntrneq = Ntrn*O         % 1678 Number of training equations
% Nw = (I+1)*H+(H+1)*O
% Ntrneq >= Nw ==> H <= Hmax <= Hub
Hub = floor((Ntrneq-O)/(I+O+1)) % 335 Upper bound for Ntrneq >= Nw
H   = round(Hub/10)             % 34 Stay well below the upper bound by dividing by 10
net = fitnet(H); % Fit a network with h nodes
[net tr y e ] = train(net,x,t); % Train network
NMSE = mse(e)/MSEref  % 0.0025 normalized MSE
Rsq  = 1-NMSE         % 0.9975 R-squared. Fraction of target variance modeled by net

Greg Heath 2018-9-26

在 MATLAB Online 中打开

 KAE: You say you typically run through a reasonable 
value, here H=34, before seeking to minimize H. 
What do the 'reasonable value' results tell you? 
Maybe a good choice of a R-square threshold to use 
later in the H minimization, so for your example we 
might require the smallest H with Rsq>= 0.9975?
 GREG: NO.
 My training goal for all of the MATLAB examples has 
always been the more realistic 
 NMSE = mse(t-y)/mean(var(t',1)) <= 0.01
 or , equivalently
 Rsq = 1- NMSE >= 0.99
 That is not to say better values can not be acheived. 
Just  that this is a more reasonable way to begin 
before trying to optimize the final result.
 %Using the command 
 >> help nndatasets
 % yields
   simplefit_dataset - Simple fitting dataset.
   abalone_dataset   - Abalone shell rings dataset.
   bodyfat_dataset   - Body fat percentage dataset.
   building_dataset  - Building energy dataset.
   chemical_dataset  - Chemical sensor dataset.
   cho_dataset       - Cholesterol dataset.
   engine_dataset    - Engine behavior dataset.
   vinyl_dataset     - Vinyl bromide dataset.
 % then 
 >>  SIZE1 = size(simplefit_dataset)    
   SIZE2 = size(abalone_dataset)      
   SIZE3 = size(bodyfat_dataset)      
   SIZE4 = size(building_dataset)     
   SIZE5 = size(chemical_dataset)     
   SIZE6 = size(cho_dataset)           
   SIZE7 = size(engine_dataset) 
   SIZE8 = size(vinyl_dataset)   
 % yields
 SIZE1 =   1     94
SIZE2 =   8   4177
SIZE3 =  13    252
SIZE4 =  14   4208
SIZE5 =   8    498
SIZE6 =  21    264
SIZE7 =   2   1199
SIZE8 =  16  68308

% Therefore, FOR OBVIOUS REASONS, most of my MATLAB tutorial-type investigations are restricted to N < 500 (i.e., simplefit, bodyfat, chemical and cho)

Hope this helps.

Greg

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Answer 1

Greg Heath 2018-9-23

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/417531-using-r-2-results-to-optimize-number-of-neurons-in-hidden-layer#answer_337953

在 MATLAB Online 中打开

The answer is H = 12. H >= 13 is overfitting.

If that makes you queasy, average the output of the H=12 net with the 2 with H = 13.

NOTE: Overfitting is not "THE" problem. "THE PROBLEM" is:

OVERTRAINING AN OVERFIT NET.

Hope this helps.

*Thank you for formally accepting my answer*

Greg

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Greg Heath 2018-9-26

编辑：Greg Heath 2018-9-26

There are TWO MAINPOINTS

1. H <= Hub PREVENTS OVERFITTING, i.e. having more unknown weights than training equations.

2. min(H) subject to R2 > 0.99 STABILIZES THE SOLUTION by REDUCING the uncertainty in weight estimates.

Hope this helps.

Greg

KAE 2018-9-26

Thanks for the education!

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