add internal parts of a vector

Dear all, I have a problem with the sum of internal parts of a vector.I'll explain my problem below.I am analyzing a series of data, after having selected the vector (vectorA) with the quantities I need, I have to add the parts in which the terms equal to 1 follow each other. Example:

vectorA=[1;1;1;1;1;3;4;8;1;1;1;2;9;1;1;1;1;4;1;1]

I have to add only the 1 values (in groups), obtaining a vectorB which has as many terms as the groups of 1:

vectorB=[5;3;4;2].

In short, I need to know how many groups of terms equal to 1 I have and their sum. I tried with if loop, but I didn't succeed.

Thanks in advance for the help!

Answer 1

[Dyuman Joshi]

Here is a simple way of dividing the data into groups of 1 and counting the number of elements for each sub-group -

%Converted to a row vector for displaying the values in a single line
vectorA=[1;1;1;1;1;3;4;8;1;1;1;2;9;1;1;1;1;4;1;1].';

%Values equal to 1
idx = vectorA == 1

idx = 1×20 logical array
   1   1   1   1   1   0   0   0   1   1   1   0   0   1   1   1   1   0   1   1

%Convert the logical vector to a char vector
vec = sprintf('%d', idx)

vec = '11111000111001111011'

%split the char vector with '0' as delimiter
vec = strsplit(vec, '0')

vec = 1×4 cell array
    {'11111'}    {'111'}    {'1111'}    {'11'}

%Get the length of each sub-string
out = cellfun('length', vec)

out = 1×4

5 3 4 2

In case the data does not start or end with 1, there will be empty sub-strings present after splitting, as the first or the last character will be a delimiter; then remove the corresponding values -

out = out(out~=0)

out = 1×4

5 3 4 2

Answer 2

[Stephen23]

A = [1;1;1;1;1;3;4;8;1;1;1;2;9;1;1;1;1;4;1;1]

A = 20×1

1 1 1 1 1 3 4 8 1 1

X = A==1;
Y = cumsum([1;diff(X)>0]);
Z = accumarray(Y,X)

Z = 4×1

5 3 4 2

Answer 3

[Hassaan]

vectorA = [1; 1; 1; 1; 1; 3; 4; 8; 1; 1; 1; 2; 9; 1; 1; 1; 1; 4; 1; 1];
vectorB = sum_consecutive_ones(vectorA);
disp(vectorB)

     5
     3
     4
     2

function vectorB = sum_consecutive_ones(vectorA)
    % Ensure vectorA is a column vector
    vectorA = vectorA(:);
    
    % Initialize the count and the result vector
    count = 0;
    vectorB = [];
    
    % Loop over the elements of vectorA
    for i = 1:length(vectorA)
        if vectorA(i) == 1
            count = count + 1; % Increment count if the current element is 1
        elseif count > 0
            vectorB(end + 1, 1) = count; % Add the count to vectorB if the sequence ends
            count = 0; % Reset the count
        end
    end
    
    % If the last element of vectorA is 1, we need to add the count to vectorB
    if count > 0
        vectorB(end + 1, 1) = count;
    end
end

---------------------------------------------------------------------------------------------------------------------------------------------------------

If you find the solution helpful and it resolves your issue, it would be greatly appreciated if you could accept the answer. Also, leaving an upvote and a comment are also wonderful ways to provide feedback.

Professional Interests

• Technical Services and Consulting
• Embedded Systems | Firmware Developement | Simulations
• Electrical and Electronics Engineering

Feel free to contact me.

Answer 4

[Shivam]

Hi,

From the provided information, I understand that you want to identify and sum the consecutive ones in a vector and then record the length of each sequence of consecutive ones into a new vector.

You can follow the below workaround to achieve the goal:

% Initialize vectorA with specific values
vectorA = [1; 1; 1; 1; 1; 3; 4; 8; 1; 1; 1; 2; 9; 1; 1; 1; 1; 4; 1; 1];
n = length(vectorA);
% Initialize an empty array to hold the counts of consecutive ones
vectorB = [];
% Initialize a counter for consecutive ones
countOne = 0;
for i = 1:n
    if vectorA(i) == 1
        countOne = countOne + 1; % Increment the counter for consecutive ones
        % If the current element is not 1 and there is a non-zero count of ones
    elseif countOne
        % Append the count of consecutive ones and 
        % reset the counterOne to 0
        vectorB = [vectorB countOne];
        countOne = 0;
    end
end
% After the loop, if there is a non-zero count of ones left
if countOne
    % Append the remaining count of consecutive ones to vectorB
    vectorB = [vectorB countOne];
end
disp(vectorB);

     5     3     4     2

I hope it helps.

Thanks