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I am tuning an SVM using a for loop to search in the range of hyperparameter's space. The svm model learned contains the following fields
SVMModel: [1×1 ClassificationSVM]
C: 2
FeaturesIdx: [4 6 8]
Score: 0.0142
Question1) What is the meaning of the field 'score' and its utility?
Question2) I am tuning the BoxConstraint, C value. Let, the number of features be denoted by the variable featsize. The variable gridC will contain the search space which can start from any value say 2^-5, 2^-3, to 2^15 etc. So, gridC = 2.^(-5:2:15). I cannot understand if there is a way to select the range?
1. score had been documented in here, which says:
Classification Score
The SVM classification score for classifying observation x is the signed distance from x to the decision boundary ranging from -∞ to +∞.
A positive score for a class indicates that x is predicted to be in
that class. A negative score indicates otherwise.
In two class cases, if there are six observations, and the predict function gave us some score value called TestScore, then we could determine which class does the specific observation ascribed by:
TestScore=[-0.4497 0.4497
-0.2602 0.2602;
-0.0746 0.0746;
0.1070 -0.1070;
0.2841 -0.2841;
0.4566 -0.4566;];
[~,Classes] = max(TestScore,[],2);
In the two-class classification, we can also use find(TestScore > 0) instead, and it is clear that the first three observations are belonging to the second class, and the 4th to 6th observations are belonging to the first class.
In multiclass cases, there could be several scores > 0, but the code max(scores,[],2) is still validate. For example, we could use the code (from here, an example called Find Multiple Class Boundaries Using Binary SVM) following to determine the classes of the predict Samples.
for j = 1:numel(classes);
[~,score] = predict(SVMModels{j},Samples);
Scores(:,j) = score(:,2); % Second column contains positive-class scores
end
[~,maxScore] = max(Scores,[],2);
Then the maxScore will denote the predicted classes of each sample.
2. The BoxConstraint denotes C in the SVM model, so we can train SVMs in different hyperparameters and select the best one by something like:
gridC = 2.^(-5:2:15);
for ii=1:length(gridC)
SVModel = fitcsvm(data3,theclass,'KernelFunction','rbf',...
'BoxConstraint',gridC(ii),'ClassNames',[-1,1]);
%if (%some constraints were meet)
% %save the current SVModel
%end
end
Note: Another way to implement this is using libsvm, a fast and easy-to-use SVM toolbox, which has the interface of MATLAB.
I'm working on gait recognition problem, the aim of this study is to be used for user authentication
I have data of 36 users
I've successfully extracted 143 features for each sample (or example) which are (36 rows and 143 columns) for each user
( in other words, I have 36 examples and 143 features are extracted for each example. Thus a matrix called All_Feat of 36*143 has been created for each individual user).
By the way, column represents the number of the extracted features and row represents the number of samples (examples) for each feature.
Then I have divided the data into two parts, training and testing (the training matrix contains 25 rows and 143 columns, while the testing Matrix contains 11 rows and 143 columns).
Then, for each user, I divided the matrix (All_Feat) into two matrixes ( Training matrix, and Test matrix ).
The training matrix contains ( 25 rows (examples) and 143 columns), while the testing matrix has (11 rows and 143 columns).
I'm new to classification and this stuff
I'd like to use machine learning (Neural Network) for classifying these features.
Therefore, the first step I need to create a reference template for each user ( which called training phase)
this can be done by training the classifier with the user's features (data) and the remaining Users as well (35 users are considered as imposters).
Based on what I have read, training Neural Network requires two classes, the first class contains all the training data of genuine user (e.g. User1) and labelled with 1 , while the second class has the training data of imposters labelled as 0 (which is binary classification, 1 for the authorised user and 0 for imposters).
**now my question is: **
1- i dont know how to create these classes!
2- For example, if I want to train Neural Network for User1, I have these variables, input and target. what should I assign to these variables?
should input= Training matrix of User1 and Training matrixes of User2, User3,.....User35 ?
Target=what should i assign to this matrix?
I really appreciate any help!
Try this: https://es.mathworks.com/help/nnet/gs/classify-patterns-with-a-neural-network.html
A few notes:
You said that, for each user, you have extracted 136 features. It sounds like you have only one repetition for each user (i.e. the user has tried-used the system once). However, I don't know the source of your data, but I dunno that it hasn't got some type of randomness. You mention gait analysis, and that sounds like that the recorded data of a given one user will be different each time that user uses the system. In other words: the user uses your system, you capture the data, you extract the 136 features (numbers); then, the user uses again the system, but the extracted 136 features will be slightly different. Therefore, you should get several examples for each user to train the classifier. In terms of "matlab matrix" your matrix should have one COLUMN for each example, and 136 rows (each of you features). Since you should have several repetitions for each user (for example 10 times), your big matrix should be something like: 136 rows x 360 columns.
You should "create" one new neural network for each user. Given a user (for example User4), you create a dataset (a new matrix) with samples of that user, and samples of several other users (User1, User3, User5...). You do a binary classification (cases: "user4" against "other users"). After training, it would be advisable to test the classifier with data of other users whose data was not present during the training phase (for example User2 and others). Since you are doing a binary classification your matrices should be somthing like follows:
Example, you have 10 trials (examples) of each user. You want to create a neural network to detect the user User1. The matrix should be like:
(notation cU1_t1 means: column with features of user 1, trial 1)
input_matrix = [cU1_t1; cU1_t2; ...; cU1_t10; cU2_t1; ...; cU36_t10]
The target matrix should be like:
target = a matrix whose 10 first columns are [ 1, 0], and the other 350 columns are [0, 1]. That means that the first 10 columns are of type A, and the others of type B. In this case "type A" means "User1", and "type B" means "Not User1".
Then, you should segment the data (train data, validation data, test data) to train the nerual network and so on. Remember to save some users just for the testing phase, for example, the train matrix should not have any of the columns of five users: user2, user6, user7, user10, user20 (50 columns).
I think you get the idea.
Regards.
************ UPDATE: ******************************
This example assumes that the user selects/indicates its name and then the system uses the neural network to authenticate the user (like a password). I will give you an small example with random numbers.
Let's say you have recorded data from 15 users (but in the future you will have more). You record "gait data" from them when they do something with your recording device. From the recorded signals you extract some features, let's say you extract 5 features (5 numbers). Hence, everytime a user uses the machine you get 5 numbers. Even if user is the same, the 5 numbers will be different each time, because the recorded signals have some randomness. Therefore, to train the neural network you have to have several examples of each user. Let's say that you have 18 repetitions performed by each user.
To sum up this example:
There are 15 users available for the experiment.
Each time the user uses the system you record 5 numbers (features). You get a feature vector. In matlab it will be a COLUMN.
For the experiment each user has performed 18 repetitions.
Now you have to create one neural network for each user. To that end, you have to construct several matrices.
Let's say you want to create the neural network (NN) of user 2 (U2). The NN will classify the feature vectors in 2 classes: U2 and NotU2. Therefore, you have to train and test the NN with examples of this. The group NotU2 represents any other user that it is not U2, however, you should NOT train the NN with data of every other user that you have in your experiment. This will be cheating (think that you can't have data from every user in the world). Therefore, to create the train dataset you will exclude all the repetitions of some users to test the NN during the training (validation dataset) and after the trainning (test dataset). For this example we will use users {U1,U3,U4} for validation, and users {U5,U6,U7} for testing.
Therefore you construct the following matrices:
Train input matrix
It wil have 12 examples of U2 (70% more or less) and every example of users {U8,U9,...,U14,U15}. Each example is a column, hence, the train matrix will be a matrix of 5 rows and 156 columns (12+8*18). I will order it as follows: [U2_ex1, U2_ex2, ..., U2_ex12, U8_ex1, U8_ex2, ..., U8_ex18, U9_ex1, ..., U15_ex1,...U15_ex18]. Where U2_ex1 represents a column vector with the 5 features obtained of User 2 during the repetition/example number 1.
-- Target matrix of train matrix. It is a matrix of 2 rows and 156 columns. Each column j represents the correct class of the example j. The column is formed by zeros, and it has a 1 at the row that indicates the class. Since we have only 2 classes the matrix has only 2 rows. I will say that class U2 will be the first one (hence the column vector for each example of this class will be [1 0]), and the other class (NotU2) will be the second one (hence the column vector for each example of this class will be [0 1]). Obviously, the columns of this matrix have the same order than the train matrix. So, according to the order that I have used, the target matrix will be:
12 columns [1 0] and 144 columns [0 1].
Validation input matrix
It will have 3 examples of U2 (15% more or less) and every example of users [U1,U3,U4]. Hence, this will be a matrix of 6 rows and 57 columns ( 3+3*18).
-- Target matrix of validation matrix: A matrix of 2 rows and 57 columns: 3 columns [1 0] and 54 columns [0 1].
Test input matrix
It will have the remaining 3 examples of U2 (15%) and every example of users [U5,U6,U7]. Hence, this will be a matrix of 6 rows and 57 columns (3+3*18).
-- Target matrix of test matrix: A matrix of 2 rows and 57 columns: 3 columns [1 0] and 54 columns [0 1].
IMPORTANT. The columns of each matrix should have a random order to improve the training. That is, do not put all the examples of U2 together and then the others. For this example I have put them in order for clarity. Obviously, if you change the order of the input matrix, you have to use the same order in the target matrix.
To use MATLAB you will have to to pass two matrices: the inputMatrix and the targetMatrix. The inputMatrix will have the train,validation and test input matrices joined. And the targetMatrix the same with the targets. So, the inputMatrix will be a matrix of 6 rows and 270 columns. The targetMatrix will have 2 rows and 270 columns. For clarity I will say that the first 156 columns are the trainning ones, then the 57 columns of validation, and finally 57 columns of testing.
The MATLAB commands will be:
% Create a Pattern Recognition Network
hiddenLayerSize = 10; %You can play with this number
net = patternnet(hiddenLayerSize);
%Specify the indices of each matrix
net.divideFcn = 'divideind';
net.divideParam.trainInd = [1: 156];
net.divideParam.valInd = [157:214];
net.divideParam.testInd = [215:270];
% % Train the Network
[net,tr] = train(net, inputMatrix, targetMatrix);
In the open window you will be able to see the performance of your neural network. The output object "net" is your neural network trained. You can use it with new data if you want.
Repeat this process for each other user (U1, U3, ...U15) to obtain his/her neural network.
I have question please; concerning cross validation, for me the cross-validation is used to find the best parameters.
but I did not understand the role of this function "crossvalind":Generate cross-validation indices, it just takes a data set without model, like in this exemple :
load fisheriris
[g gn] = grp2idx(species);
[trainIdx testIdx] = crossvalind('HoldOut', species, 1/3);
crossvalind() function splits your data in two groups: the training set and the cross-validation set.
By your example:
[trainIdx testIdx] = crossvalind('HoldOut', size(species,1), 1/3); means split the data in species (2/3 in the training set and 1/3 in the cross-validation set).
Supposing that your data is like:
species=[datarow1;datarow2;datarow3;datarow4;datarow5;datarow6] then
trainIdx would be like [1;1;0;1;1;0] and testIdx would be like [0;0;1;0;0;1] meaning that from the 6 total elements in our set crossvalind function assigned 4 to the train set and 2 to the cross-validation set. Of course this is a random assignment meaning that the zero and ones indices will vary every time you call the function but the proportion between them will be fixed and trainIdx + testIdx will always be ones(size(species,1),1)
crossvalind('LeaveMout',size(species,1),2) would be exactly the same as crossvalind('HoldOut', size(species,1), 1/3) in this particular case. In the 'HoldOut' format you provide parameter P which takes values from 0 to 1 (like 1/3 in the example above) while with the option 'LeaveMout' you provide integer M like 2 samples from the 6 total or like 2000 samples from the 10000 total samples in your dataset. In case of 'Resubstitution': crossvalind('Resubstitution', size(species,1), [1/3,2/3]) would be yet the same but here you also have the option of let's say [1/3,3/4] meaning that some samples can be on both the train and cross-validation sets, or even [1,1] which means that all the samples are used in both sets (trainIdx=testIdx=[1;1;1;1;1;1] in the above example). I strongly suggest to type help crossvalind and take a look at the help file which is always a lot more detailed and helpful than i could ever be.
I have a dataset 6x1000 of binary data (6 data points, 1000 boolean dimensions).
I perform cluster analysis on it
[idx, ctrs] = kmeans(x, 3, 'distance', 'hamming');
And I get the three clusters. How can I visualize my result?
I have 6 rows of data each having 1000 attributes; 3 of them should be alike or similar in a way. Applying clustering will reveal the clusters. Since I know the number of clusters
I only need to find similar rows. Hamming distance tell us the similarity between rows and the result is correct that there are 3 clusters.
[EDIT: for any reasonable data, kmeans will always finds asked number
of clusters]
I want to take that knowledge
and make it easily observable and understandable without having to write huge explanations.
Matlab's example is not suitable since it deals with numerical 2D data while my questions concerns n-dimensional categorical data.
The dataset is here http://pastebin.com/cEWJfrAR
[EDIT1: how to check if clusters are significant?]
For more information please visit the following link:
https://chat.stackoverflow.com/rooms/32090/discussion-between-oleg-komarov-and-justcurious
If the question is not clear ask, for anything you are missing.
For representing the differences between high-dimensional vectors or clusters, I have used Matlab's dendrogram function. For instance, after loading your dataset into the matrix x I ran the following code:
l = linkage(a, 'average');
dendrogram(l);
and got the following plot:
The height of the bar that connects two groups of nodes represents the average distance between members of those two groups. In this case it looks like (5 and 6), (1 and 2), and (3 and 4) are clustered.
If you would rather use the hamming distance rather than the euclidian distance (which linkage does by default), then you can just do
l = linkage(x, 'average', {'hamming'});
although it makes little difference to the plot.
You can start by visualizing your data with a 'barcode' plot and then labeling rows with the cluster group they belong:
% Create figure
figure('pos',[100,300,640,150])
% Calculate patch xy coordinates
[r,c] = find(A);
Y = bsxfun(#minus,r,[.5,-.5,-.5, .5])';
X = bsxfun(#minus,c,[.5, .5,-.5,-.5])';
% plot patch
patch(X,Y,ones(size(X)),'EdgeColor','none','FaceColor','k');
% Set axis prop
set(gca,'pos',[0.05,0.05,.9,.9],'ylim',[0.5 6.5],'xlim',[0.5 1000.5],'xtick',[],'ytick',1:6,'ydir','reverse')
% Cluster
c = kmeans(A,3,'distance','hamming');
% Add lateral labeling of the clusters
nc = numel(c);
h = text(repmat(1010,nc,1),1:nc,reshape(sprintf('%3d',c),3,numel(c))');
cmap = hsv(max(c));
set(h,{'Background'},num2cell(cmap(c,:),2))
Definition
The Hamming distance for binary strings a and b the Hamming distance is equal to the number of ones (population count) in a XOR b (see Hamming distance).
Solution
Since you have six data strings, so you could create a 6 by 6 matrix filled with the Hamming distance. The matrix would be symetric (distance from a to b is the same as distance from b to a) and the diagonal is 0 (distance for a to itself is nul).
For example, the Hamming distance between your first and second string is:
hamming_dist12 = sum(xor(x(1,:),x(2,:)));
Loop that and fill your matrix:
hamming_dist = zeros(6);
for i=1:6,
for j=1:6,
hamming_dist(i,j) = sum(xor(x(i,:),x(j,:)));
end
end
(And yes this code is a redundant given the symmetry and zero diagonal, but the computation is minimal and optimizing not worth the effort).
Print your matrix as a spreadsheet in text format, and let the reader find which data string is similar to which.
This does not use your "kmeans" approach, but your added description regarding the problem helped shaping this out-of-the-box answer. I hope it helps.
Results
0 182 481 495 490 500
182 0 479 489 492 488
481 479 0 180 497 517
495 489 180 0 503 515
490 492 497 503 0 174
500 488 517 515 174 0
Edit 1:
How to read the table? The table is a simple distance table. Each row and each column represent a series of data (herein a binary string). The value at the intersection of row 1 and column 2 is the Hamming distance between string 1 and string 2, which is 182. The distance between string 1 and 2 is the same as between string 2 and 1, this is why the matrix is symmetric.
Data analysis
Three clusters can readily be identified: 1-2, 3-4 and 5-6, whose Hamming distance are, respectively, 182, 180, and 174.
Within a cluster, the data has ~18% dissimilarity. By contrast, data not part of a cluster has ~50% dissimilarity (which is random given binary data).
Presentation
I recommend Kohonen network or similar technique to present your data in, say, 2 dimensions. In general this area is called Dimensionality reduction.
I you can also go simpler way, e.g. Principal Component Analysis, but there's no quarantee you can effectively remove 9998 dimensions :P
scikit-learn is a good Python package to get you started, similar exist in matlab, java, ect. I can assure you it's rather easy to implement some of these algorithms yourself.
Concerns
I have a concern over your data set though. 6 data points is really a small number. moreover your attributes seem boolean at first glance, if that's the case, manhattan distance if what you should use. I think (someone correct me if I'm wrong) Hamming distance only makes sense if your attributes are somehow related, e.g. if attributes are actually a 1000-bit long binary string rather than 1000 independent 1-bit attributes.
Moreover, with 6 data points, you have only 2 ** 6 combinations, that means 936 out of 1000 attributes you have are either truly redundant or indistinguishable from redundant.
K-means almost always finds as many clusters as you ask for. To test significance of your clusters, run K-means several times with different initial conditions and check if you get same clusters. If you get different clusters every time or even from time to time, you cannot really trust your result.
I used a barcode type visualization for my data. The code which was posted here earlier by Oleg was too heavy for my solution (image files were over 500 kb) so I used image() to make the figures
function barcode(A)
B = (A+1)*2;
image(B);
colormap flag;
set(gca,'Ydir','Normal')
axis([0 size(B,2) 0 size(B,1)]);
ax = gca;
ax.TickDir = 'out'
end
I have production (q) values from 4 different methods stored in the 4 matrices. Each of the 4 matrices contains q values from a different method as:
Matrix_1 = 1 row x 20 column
Matrix_2 = 100 rows x 20 columns
Matrix_3 = 100 rows x 20 columns
Matrix_4 = 100 rows x 20 columns
The number of columns indicate the number of years. 1 row would contain the production values corresponding to the 20 years. Other 99 rows for matrix 2, 3 and 4 are just the different realizations (or simulation runs). So basically the other 99 rows for matrix 2,3 and 4 are repeat cases (but not with exact values because of random numbers).
Consider Matrix_1 as the reference truth (or base case ). Now I want to compare the other 3 matrices with Matrix_1 to see which one among those three matrices (each with 100 repeats) compares best, or closely imitates, with Matrix_1.
How can this be done in Matlab?
I know, manually, that we use confidence interval (CI) by plotting the mean of Matrix_1, and drawing each distribution of mean of Matrix_2, mean of Matrix_3 and mean of Matrix_4. The largest CI among matrix 2, 3 and 4 which contains the reference truth (or mean of Matrix_1) will be the answer.
mean of Matrix_1 = (1 row x 1 column)
mean of Matrix_2 = (100 rows x 1 column)
mean of Matrix_3 = (100 rows x 1 column)
mean of Matrix_4 = (100 rows x 1 column)
I hope the question is clear and relevant to SO. Otherwise please feel free to edit/suggest anything in question. Thanks!
EDIT: My three methods I talked about are a1, a2 and a3 respectively. Here's my result:
ci_a1 =
1.0e+008 *
4.084733001497999
4.097677503988565
ci_a2 =
1.0e+008 *
5.424396063219890
5.586301025525149
ci_a3 =
1.0e+008 *
2.429145282593182
2.838897116739112
p_a1 =
8.094614835195452e-130
p_a2 =
2.824626709966993e-072
p_a3 =
3.054667629953656e-012
h_a1 = 1; h_a2 = 1; h_a3 = 1
None of my CI, from the three methods, includes the mean ( = 3.454992884900722e+008) inside it. So do we still consider p-value to choose the best result?
If I understand correctly the calculation in MATLAB is pretty strait-forward.
Steps 1-2 (mean calculation):
k1_mean = mean(k1);
k2_mean = mean(k2);
k3_mean = mean(k3);
k4_mean = mean(k4);
Step 3, use HIST to plot distribution histograms:
hist([k2_mean; k3_mean; k4_mean]')
Step 4. You can do t-test comparing your vectors 2, 3 and 4 against normal distribution with mean k1_mean and unknown variance. See TTEST for details.
[h,p,ci] = ttest(k2_mean,k1_mean);
EDIT : I misinterpreted your question. See the answer of Yuk and following comments. My answer is what you need if you want to compare distributions of two vectors instead of a vector against a single value. Apparently, the latter is the case here.
Regarding your t-tests, you should keep in mind that they test against a "true" mean. Given the number of values for each matrix and the confidence intervals it's not too difficult to guess the standard deviation on your results. This is a measure of the "spread" of your results. Now the error on your mean is calculated as the standard deviation of your results divided by the number of observations. And the confidence interval is calculated by multiplying that standard error with appx. 2.
This confidence interval contains the true mean in 95% of the cases. So if the true mean is exactly at the border of that interval, the p-value is 0.05 the further away the mean, the lower the p-value. This can be interpreted as the chance that the values you have in matrix 2, 3 or 4 come from a population with a mean as in matrix 1. If you see your p-values, these chances can be said to be non-existent.
So you see that when the number of values get high, the confidence interval becomes smaller and the t-test becomes very sensitive. What this tells you, is nothing more that the three matrices differ significantly from the mean. If you have to choose one, I'd take a look at the distributions anyway. Otherwise the one with the closest mean seems a good guess. If you want to get deeper into this, you could also ask on stats.stackexchange.com
Your question and your method aren't really clear :
Is the distribution equal in all columns? This is important, as two distributions can have the same mean, but differ significantly :
is there a reason why you don't use the Central Limit Theorem? This seems to me like a very complex way of obtaining a result that can easily be found using the fact that the distribution of a mean approaches a normal distribution where sd(mean) = sd(observations)/number of observations. Saves you quite some work -if the distributions are alike! -
Now if the question is really the comparison of distributions, you should consider looking at a qqplot for a general idea, and at a 2-sample kolmogorov-smirnov test for formal testing. But please read in on this test, as you have to understand what it does in order to interprete the results correctly.
On a sidenote : if you do this test on multiple cases, make sure you understand the problem of multiple comparisons and use the appropriate correction, eg. Bonferroni or Dunn-Sidak.