Create vector with unique values

I need to create a vector of length 5000 in the interval from 1 to 2 with unique values ​​(so that there are no repetitions), is it possible to do this? (the randi command gives me the values, but there appear repetitions)

 Respuesta aceptada

David Hill
David Hill el 31 de Mzo. de 2022
Editada: David Hill el 31 de Mzo. de 2022
v=1+rand(1,5000);

1 comentario

Bruno Luong
Bruno Luong el 31 de Mzo. de 2022
Editada: Bruno Luong el 31 de Mzo. de 2022
You can't be sure there is no repetition, especially consider the number of floating point numbers in (0,1) and generate by rand() on a computer are finite (but large), but I admit the chance is tiny.

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Más respuestas (2)

Bruno Luong
Bruno Luong el 31 de Mzo. de 2022
Editada: Bruno Luong el 31 de Mzo. de 2022
Rejection method, it likely needs a single iteration
n = 5000;
while true
r = unique(1+rand(1,round(n*1.1)));
p = length(r);
if p >= n
r = r(randperm(p,n));
break
end
end
r
r = 1×5000
1.0939 1.1626 1.1336 1.1551 1.2421 1.0354 1.5319 1.3017 1.2257 1.9326 1.8532 1.9819 1.9699 1.2436 1.2587 1.9323 1.9716 1.1154 1.2823 1.7038 1.7327 1.1640 1.9592 1.9968 1.1904 1.6269 1.1277 1.5514 1.2853 1.3155
% check
all(r>=1 & r<=2)
ans = logical
1
length(unique(r))==length(r)
ans = logical
1

4 comentarios

Les Beckham
Les Beckham el 31 de Mzo. de 2022
Editada: Les Beckham el 31 de Mzo. de 2022
I think this is the most correct approach.
Just curious, though, @Bruno Luong, why do the randperm "scrambling" of r? Couldn't you just take the first n elements since r is already random?
Actually, I forgot that unique sorts by default. One could use
r = unique(1+rand(1,round(n*1.1)), 'stable');
to avoid the sorting and then just truncate the result to the first n elements.
Bruno Luong
Bruno Luong el 31 de Mzo. de 2022
Editada: Bruno Luong el 31 de Mzo. de 2022
Yes, you point correctly unique sort the random stream.
+1 Good point alsoo using 'stable' option and avoid randperm.
Here is complete code with modification suggested by @Les Beckham
n = 5000;
while true
r = unique(1+rand(1,round(n*1.1)),'stable');
p = length(r);
if p >= n
r = r(1:n);
break
end
end

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Bruno Luong
Bruno Luong el 31 de Mzo. de 2022
Editada: Bruno Luong el 31 de Mzo. de 2022
% I'm sure there is no repetition but the set of values is not random
r = 1+randperm(5000)/5000;
% check
all(r>=1 & r<=2)
length(unique(r))==length(r)

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