Is there a smarter way to create a large vector, with repeated numbers in a diminishing quantity?

20 views (last 30 days)
I need the vector to begin with 144 evenly spaced '2's, then 143 '3's, then 142 '4's, etc. I see the brute force method, but it doesn't seem very smart. Is there a way to get Matlab to create this vector for me?
thx

Accepted Answer

Sean de Wolski
Sean de Wolski on 3 Jun 2011
A one liner:
n = 144;
v = sort(nonzeros(triu(bsxfun(@times,(2:n+1).',ones(1,n)))));
or
v = sort(nonzeros(triu(toeplitz(2:n+1))));
Golf!
v=hankel(n+1:-1:2);
v=sort(v(~~v))
%33 characters
transpose it if you want a row vector.

More Answers (4)

Friedrich
Friedrich on 3 Jun 2011
I think the repmat command can help:
[repmat(2,144,1);repmat(3,143,1);repmat(4,142,1)]
  2 Comments
Igor
Igor on 3 Jun 2011
no :) 144 times?
n=144;
A1=(1:n)'*ones(1,n);
A2=tril(A1);
A3=reshape(A2...);
find... all zero elements
Friedrich
Friedrich on 3 Jun 2011
Upps^^, than this way:
start_repeat = 144;
start_number = 2;
final_size = (start_repeat+1)*start_repeat / 2;
final_vec = zeros(final_size,1);
index = 1;
for i=start_repeat:-1:1
final_vec(index:index+i-1,1) = repmat(start_number+start_repeat-i,i,1);
index = index + i;
end

Sign in to comment.


Igor
Igor on 3 Jun 2011
[1,2,2,3,3,3,4,4,4,4,...] similar task
n=1:10 trinv(n) = floor((1+sqrt(1+8*n))/2)
(<http://oeis.org/A002024>)

Oleg Komarov
Oleg Komarov on 3 Jun 2011
Using rude for run length decoding:
n = 144;
val = 1:n;
len = val;
Out = rude(len,val); % rude on fex

Andrei Bobrov
Andrei Bobrov on 3 Jun 2011
n = 144;
vm = triu(repmat(n+1:-1:1,n+1,1));
vm(vm>0);

Categories

Find more on Elementary Math in Help Center and File Exchange

Community Treasure Hunt

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

Start Hunting!