How do I generate a vector of ordered integer without a loop or in faster way

1 Ansicht (letzte 30 Tage)
I have to generate a vector 0123123451234567123456789 and the quickest idea was given by cyclist
M=[0];
for q=3:2:9
M=[M,1:q];
end
which is very slow. Does anyone have idea to make it faster? Thanks alot

Antworten (2)

dpb
dpb am 25 Nov. 2015
Prealllocate and populate instead of concatenating for starters...
M=zeros(1,1+sum(q));
i1=2;
for q=3:2:9
i2=i1-1+q;
M(i1:i2)=[1:q];
i1=i2+1;
end
I gotta' run at the moment but a kreative use of
doc kron % _might_ bring nirvana, not sure otomh; haven't had time to think it thru...
  3 Kommentare
Oluwaseun Lijoka
Oluwaseun Lijoka am 25 Nov. 2015
I also think matlab will complain about the first line M=zeros(1,1+sum(q)); where q is not defined.
dpb
dpb am 25 Nov. 2015
Well, you have to define the range, yes...

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Stephen23
Stephen23 am 25 Nov. 2015
Bearbeitet: Stephen23 am 25 Nov. 2015
You can adapt my answer to your other question:
X = floor(sqrt(0:(p+1)^2-1));
Y = diff(X);
Y(logical(Y)) = 0:-2:-2*nnz(Y)+1;
Y(Y==0) = 1;
Z = [0,cumsum(Y)];
This is hundreds of times faster than your original code (p=10000):
Elapsed time is 3.095720 seconds.
Elapsed time is 0.000041 seconds.
Note that as p gets larger it will reach your memory limit very quickly: each magnitude increase in p causes a two-magnitude increase in the number of elements of the output vector.

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