how to create for loop in below formula

% Define the LP values
LP = [LP1, LP2, LP3, LP4, LP5, LP6];
% Initialize the array to store the results
Lc = zeros(1, 6);
% Loop through the LP values and calculate the Lc values
for i = 1:6
    if i == 1
        Lc(i) = LP(i);
    else
        Lc(i) = LP(i) - sum(Lc(1:i-1));
    end
end
%Hope this was helpful

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Answer 3

Stephen23 2023-6-16

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1983984-how-to-create-for-loop-in-below-formula#answer_1256984

在 MATLAB Online 中打开

Your inefficient approach (with anti-pattern numbered variable names):

LP1 = rand;
LP2 = rand;
LP3 = rand;
LP4 = rand;
LP5 = rand;
LP6 = rand;
Lc1 = LP1
Lc1 = 0.0735
Lc2 = LP2-Lc1
Lc2 = 0.2472
Lc3 = LP3-(Lc1+Lc2)
Lc3 = 0.6106
Lc4 = LP4-(Lc1+Lc2+Lc3)
Lc4 = -0.2612
Lc5 = LP5-(Lc1+Lc2+Lc3+Lc4)
Lc5 = 0.0396
Lc6 = LP6-(Lc1+Lc2+Lc3+Lc4+Lc5)
Lc6 = -0.0814

The MATLAB approach (with vectors):

LP = [LP1,LP2,LP3,LP4,LP5,LP6];
Lc = LP;
for k = 2:numel(LP)
    Lc(k) = LP(k)-sum(Lc(1:k-1));
end
disp(Lc)
    0.0735    0.2472    0.6106   -0.2612    0.0396   -0.0814

The best approach:

Lc = [LP(1),diff(LP)]
Lc = 1×6
    0.0735    0.2472    0.6106   -0.2612    0.0396   -0.0814

Why do you keep deleting and then reposting this exact question? Or were those classmates of yours?

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Answer 4

Malay Agarwal 2023-6-16

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1983984-how-to-create-for-loop-in-below-formula#answer_1257009

编辑：Malay Agarwal 2023-6-16

在 MATLAB Online 中打开

Hi, Devaki

You can use the following code, which generalizes it to a vector of any size:

% Declare an array of the variables
% This could be an arbitrary number of variables
% Here, I declare it as a random vector of 6 variables
LP = rand(1, 6);
% Initialize the result array with zeros
Lc = zeros(size(LP));
% Since Lc1 = LP1, set Lc(1) to the first value in LP(1)
Lc(1) = LP(1);
[rows, cols] = size(LP);
% Loop and set the remaining values
for i=2:cols
    % sum(Lc(1:i - 1)) is the sum of the first i - 1 elements in Lc
    Lc(i) = LP(i) - sum(Lc(1:i-1));
end