How to Update value of some matriks element with looping?

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Eddy Iswardi 2020-3-27

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https://ww2.mathworks.cn/matlabcentral/answers/513358-how-to-update-value-of-some-matriks-element-with-looping

编辑： Les Beckham 2020-3-28

I have condition for my matrix. Matrix x1,x2, and y. When element of y is less than 10, the y element will update will sum of x1 and x2 until the element is 10 or more. This is my code.

clc;
clear;
x1=[1 2 3 4 5 6 7 8 9 10];
x2=[1 2 3 4 5 6 7 8 9 10];
y=[10 9 11 10 11 11 9 10 12 12];
while y<10
    y=x1+x2;
end
y

And then, if the problem obove is clear. I want to know how many sum (in this case, how many iteration) to achieve values of 10 or more of each element

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Ameer Hamza 2020-3-27

编辑：Ameer Hamza 2020-3-27

You said it make all elements of y 10 or more but then you said

[10 9 11 10 11 11 9 10 12 12] become [10 12 11 10 11 11 9 10 12 12]

The second vector still has 9.

Eddy Iswardi 2020-3-28

Yeah. But it's only an example. y(7) will be update to, some ways with y(2). So the vector will get [10 12 11 10 11 11 14 10 12 12]

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采纳的回答

Ameer Hamza 2020-3-28

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https://ww2.mathworks.cn/matlabcentral/answers/513358-how-to-update-value-of-some-matriks-element-with-looping#answer_422469

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Try this:

x1=[1 2 3 4 5 6 7 8 9 10];
x2=[1 2 3 4 5 6 7 8 9 10];
y=[10 9 11 10 11 11 9 10 12 12];
k = 10;
x12 = x1 + x2;
mask = y < k;
y(mask) = ceil(k./x12(mask)).*x12(mask);

Result:

y =
    10    12    11    10    11    11    14    10    12    12

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Eddy Iswardi 2020-3-28

thank you

Les Beckham 2020-3-28

编辑：Les Beckham 2020-3-28

在 MATLAB Online 中打开

For this part of your question: "I want to know how many sum (in this case, how many iteration) to achieve values of 10 or more of each element", the "iterations" required are given by this part of Ameer's excellent answer:

ceil(k./x12(mask))

If you do this:

iterations = zeros(size(x1));
iterations(mask) = ceil(k./x12(mask))

you will get a vector showing the number of sums required for every element of your original vector with zeros where no iterations (sums) were required.

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