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Generate a matrix with alternative positive and negative values with ones

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Hello, Any idea how to generate a matrix with ones with positive and negative values. For example
We know,
A=ones(n,1)=[1;1;1;1], if n=4
I would like a matrix like this:
A=[1;-1;1;-1]
However, n will change of size depending on the processed data.
On the other hand, I would like to generate a matrix with the following form
if n=10
A=[1;0.5;0;-0.5;-1;-.5;0;0.5;1;0.5], the n value can change depending on the uploaded data.
Thank you.

Accepted Answer

John D'Errico
John D'Errico on 21 Jul 2023
Edited: John D'Errico on 21 Jul 2023
Learn to use various tools in MATLAB. In this case, mod will help you. Forexample:
n = 4;
mod(1:n,2)
ans = 1×4
1 0 1 0
Does that get you close to what you want? You want a column. But that is easy. And you want +/-1. Also easy.
mod((1:n)',2)*2 - 1
ans = 4×1
1 -1 1 -1
Simple enough. Again, look for your target, and think of what you can do to get there.
As for the second case, it is not clear what general pattern you expect in there. If n was larger, would the step still be 0.5? Would this be a periodic function? Or a simple V-shape?
  2 Comments
Jorge Luis Paredes Estacio
Thank you very much, concering to your question. It could be like a random signal (interms of amplitudes), not necesarily a V shape.
John D'Errico
John D'Errico on 21 Jul 2023
Edited: John D'Errico on 21 Jul 2023
A v-shape is most simply achieved using abs. Again, look for something that gets you close to your target. For example:
n = 9;
abs((1:n) - (n+1)/2)
ans = 1×9
4 3 2 1 0 1 2 3 4
Now we can scale those numbers. and put in a shift.
abs((1:n) - (n+1)/2)/2
ans = 1×9
2.0000 1.5000 1.0000 0.5000 0 0.5000 1.0000 1.5000 2.0000
V = abs((1:n) - (n+1)/2)/2 - 1
V = 1×9
1.0000 0.5000 0 -0.5000 -1.0000 -0.5000 0 0.5000 1.0000
plot(V,'-o')

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More Answers (2)

Dyuman Joshi
Dyuman Joshi on 21 Jul 2023
n = 10;
vec = (0:n-1)';
%Array A
A = cospi(vec)
A = 10×1
1 -1 1 -1 1 -1 1 -1 1 -1
%Array B
vec0 = vec+4;
B = 4*abs(vec0/8-floor(vec0/8+0.5))-1
B = 10×1
1.0000 0.5000 0 -0.5000 -1.0000 -0.5000 0 0.5000 1.0000 0.5000

Steven Lord
Steven Lord on 21 Jul 2023
Another approach:
r = 4;
c = 5;
A = (-1).^((1:r).' + (1:c))
A = 4×5
1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1

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