Say I create a neural network to separate classes:
X1; %Some data in Class 1 100x2
X2; %Some data in Class 2 100x2
classInput = [X1;X2];
negative = zeros(N, 1);
positive = ones(N,1);
classTarget = [positive negative; negative positive];
net = feedforwardnet(20);
net = configure(net, classInput, classTarget);
net = train(net, classInput, classTarget);
%output of training data
output = net(classInput);
I can plot the classes and they are correctly separated:
figure();
hold on
style = {'ro' 'bx'};
for i=1:(2*N)
plot(classInput(i,1),classInput(i,2), style{round(output(i,1))+1});
end
However, how can I apply the network that's just been trained to unseen data? There must be a model which is generated by the network that can be applied to new data?
EDIT: Using sim:
Once the network is trained, if I use sim on the training data:
[Z,Xf,Af] = sim(net,classInput);
The result is as expected. But this only works if the input is of the same size. If for example I want to evalute an individual data point:
[Z1,Xf,Af] = sim(net,[1,2]);
size(Z) == size(Z1), but this clearly doesn't make sense? Surely I can evaluate a single data point?
I'm the OP,
I had assumed that the rows of the input matrices were the data samples and the columns were the "categories", this is the other way around. Transposing the matrices before inputting them to the train() function fixes this.
Related
I have trained my Neural network model using MATLAB NN Toolbox. My network has multiple inputs and multiple outputs, 6 and 7 respectively, to be precise. I would like to clarify few questions based on it:-
The final regression plot showed at the end of the training shows a very good accuracy, R~0.99. However, since I have multiple outputs, I am confused as to which scatter plot does it represent? Shouldn't we have 7 target vs predicted plots for each of the output variable?
According to my knowledge, R^2 is a better method of commenting upon the accuracy of the model, whereas MATLAB reports R in its plot. Do I treat that R as R^2 or should I square the reported R value to obtain R^2.
I have generated the Matlab Script containing weight, bias and activation functions, as a final Result of the training. So shouldn't I be able to simply give my raw data as input and obtain the corresponding predicted output. I gave the exact same training set using the indices Matlab chose for training (to cross check), and plotted the predicted output vs actual output, but the result is not at all good. Definitely, not along the lines of R~0.99. Am I doing anything wrong?
code:
function [y1] = myNeuralNetworkFunction_2(x1)
%MYNEURALNETWORKFUNCTION neural network simulation function.
% X = [torque T_exh lambda t_Spark N EGR];
% Y = [O2R CO2R HC NOX CO lambda_out T_exh2];
% Generated by Neural Network Toolbox function genFunction, 17-Dec-2018 07:13:04.
%
% [y1] = myNeuralNetworkFunction(x1) takes these arguments:
% x = Qx6 matrix, input #1
% and returns:
% y = Qx7 matrix, output #1
% where Q is the number of samples.
%#ok<*RPMT0>
% ===== NEURAL NETWORK CONSTANTS =====
% Input 1
x1_step1_xoffset = [-24;235.248;0.75;-20.678;550;0.799];
x1_step1_gain = [0.00353982300884956;0.00284355877067267;6.26959247648903;0.0275865874012055;0.000366568914956012;0.0533831576137729];
x1_step1_ymin = -1;
% Layer 1
b1 = [1.3808996210168685;-2.0990163849711894;0.9651733083552595;0.27000953282929346;-1.6781835509820286;-1.5110463684800366;-3.6257438832309905;2.1569498669085361;1.9204156230460485;-0.17704342477904209];
IW1_1 = [-0.032892214008082517 -0.55848270745152429 -0.0063993424771670616 -0.56161004933654057 2.7161844536020197 0.46415317073346513;-0.21395624254052176 -3.1570133640176681 0.71972178875396853 -1.9132557838515238 1.3365248285282931 -3.022721627052706;-1.1026780445896862 0.2324603066452392 0.14552308208231421 0.79194435276493658 -0.66254679969168417 0.070353201192052434;-0.017994515838487352 -0.097682677816992206 0.68844109281256027 -0.001684535122025588 0.013605622123872989 0.05810686279306107;0.5853667840629273 -2.9560683084876329 0.56713425120259764 -2.1854386350040116 1.2930115031659106 -2.7133159265497957;0.64316656469750333 -0.63667017646313084 0.50060179040086761 -0.86827897068177973 2.695456517458648 0.16822164719859456;-0.44666821007466739 4.0993786464616679 -0.89370838440321498 3.0445073606237933 -3.3015566360833453 -4.492874075961689;1.8337574137485424 2.6946232855369989 1.1140472073136622 1.6167763205944321 1.8573696127039145 -0.81922672766933646;-0.12561950922781362 3.0711045035224349 -0.6535751823440773 2.0590707752473199 -1.3267693770634292 2.8782780742777794;-0.013438026967107483 -0.025741311825949621 0.45460734966889638 0.045052447491038108 -0.21794568374100454 0.10667240367191703];
% Layer 2
b2 = [-0.96846557414356171;-0.2454718918618051;-0.7331628718025488;-1.0225195290982099;0.50307202195645395;-0.49497234988401961;-0.21817117469133171];
LW2_1 = [-0.97716474643411022 -0.23883775971686808 0.99238069915206006 0.4147649511973347 0.48504023209224734 -0.071372217431684551 0.054177719330469304 -0.25963474838320832 0.27368380212104881 0.063159321947246799;-0.15570858147605909 -0.18816739764334323 -0.3793600124951475 2.3851961990944681 0.38355142531334563 -0.75308427071748985 -0.1280128732536128 -1.361052031781103 0.6021878865831336 -0.24725687748503239;0.076251356114485525 -0.10178293627600112 0.10151304376762409 -0.46453434441403058 0.12114876632815359 0.062856969143306296 -0.0019628163322658364 -0.067809039768745916 0.071731544062023825 0.65700427778446913;0.17887084584125315 0.29122649575978238 0.37255802759192702 1.3684190468992126 0.60936238465090853 0.21955911453674043 0.28477957899364675 -0.051456306721251184 0.6519451272106177 -0.64479205028051967;0.25743349663436799 2.0668075180209979 0.59610776847961111 -3.2609682919282603 1.8824214917530881 0.33542869933904396 0.03604272669356564 -0.013842766338427388 3.8534510207741826 2.2266745660915586;-0.16136175574939746 0.10407287099228898 -0.13902245286490234 0.87616472446622717 -0.027079111747601223 0.024812287505204988 -0.030101536834009103 0.043168268669541855 0.12172932035587079 -0.27074383434206573;0.18714562505165402 0.35267726325386606 -0.029241400610813449 0.53053853235049087 0.58880054832728757 0.047959541165126809 0.16152268183097709 0.23419456403348898 0.83166785128608967 -0.66765237856750781];
% Output 1
y1_step1_ymin = -1;
y1_step1_gain = [0.114200879346771;0.145581598485951;0.000139011547272197;0.000456244862967996;2.05816254143146e-05;5.27704485488127;0.00284355877067267];
y1_step1_xoffset = [-0.045;1.122;2.706;17.108;493.726;0.75;235.248];
% ===== SIMULATION ========
% Dimensions
Q = size(x1,1); % samples
% Input 1
x1 = x1';
xp1 = mapminmax_apply(x1,x1_step1_gain,x1_step1_xoffset,x1_step1_ymin);
% Layer 1
a1 = tansig_apply(repmat(b1,1,Q) + IW1_1*xp1);
% Layer 2
a2 = repmat(b2,1,Q) + LW2_1*a1;
% Output 1
y1 = mapminmax_reverse(a2,y1_step1_gain,y1_step1_xoffset,y1_step1_ymin);
y1 = y1';
end
% ===== MODULE FUNCTIONS ========
% Map Minimum and Maximum Input Processing Function
function y = mapminmax_apply(x,settings_gain,settings_xoffset,settings_ymin)
y = bsxfun(#minus,x,settings_xoffset);
y = bsxfun(#times,y,settings_gain);
y = bsxfun(#plus,y,settings_ymin);
end
% Sigmoid Symmetric Transfer Function
function a = tansig_apply(n)
a = 2 ./ (1 + exp(-2*n)) - 1;
end
% Map Minimum and Maximum Output Reverse-Processing Function
function x = mapminmax_reverse(y,settings_gain,settings_xoffset,settings_ymin)
x = bsxfun(#minus,y,settings_ymin);
x = bsxfun(#rdivide,x,settings_gain);
x = bsxfun(#plus,x,settings_xoffset);
end
The above one is the automatically generated code. The plot which I generated to cross-check the first variable is below:-
% X and Y are input and output - same as above
X_train = X(results.info1.train.indices,:);
y_train = Y(results.info1.train.indices,:);
out_train = myNeuralNetworkFunction_2(X_train);
scatter(y_train(:,1),out_train(:,1))
To answer your question about R: Yes, you should square R to get the R^2 value. In this case, they will be very close since R is very close to 1.
The graphs give the correlation between the estimated and real (target) values. So R is the strenght of the correlation. You can square it to find the R-square.
The graph you draw and matlab gave are not the graph of the same variables. The ranges or scales of the axes are very different.
First of all, is the problem you are trying to solve a regression problem? Or is it a classification problem with 7 classes converted to numeric? I assume this is a classification problem, as you are trying to get the success rate for each class.
As for your first question: According to the literature it is recommended to use the value "All: R". If you want to get the success rate of each of your classes, Precision, Recall, F-measure, FP rate, TP Rate, etc., which are valid in classification problems. values you need to reach. There are many matlab documents for this (help ROC) and you can look at the details. All the values I mentioned and which I think you actually want are obtained from the confusion matrix.
There is a good example of this.
[x,t] = simpleclass_dataset;
net = patternnet(10);
net = train(net,x,t);
y = net(x);
[c,cm,ind,per] = confusion(t,y)
I hope you will see what you want from the "nntraintool" window that appears when you run the code.
Your other questions have already been answered. Alternatively, you can consider using a machine learning algorithm with open source software such as Weka.
I am trying to learn how to use neural networks in MATLAB and I am starting with a simple example that uses four data points that I split into two row vectors. One of them is Input and the other is Temp. The input vector is a vector from 1 to 4.
Next I run some neural network coding I found from examples. Now I would like for the neural network to predict the outcome of a sample input vector which is a row vector [5 6].
clear all
clc
Input = [1,2,3,4];
Temp = [.25,.15,.1,.07];
Smpl = [5,6]
net = newff(minmax(Input),[20,1],{'logsig','purelin','trainlm'})
net.trainparam.epochs = 500;
net.trainparam.goal = 1e-25;
net.trainparam.lr = .01;
net = train(net,Input,Temp)
TempPr = net(Input)
error = TempPr - Temp
TempPrSmpl = net(Smpl)
The row vector, TempPr, generated by the neural network exactly matches with the target vector, Temp. However, it seems that I am unable to predict values properly. For example I try to predict temperature values for inputs 5 and 6 which I expect them to be less than .07.
But instead the matlab code is returning:
TempPrSmpl =
0.3560 0.3560
Two questions:
Why is the value being returned from MATLAB greater than .07?
Why are there not two different values being returned from MATLAB (one for 5 and one for 6)?
I have implemented the Naive Bayse Classifier for multiclass but problem is my error rate is same while I increase the training data set. I was debugging this over an over but wasn't able to figure why its happening. So I thought I ll post it here to find if I am doing anything wrong.
%Naive Bayse Classifier
%This function split data to 80:20 as data and test, then from 80
%We use incremental 5,10,15,20,30 as the test data to understand the error
%rate.
%Goal is to compare the plots in stanford paper
%http://ai.stanford.edu/~ang/papers/nips01-discriminativegenerative.pdf
function[tPercent] = naivebayes(file, iter, percent)
dm = load(file);
for i=1:iter
%Getting the index common to test and train data
idx = randperm(size(dm.data,1))
%Using same idx for data and labels
shuffledMatrix_data = dm.data(idx,:);
shuffledMatrix_label = dm.labels(idx,:);
percent_data_80 = round((0.8) * length(shuffledMatrix_data));
%Doing 80-20 split
train = shuffledMatrix_data(1:percent_data_80,:);
test = shuffledMatrix_data(percent_data_80+1:length(shuffledMatrix_data),:);
%Getting the label data from the 80:20 split
train_labels = shuffledMatrix_label(1:percent_data_80,:);
test_labels = shuffledMatrix_label(percent_data_80+1:length(shuffledMatrix_data),:);
%Getting the array of percents [5 10 15..]
percent_tracker = zeros(length(percent), 2);
for pRows = 1:length(percent)
percentOfRows = round((percent(pRows)/100) * length(train));
new_train = train(1:percentOfRows,:);
new_train_label = train_labels(1:percentOfRows);
%get unique labels in training
numClasses = size(unique(new_train_label),1);
classMean = zeros(numClasses,size(new_train,2));
classStd = zeros(numClasses, size(new_train,2));
priorClass = zeros(numClasses, size(2,1));
% Doing the K class mean and std with prior
for kclass=1:numClasses
classMean(kclass,:) = mean(new_train(new_train_label == kclass,:));
classStd(kclass, :) = std(new_train(new_train_label == kclass,:));
priorClass(kclass, :) = length(new_train(new_train_label == kclass))/length(new_train);
end
error = 0;
p = zeros(numClasses,1);
% Calculating the posterior for each test row for each k class
for testRow=1:length(test)
c=0; k=0;
for class=1:numClasses
temp_p = normpdf(test(testRow,:),classMean(class,:), classStd(class,:));
p(class, 1) = sum(log(temp_p)) + (log(priorClass(class)));
end
%Take the max of posterior
[c,k] = max(p(1,:));
if test_labels(testRow) ~= k
error = error + 1;
end
end
avgError = error/length(test);
percent_tracker(pRows,:) = [avgError percent(pRows)];
tPercent = percent_tracker;
plot(percent_tracker)
end
end
end
Here is the dimentionality of my data
x =
data: [768x8 double]
labels: [768x1 double]
I am using Pima data set from UCI
What are the results of your implementation of the training data itself? Does it fit it at all?
It's hard to be sure but there are couple things that I noticed:
It is important for every class to have training data. You can't really train a classifier to recognize a class if there was no training data.
If possible number of training examples shouldn't be skewed towards some of classes. For example if in 2-class classification number of training and cross validation examples for class 1 constitutes only 5% of the data then function that always returns class 2 will have error of 5%. Did you try checking precision and recall separately?
You're trying to fit normal distribution to each feature in a class and then use it for posterior probabilities. I'm not sure how it plays out in terms of smoothing. Could you try to re-implement it with simple counting and see if it gives any different results?
It also could be that features are highly redundant and bayes method overcounts probabilities.
I am trying to implement Naive Bayes Classifier using a dataset published by UCI machine learning team. I am new to machine learning and trying to understand techniques to use for my work related problems, so I thought it's better to get the theory understood first.
I am using pima dataset (Link to Data - UCI-ML), and my goal is to build Naive Bayes Univariate Gaussian Classifier for K class problem (Data is only there for K=2). I have done splitting data, and calculate the mean for each class, standard deviation, priors for each class, but after this I am kind of stuck because I am not sure what and how I should be doing after this. I have a feeling that I should be calculating posterior probability,
Here is my code, I am using percent as a vector, because I want to see the behavior as I increase the training data size from 80:20 split. Basically if you pass [10 20 30 40] it will take that percentage from 80:20 split, and use 10% of 80% as training.
function[classMean] = naivebayes(file, iter, percent)
dm = load(file);
for i=1:iter
idx = randperm(size(dm.data,1))
%Using same idx for data and labels
shuffledMatrix_data = dm.data(idx,:);
shuffledMatrix_label = dm.labels(idx,:);
percent_data_80 = round((0.8) * length(shuffledMatrix_data));
%Doing 80-20 split
train = shuffledMatrix_data(1:percent_data_80,:);
test = shuffledMatrix_data(percent_data_80+1:length(shuffledMatrix_data),:);
train_labels = shuffledMatrix_label(1:percent_data_80,:)
test_labels = shuffledMatrix_data(percent_data_80+1:length(shuffledMatrix_data),:);
%Getting the array of percents
for pRows = 1:length(percent)
percentOfRows = round((percent(pRows)/100) * length(train));
new_train = train(1:percentOfRows,:)
new_trin_label = shuffledMatrix_label(1:percentOfRows)
%get unique labels in training
numClasses = size(unique(new_trin_label),1)
classMean = zeros(numClasses,size(new_train,2));
for kclass=1:numClasses
classMean(kclass,:) = mean(new_train(new_trin_label == kclass,:))
std(new_train(new_trin_label == kclass,:))
priorClassforK = length(new_train(new_trin_label == kclass))/length(new_train)
priorClassforK_1 = 1 - priorClassforK
end
end
end
end
First, compute the probability of evey class label based on frequency counts. For a given sample of data and a given class in your data set, you compute the probability of evey feature. After that, multiply the conditional probability for all features in the sample by each other and by the probability of the considered class label. Finally, compare values of all class labels and you choose the label of the class with the maximum probability (Bayes classification rule).
For computing conditonal probability, you can simply use the Normal distribution function.
I'm currently working on classifying images with different image-descriptors. Since they have their own metrics, I am using precomputed kernels. So given these NxN kernel-matrices (for a total of N images) i want to train and test a SVM. I'm not very experienced using SVMs though.
What confuses me though is how to enter the input for training. Using a subset of the kernel MxM (M being the number of training images), trains the SVM with M features. However, if I understood it correctly this limits me to use test-data with similar amounts of features. Trying to use sub-kernel of size MxN, causes infinite loops during training, consequently, using more features when testing gives poor results.
This results in using equal sized training and test-sets giving reasonable results. But if i only would want to classify, say one image, or train with a given amount of images for each class and test with the rest, this doesn't work at all.
How can i remove the dependency between number of training images and features, so i can test with any number of images?
I'm using libsvm for MATLAB, the kernels are distance-matrices ranging between [0,1].
You seem to already have figured out the problem... According to the README file included in the MATLAB package:
To use precomputed kernel, you must include sample serial number as
the first column of the training and testing data.
Let me illustrate with an example:
%# read dataset
[dataClass, data] = libsvmread('./heart_scale');
%# split into train/test datasets
trainData = data(1:150,:);
testData = data(151:270,:);
trainClass = dataClass(1:150,:);
testClass = dataClass(151:270,:);
numTrain = size(trainData,1);
numTest = size(testData,1);
%# radial basis function: exp(-gamma*|u-v|^2)
sigma = 2e-3;
rbfKernel = #(X,Y) exp(-sigma .* pdist2(X,Y,'euclidean').^2);
%# compute kernel matrices between every pairs of (train,train) and
%# (test,train) instances and include sample serial number as first column
K = [ (1:numTrain)' , rbfKernel(trainData,trainData) ];
KK = [ (1:numTest)' , rbfKernel(testData,trainData) ];
%# train and test
model = svmtrain(trainClass, K, '-t 4');
[predClass, acc, decVals] = svmpredict(testClass, KK, model);
%# confusion matrix
C = confusionmat(testClass,predClass)
The output:
*
optimization finished, #iter = 70
nu = 0.933333
obj = -117.027620, rho = 0.183062
nSV = 140, nBSV = 140
Total nSV = 140
Accuracy = 85.8333% (103/120) (classification)
C =
65 5
12 38