An easy way for adding on pervious row that expands

Question

Yaser Khojah 2022-3-7

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https://ww2.mathworks.cn/matlabcentral/answers/1665894-an-easy-way-for-adding-on-pervious-row-that-expands

评论： Yaser Khojah 2022-3-8

I have two questions that are brought from my codes where I need to perform to jobs as below. Is there an easy approach as I have large data.

Thanks for the help!

% First part
x1      = [100; 200; 50; 100];
Muliple = 0.1;
 x1(1)                     .* Muliple     = 10; % this will be sent to the following row
(x1(2)  - 10)              .* Muliple     = 19; % this will be sent while adding the previous row (10 + 19)
(x1(3)  - (10 + 19))       .* Muliple     = 2.1; % (
(x1(4)  - (10 + 19 + 2.1)) .* Muliple     = 21.1;
% Second part
%Another Part I need help please how to divide x1 by 2 and send each half to the following row while adding 
%for example 
x2= [50; 50 + 100; 100 + 25; 25 + 50; 50];

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Answer 1

Davide Masiello 2022-3-7

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https://ww2.mathworks.cn/matlabcentral/answers/1665894-an-easy-way-for-adding-on-pervious-row-that-expands#answer_912324

编辑：Davide Masiello 2022-3-7

在 MATLAB Online 中打开

x = [100; 200; 50; 100];
Muliple = 0.1;
x1 = zeros(size(x));
x1(1) = x(1)*Muliple;
for i = 2:length(x) 
    x1(i) = (x(i)- sum(x1(1:i-1)))*Muliple; 
end
x2 = [x/2;0]+[0;x/2];

This yields

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Yaser Khojah 2022-3-8

Thanks so much for the code. :)

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Answer 2

Max Alger-Meyer 2022-3-7

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1665894-an-easy-way-for-adding-on-pervious-row-that-expands#answer_912329

在 MATLAB Online 中打开

First part (note that this follows the algorithm you described but the fourth that you listed for the first part is wrong):

x1 = [100; 200; 50; 100];
Multiple = 0.1;
x2 = zeros(size(x1));
for i = 1:numel(x2)
    if i > 1
        x2(i) = (x1(i)-sum(x2(1:(i-1))))*Multiple;
    else
        x2(i) = x1(i)*Multiple;
    end
end
x2
x2 = 4×1
   10.0000
   19.0000
    2.1000
    6.8900

Second Part:

x2 = (x1)/2;
x3 = x2;
for i = 1:numel(x2)
    if i > 1
        x3(i) = x2(i) + x2(i-1);
    end
end
x3(end+1) = x2(end);
x3
x3 = 5×1
    50
   150
   125
    75
    50