Create vector with unique values

10 vues (au cours des 30 derniers jours)
Lev Mihailov
Lev Mihailov le 31 Mar 2022
Commenté : Bruno Luong le 31 Mar 2022
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
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.
Bruno Luong
Bruno Luong le 31 Mar 2022
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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