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

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Adam Quintero
Adam Quintero 2011-6-3
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

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采纳的回答

Sean de Wolski
Sean de Wolski 2011-6-3
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.

更多回答（4 个）

Friedrich
Friedrich 2011-6-3
I think the repmat command can help:
[repmat(2,144,1);repmat(3,143,1);repmat(4,142,1)]
  2 个评论
Friedrich
Friedrich 2011-6-3
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

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Igor
Igor 2011-6-3
[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 2011-6-3
Using rude for run length decoding:
n = 144;
val = 1:n;
len = val;
Out = rude(len,val); % rude on fex
Andrei Bobrov
Andrei Bobrov 2011-6-3
n = 144;
vm = triu(repmat(n+1:-1:1,n+1,1));
vm(vm>0);

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