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)

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David Hill
David Hill am 31 Mär. 2022
Bearbeitet: David Hill am 31 Mär. 2022

0 Stimmen

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

2 Stimmen

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

Les Beckham
Les Beckham am 31 Mär. 2022
Bearbeitet: Les Beckham am 31 Mär. 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 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.
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

1 Stimme

% 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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