Quickest way for alternate indexing a vector
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Nicolas Douillet
il 21 Dic 2023
Commentato: Matt J
il 21 Dic 2023
Hey,
I am looking for the quickest way to create a vector like this :
u = [5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 53 55]
In which steps 2 and 4 alternate as you can see.
At the moment I manage to do it like this for instance :
u = 1+cumsum(repmat([4 2],[1 9]))
Where n is a given limit value. But this is too slow to me.
Would you know a way to get the same result but quicker ? Perhaps using only Matlab semi colon operator and / or basic math (+,*) operations ? EDIT : especially when u has a lot of elements.
Thank you.
Cheers,
Nicolas
2 Commenti
Mathieu NOE
il 21 Dic 2023
hello
IMHO this seems not very slow
n = 1e6;
tic
u = 1+cumsum(repmat([4 2],[1 floor(n/6)]));
toc
size(u)
Risposta accettata
Stephen23
il 21 Dic 2023
u = [5,7,11,13,17,19,23,25,29,31,35,37,41,43,47,49,53,55]
v = 5:2:55;
v(3:3:end) = []
4 Commenti
Più risposte (4)
Dyuman Joshi
il 21 Dic 2023
n = 9;
u = [5:6:6*(n-1)+5; 7:6:6*(n-1)+7];
u = reshape(u, 1, [])
0 Commenti
Bruno Luong
il 21 Dic 2023
Modificato: Bruno Luong
il 21 Dic 2023
Try to do something clever (1st and 2nd methods) but it is slower than your code 53rd method). A small modificationof your code seems to be the most efficient (last method)
test
function test
n=1000000; % array length
tic
u=(5:3:(n-1)*3+5)-mod(0:n-1,2);
toc % Elapsed time is 0.010612 seconds.
tic
u=0:n-1;
u=5+3*u-mod(u,2);
toc % Elapsed time is 0.009249 seconds.
tic % Your method
u = 1+cumsum(repmat([4 2],[1 n/2]));
toc % Elapsed time is 0.002047 seconds.
tic % sligh modification method
u = repmat([4 2],[1 n/2]);
u(1)=5;
u = cumsum(u);
toc % Elapsed time is 0.001861 seconds.
end
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