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)

 Réponse acceptée

David Hill
David Hill le 31 Mar 2022
Modifié(e) : David Hill le 31 Mar 2022
v=1+rand(1,5000);

1 commentaire

Bruno Luong
Bruno Luong le 31 Mar 2022
Modifié(e) : Bruno Luong le 31 Mar 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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Plus de réponses (2)

Bruno Luong
Bruno Luong le 31 Mar 2022
Modifié(e) : Bruno Luong le 31 Mar 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 commentaires

Les Beckham
Les Beckham le 31 Mar 2022
Modifié(e) : Les Beckham le 31 Mar 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 le 31 Mar 2022
Modifié(e) : Bruno Luong le 31 Mar 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 le 31 Mar 2022
Modifié(e) : Bruno Luong le 31 Mar 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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