How to Update value of some matriks element with looping?
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Eddy Iswardi
el 27 de Mzo. de 2020
Editada: Les Beckham
el 28 de Mzo. de 2020
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
4 comentarios
Ameer Hamza
el 27 de Mzo. de 2020
Editada: Ameer Hamza
el 27 de Mzo. de 2020
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.
Respuesta aceptada
Ameer Hamza
el 28 de Mzo. de 2020
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
2 comentarios
Les Beckham
el 28 de Mzo. de 2020
Editada: Les Beckham
el 28 de Mzo. de 2020
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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