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How can I make 1 by length x array of random numbers from -pi to pi ??

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How can I make 1 by length x array of random numbers from -pi to pi ??

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Jan
Jan am 1 Nov. 2021
See:
help rand
% Generate values from the uniform distribution on the interval (a, b):
r = a + (b - a) .* rand(x, 1);
Now set a=-pi and b=p:
r = -pi + 2*pi * rand(x, 1);
Actually we do not solve homework questions in the forum. But here I hope you see, how useful the hint is to take a look into the documentation. For questions concerning a command, start with help() and doc(). If this does not help directly, look at the bottom in the "See also" line, which contains other similar commands.

Weitere Antworten (1)

M.MUSBA Elhadid
M.MUSBA Elhadid am 31 Okt. 2021
Bearbeitet: M.MUSBA Elhadid am 31 Okt. 2021
x = ones(1,100); a = [rand(1,length(x))-0.5]+linspace(-pi+0.5,pi-0.5,length(x));
  2 Kommentare
John D'Errico
John D'Errico am 1 Nov. 2021
Bearbeitet: John D'Errico am 1 Nov. 2021
I'm sorry, but do you realize this solution does not generate uniformly distributed random numbers? Worse, it requires far more effort than the true solution.
Does it work? Is it even remotely close? Perhaps they only way to prove it is to do a large simulation.
a = zeros(1000,100);
for i = 1:1000
x = ones(1,100); a(i,:) = [rand(1,length(x))-0.5]+linspace(-pi+0.5,pi-0.5,length(x));
end
hist(a(:),1000)
Does that look even remotely uniformly distributed? See the answer by Jan for a far better solution. And even worse, you are trying to do what is surely someone's homework assignment. We do not do homework assignments for students on this site.
M.MUSBA Elhadid
M.MUSBA Elhadid am 1 Nov. 2021
thank you for the advice. I will check next times.

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