MATLAB Answers

Random number with no repeats in set matrix

117 views (last 30 days)
Hello again.
I want to make a matrix with random numbers that do not repeat in each row in a k X p matrix
and can be modified depending on inputs.
So far i have a code:
clc
k=input('# rows ');
p=input('# columns')
S=randperm(p);
S=S(1:(k*p));
S=reshape(S,k,p)
This works some times but not all the times.
Please help?

  4 Comments

Show 1 older comment
Rainaire Hansford
Rainaire Hansford on 19 Apr 2019
Ok so I decided to do this:
clc
k=input('# of rows ')
s=randi(100,[k,10])
Now I know numbers will repeat.
Is there a way to make a code detect repeated numbers?
Walter Roberson
Walter Roberson on 20 Apr 2019
hasdup = any( sum(bsxfun(@eq, s, permute(unique(s), [2 3 1])),2) > 1, 3);
Now hasdup is true for the rows of s which contain duplicate values within the row (and so you will need to regenerate the row.)
It is not clear to me why you do not use the sort-based code that I posted: it is guaranteed that the rows of S will contain unique integers in the range 1 to p with no need to regenerate.
Rainaire Hansford
Rainaire Hansford on 20 Apr 2019
it is because i wanted to change the range any time so far it only give me number 1 to 5 at random.
How do i get it to give me number from 1 to 100 but I want a k by p matrix?

Sign in to comment.

Accepted Answer

Walter Roberson
Walter Roberson on 14 Apr 2019
Old trick that randperm() used to implement:
[~, S] = sort(rand(k, p), 2);

  14 Comments

Rainaire Hansford
Rainaire Hansford on 23 May 2019
Ok so I got this code it randomize the order between 1 and 5.
clc
k=1;
s=5;
w=1;
p=5;
S=randperm(s);
S=S(1:(k*5));
S=reshape(S,k,5)
[~, Stemp] = sort(rand(w, s), 2);
winner = Stemp(:, 1:p)
any( ismember(sort(S,2), sort(winner,2), 'rows') )
When I run it I get:
S =
2 1 5 4 3
cost = 2
winner =
2 1 5 3 4
ans = 1
Does "ans=1" mean that all of them match or only one of them match?
If it means all of them match is there a way to see if less then all of them match and is there a way to make the code check all rows?
Rainaire Hansford
Rainaire Hansford on 23 May 2019
Also this code will not allow me to run mutliple rows for S. When I change "k" to 2 or more it will give a error saying Im out of bound or something like so.
Rainaire Hansford
Rainaire Hansford on 4 Jun 2019
What can I do to make my code find any amount of matching numbers and not all of them?
Please provide example. Thank you

Sign in to comment.

More Answers (2)

Steven Lord
Steven Lord on 21 Jun 2019
You have a requirement that no row may contain the same number twice. Can different rows in the same matrix contain the same number? So in the example below, A would not be acceptable (the second row has two 3's) but B would be acceptable (the first and second rows each have 2)?
A = [1 2; 3 3];
B = [1 2; 2 3];
If so, call randperm once for each row, telling it to generate size(B, 2) numbers from 1 to your upper limit.
nrows = 10;
ncols = 7;
maxvalue = 100;
A = zeros(nrows, ncols);
for whichrow = 1:nrows
A(whichrow, :) = randperm(maxvalue, ncols);
end
If you call rng default immediately before calling that code the resulting matrix should have three rows that contain the number 4 and three rows that contain 89, but that neither 4 nor 89 repeat in the same row.

  0 Comments

Sign in to comment.


Rainaire Hansford
Rainaire Hansford on 21 Jun 2019
What can I do to make my code find any amount of matching numbers and not all of them?
Please provide example. Thank you

  7 Comments

Show 4 older comments
Walter Roberson
Walter Roberson on 28 Jun 2019
By the way, there are vectorized solutions that can work when the amount of data is not big.
Walter Roberson
Walter Roberson on 28 Jun 2019
You can use reshape and bsxfun to compare each member of one array to each member of another, resulting in a k by s by w by p array. You then any() along one of the dimensions and sum along a different dimension and squeeze to get a 2d array of results.

Sign in to comment.

Products


Release

R2018a

Community Treasure Hunt

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

Start Hunting!