How to produce data randomly from a given matrix without changing the dimension

4 visualizzazioni (ultimi 30 giorni)
I have some data for eg
A=[1;3;5;7;2]
Now i have to produced all these data randomly. Is there any matlab command to produce these data randomly without changing the dimension...Please help...

Risposta accettata

Jan
Jan il 6 Apr 2014
Modificato: Jan il 6 Apr 2014
Perhaps you mean something like this:
A = [1;3;5;7;2]
for k = 1:10
A = A(randperm(length(A)))
end

Più risposte (2)

Image Analyst
Image Analyst il 6 Apr 2014
Yes. It's a 5 row column vector. Of course it's possible to produce it, it just depends on what you mean by randomly. Do you just want to keep using randi(7,5) until your number match those? Like this:
A=[1;3;5;7;2]
% Use a Monte Carlo approach to generate A randomly
count = 1;
while count < 10000000
rv = randi(7, [5, 1]);
difference = abs(rv-A);
if max(difference) == 0
break;
end
count = count + 1;
end
message = sprintf('Found it after %d iterations', count);
uiwait(helpdlg(message));
Please clarify.
  2 Commenti
Tinkul
Tinkul il 6 Apr 2014
Suppose A=[1;3;5;7;2] I mentioned randomly, which means
A1=[1;7;3;5;2] or A1=[2;5;1;3;7] etc.etc.. That means the row of A1 matrix have the value which is also in A but in different row.
Image Analyst
Image Analyst il 7 Apr 2014
This would have been helpful to include in your original post, don't you think? But it looks like Jan's code does what you want.

Accedi per commentare.


Jos (10584)
Jos (10584) il 7 Apr 2014
Do you want a random permutation or a random derangement (i.e., none of the elements stays in place, as suggested by your comment to Image Analyst)?
% random permutation
B = A(randperm(numel(A))
% random derangement
B = A ;
if numel(A)>1
while any(B==A) % a rejection method
B = A(randperm(numel(A)) ;
end
else
disp('No derangement possible.')
end

Categorie

Scopri di più su Loops and Conditional Statements in Help Center e File Exchange

Tag

Non è stata ancora inserito alcun tag.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by