Finding a specific pattern in data

Hello all,

I would like to create a routine to identity if a specifi pattern exists in may data.

So far I did this:

data=somematrix;
% Define the pattern to search for
pattern = [3, 4, 5];
% Initialize a flag to check if the pattern is found
patternFound = false;
% Iterate through the data with a sliding window
for i = 1:(length(data) - length(pattern) + 1)
% Extract a subsequence of the same length as the pattern
subsequence = data(i:i+length(pattern)-1);
% Compare the subsequence with the pattern
if isequal(subsequence, pattern)
patternFound = true;
break; % Exit the loop if the pattern is found
end
end
% Display the result
if patternFound
disp('Pattern Found in the Data.');
else
disp('Pattern Not Found in the Data.');
end

This works when the data is a vector but is not working if instead a vector I'm working with a matrix.

Thanks in advance.

[Dyuman Joshi]

How should the pattern be detected in the matrix?

Horizontally? Vertically? Diagonally? Or reverse of the orientations mentioned?

Answer 1

[Johnny]

Hello Ricardo,

You can use the ismember function (https://www.mathworks.com/help/matlab/ref/double.ismember.html) combined with logical operations.

patternFound = false ;
% Check if pattern is a subset of data
if size(data,2) < size(pattern,2)
     % pattern cannot be found: column number in pattern is greater than column number in data
    patternFound = false ;
elseif size(data,2) == size(pattern,2)
    % Same number of column between pattern and data
    patternFound = all(ismember(pattern, data, 'rows'));
else
    % column number in data is greater than column number in pattern
    % Extract submatrices from data, to have the same number of columns
    % for comparison and loop through the different combinations.
    for k = 1:(size(data,2)-size(pattern,2)+1)
        patternFound = all(ismember(pattern, data(:, k:(k+size(pattern, 2)-1)), 'rows'));
        if patternFound
            break
        end
    end
end
%
if patternFound
    disp('Pattern Found in the Data.');
else
    disp('Pattern Not Found in the Data.');
end

[Ricardo Duarte]

Thank you @Johnny this helped, but I still have a question.

How can I introduce a kind of margin in the pattern identification. Let's say I don't have exactly the same pattern but a very close one (margin of 5% of correspondence).

Thank you in advance.

[Dyuman Joshi]

So for the pattern - [3 4 5]

The accepted values should be - [3+- 5% 4+-5% 5+-5%] ?

Also, the pattern should only be matched with the same orientation? or any other/specific orientation(s)?

[Ricardo Duarte]

Hello @Dyuman Joshi

Yes, the accepted pattern shoud be [3+- 5% 4+-5% 5+-5%] .

I don't understand what you mean by orientation.

Thank you.

[Johnny]

Depending on the dataset you have, you can use ismembertol instead of ismember. You can refer to the documentation for more information.

Another approach is to update your initial code to work on matrices.

% Check if pattern is a subset of data
tol = 0.05 ; % Update the tolerance as needed
patternFound = false ;
if size(data,2) < size(pattern,2)
    patternFound = false ;
elseif size(data,2) == size(pattern,2)
    lowerLimit = data >= (1-tol)*pattern ;
    upperLimit = data <= (1+tol)*pattern ;
    patternFound = all(upperLimit & lowerLimit,"all") ;
else
    for k = 1:(size(data,1)-size(pattern,1)+1)
        for l = 1:(size(data,2)-size(pattern,2)+1)
            % Extract a subdata matrix which dimensions match the pattern
            subdata = data(k:(k+size(pattern,1)-1),l:(l+size(pattern,2)-1)) ;
            % Use logical operations to determine if the data is within
            % range
            lowerLimit = subdata >= (1-tol)*pattern ;
            upperLimit = subdata <= (1+tol)*pattern ;
            patternFound = all(upperLimit & lowerLimit,"all") ;
            if patternFound
                break
            end
        end
        if patternFound
            break % double break to exit nested loops
        end
    end
end
%
if patternFound
    disp('Pattern Found in the Data.');
else
    disp('Pattern Not Found in the Data.');
end

For the orientation, I believe @Dyuman Joshi is asking whether you want to look for the pattern as it is, or if you also want to search for its transpose form for example. Note that the code above only works for the same orientation.

[Dyuman Joshi]

@Ricardo Duarte, see the below example -

arr = [3 4 6
       4 4 6
       5 6 5];

pattern = [3 4 5];

In the array, the pattern is found - but vertically and diagonally, which is not the same the orientation as of the pattern, that is horizontal.

So, here, should the pattern be considered to be found, or not?

[Ricardo Duarte]

Thank you both @Johnny and @Dyuman Joshi,

The purpose is to find the pattern of a biological signal which is kind as wave. So in this case I think that diagonally could be the best option.

Thank you