splitting a vector into separate vectors
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Hi Matlab community,
I have a vector like 
V = [2 2 2 2 2 4 4 4 7 7 8 9]
I want to separate this vector into 
V1 = [2 2 2 2 2 ], V2=[4 4 4 ], V3=[7 7 ], V4=8, V5=9
Suppose that I do not know the value in vector V. In each iteration, I have a new vector. 
Thanks 
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採用された回答
  Stephen23
      
      
 2024 年 9 月 30 日
        
      編集済み: Stephen23
      
      
 2024 年 9 月 30 日
  
      A much better approach using a cell array:
V = [2,2,2,2,2,4,4,4,7,7,8,9]
L = diff(find([1,diff(V),1]))
C = mat2cell(V,1,L)
3 件のコメント
  Umeshraja
 2024 年 9 月 30 日
				You can use indexing to extract the individual arrays and then use functions like 'length' or 'numel' to determine the number of elements in each array. 
firstArray = C{1};
numElementsInFirstArray = numel(firstArray);
Please refer to the following documentation to know more
  Stephen23
      
      
 2024 年 9 月 30 日
				
      編集済み: Stephen23
      
      
 2024 年 9 月 30 日
  
			"Now, how can I access the resutl? I mean how can I understand how many elements are there in the first array? and second array? I mean, how many 2 we have here?"
My answer already tells you that, take a look at the variable L, it gives you the lengths of each vector.
You can access the arrays using very basic MATLAB indexing, eg. the 2nd vector:
C{2} % the vector
L(2) % its length
You do not need to call an extra function like LENGTH or NUMEL, because the information is already there.
その他の回答 (2 件)
  Steven Lord
    
      
 2024 年 9 月 30 日
        Can you dynamically create variables with numbered names like V1, V2, V3, etc.? Yes.
Should you do this? The general consensus is no. That Discussions post explains why this is generally discouraged and offers several alternative approaches.
  Umeshraja
 2024 年 9 月 30 日
        If you are working with large 1D vectors. using a binary search might be a good option to optimize time complexity.
Below is the sample code:
function separatedVectors = separateUsingBinarySearch(V)
    % Initialize an empty cell array to store the separated vectors
    separatedVectors = {};
    n = length(V);   
    if n == 0
        return;
    end
    startIdx = 1; 
    % Continue until the start index exceeds the length of the vector
    while startIdx <= n
        value = V(startIdx);  
        endIdx = findEndIndex(V, startIdx, n, value);
        % Store the current group of identical elements in the cell array
        separatedVectors{end+1} = V(startIdx:endIdx);
        startIdx = endIdx + 1;
    end
end
function endIdx = findEndIndex(V, startIdx, n, value)
    low = startIdx; 
    high = n; 
    % Perform binary search to find the last occurrence of 'value'
    while low < high
        mid = floor((low + high + 1) / 2); 
        if V(mid) == value
            low = mid; % Move the lower bound up if the mid value matches
        else
            high = mid - 1; % Move the upper bound down if the mid value does not match
        end
    end
    endIdx = low; 
end
% Example usage
V = [2 2 2 2 2 4 4 4 7 7 8 9 9 9 9]; 
separatedVectors = separateUsingBinarySearch(V);
% Display the results
for i = 1:length(separatedVectors)
    fprintf('[%s]\n', num2str(separatedVectors{i}));
end
The function separateUsingBinarySearch efficiently divides a sorted vector V into sub-vectors, each containing consecutive identical elements. For each starting index, findEndIndex uses binary search to find the last occurrence of the current value efficiently. The sub-vector from startIdx to endIdx is extracted and added to separatedVectors. The process repeats by updating startIdx to just after endIdx for the next group.
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