Create vector with unique values

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Lev Mihailov
Lev Mihailov am 31 Mär. 2022
Kommentiert: Bruno Luong am 31 Mär. 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)

Akzeptierte Antwort

David Hill
David Hill am 31 Mär. 2022
Bearbeitet: David Hill am 31 Mär. 2022
v=1+rand(1,5000);
  1 Kommentar
Bruno Luong
Bruno Luong am 31 Mär. 2022
Bearbeitet: Bruno Luong am 31 Mär. 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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Weitere Antworten (2)

Bruno Luong
Bruno Luong am 31 Mär. 2022
Bearbeitet: Bruno Luong am 31 Mär. 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 Kommentare
Bruno Luong
Bruno Luong am 31 Mär. 2022
Bearbeitet: Bruno Luong am 31 Mär. 2022
Yes, you point correctly unique sort the random stream.
+1 Good point alsoo using 'stable' option and avoid randperm.
Bruno Luong
Bruno Luong am 31 Mär. 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 am 31 Mär. 2022
Bearbeitet: Bruno Luong am 31 Mär. 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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