Create vector with unique values

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Lev Mihailov
Lev Mihailov 2022-3-31
评论： Bruno Luong 2022-3-31
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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采纳的回答

David Hill
David Hill 2022-3-31
编辑：David Hill 2022-3-31
v=1+rand(1,5000);
  1 个评论
Bruno Luong
Bruno Luong 2022-3-31
编辑：Bruno Luong 2022-3-31
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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更多回答（2 个）

Bruno Luong
Bruno Luong 2022-3-31
编辑：Bruno Luong 2022-3-31
Ran in:
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 个评论
显示 2更早的评论
Bruno Luong
Bruno Luong 2022-3-31
编辑：Bruno Luong 2022-3-31
Yes, you point correctly unique sort the random stream.
+1 Good point alsoo using 'stable' option and avoid randperm.
Bruno Luong
Bruno Luong 2022-3-31
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 2022-3-31
编辑：Bruno Luong 2022-3-31
% 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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