Quickest way for alternate indexing a vector

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Nicolas Douillet
Nicolas Douillet 2023-12-21
评论： Matt J 2023-12-21
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 个评论
Mathieu NOE
Mathieu NOE 2023-12-21
Ran in:
hello
IMHO this seems not very slow
n = 1e6;
tic
u = 1+cumsum(repmat([4 2],[1 floor(n/6)]));
toc
Elapsed time is 0.003157 seconds.
size(u)
ans = 1×2
1 333332
Nicolas Douillet
Nicolas Douillet 2023-12-21
编辑：Nicolas Douillet 2023-12-21
Thank you all for your interest and solutions :-)
I am going to test them, pick the best in my case, and then choose for the corresponding best answer too.
I actually realize this challenge is very difficult since the goal in my case is nothing less than beating the semi colon operator indexing speed (!)
The only way I guess would be a real/direct semi-colon alternate indexing.

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采纳的回答

Stephen23
Stephen23 2023-12-21
Ran in:
u = [5,7,11,13,17,19,23,25,29,31,35,37,41,43,47,49,53,55]
u = 1×18
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) = []
v = 1×18
5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 53 55
  4 个评论
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更多回答（4 个）

Matt J
Matt J 2023-12-21
编辑：Matt J 2023-12-21
Ran in:
n=9;c=[4;2];
s=c(1)+c(2);
c(2)=s;
u= 1 + c + (0:s:s*(n-1));
u=u(:)'
u = 1×18
5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 53 55
Dyuman Joshi
Dyuman Joshi 2023-12-21
Ran in:
n = 9;
u = [5:6:6*(n-1)+5; 7:6:6*(n-1)+7];
u = reshape(u, 1, [])
u = 1×18
5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 53 55
Matt J
Matt J 2023-12-21
编辑：Matt J 2023-12-21
Ran in:
n=9;c=[4;2];
s=c(1)+c(2);
clear u
u(2:2:2*n)=1+s:s:s*n+1;
u(1:2:2*n)=u(2:2:2*n)-c(2)
u = 1×18
5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 53 55
Bruno Luong
Bruno Luong 2023-12-21
编辑：Bruno Luong 2023-12-21
Ran in:
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
Elapsed time is 0.028802 seconds. Elapsed time is 0.024224 seconds. Elapsed time is 0.002394 seconds. Elapsed time is 0.001397 seconds.
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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