choosing n set from all possible permutations

assume I have a vector 1 by 100. I want to choose 5 possible permutations (each 1 by 100)? (hint: from all possible permutations I need five of them)...how? tnx

Answer 1

[the cyclist]

One of many possible ways to do this is using the randperm() command.

[Abed]

thanks...but lets say I have A = [2 5 7 11 13] then I need five different samples each 1 by 5, with permutation from A. how?

Answer 2

[Star Strider]

Use perms.

v = [2 5 7 11 13];

p = perms(v)

Choose any five rows of p.

[Abed]

but it does not works when v is 1 by 100. when v is 1 by 100 what should i do?

[Star Strider]

So far, you’ve ruled out randperms and perms. Seems you’ll have to write your own function.

One way might be to preallocate a (100x100) matrix, then take your 100-element vector and select 100 random elements from it without repetitions, then do that for each subsequent row of your matrix, checking that each new row is not a repetition of any previous row. (This will likely take a while to run.) Then, choose five rows.

For the row constructions, the unique function might be your friend. That will help you detect and replace any repetitions.

Just subtract your current row from previous rows to see if they're unique (although the unique function might help you there, too). If any difference is zero, repeat, if not, create the next row.

[Star Strider]

If repetitions are allowed and you only need 10 rows of 100 numbers, they’re simply random numbers.

This works:

M = randi(100, 10, 100)

Answer 3

[Image Analyst]

Try this:

clc; workspace;

% Create the original 1 by 100 vector.
v = randi(99,[1,100])
% Now get 5 new vectors which are v scrambled
% into different permutations.
for k  = 1 : 5
  newOrder = randperm(100);
  newV = v(newOrder)
end

You can have newV be a 2D array if you want:

% Create the original 1 by 100 vector.
v = randi(99,[1,100])
% Now get 5 new vectors which are v scrambled
% into different permutations.
for k  = 1 : 5
  newOrder = randperm(100);
  newV(k, :) = v(newOrder)
end

[Abed]

thank you so much...the only issue remains is this; I need these sets with replacement. These are without replacement...what can i do now?tnx

[Abed]

I did it for smaller set just for simplicity. clc clear all % Create the original 1 by 100 vector. v = [2 3 5 6 11] % Now get 5 new vectors which are v scrambled % into different permutations. for k = 1 : 5 newOrder = randperm(5); newV = v(newOrder) end

these sets are without replacement. I need with replacement. for example for this case I expect to have [11 2 2 5 5] as well.

[Image Analyst]

I don't know what with or without replacement means in this context. I just scrambled the order of the existing elements. That's what we've all suggested - isn't that what you want?