Sequence within a sequence
显示 更早的评论
Hi I want to be able to create a sequence as such
j=1111,2222,3333,4444
It is for use in a loop where j will jump up by 1 for every forth value of i.
thanks
采纳的回答
Wayne King
2012-3-28
Another way (you need Signal Processing Toolbox)
x = 1:4;
y = upsample(x,4);
h = ones(4,1);
y = filter(h,1,y);
3 个评论
Daniel Shub
2012-3-28
Hey for once my solution is faster ...
n = 1e7;
m = 10;
tic
x = 1:n;
y = upsample(x,m);
h = ones(m,1);
y = filter(h,1,y);
toc
tic
z = reshape(repmat(1:n, m, 1), n*m, 1);
toc
Elapsed time is 5.215208 seconds.
Elapsed time is 3.261213 seconds.
Daniel Shub
2012-3-28
Despite my answer being faster (and I think simpler), I like your answer better from a theoretical vantage.
Wayne King
2012-3-28
:) I don't doubt that at all!
更多回答（3 个）
Daniel Shub
2012-3-28
n = 5;
m = 3;
reshape(repmat(1:n, m, 1), n*m, 1)
Dr. Seis
2012-3-28
To beat a dead horse...
I thought I remembered a version by Walter using cumsum, but I couldn't find the link. Here's my attempt to recreate:
n = 5;
m = 3;
a = zeros(m,n);
a(1,:) = 1;
b = cumsum(reshape(a,1,numel(a)));
Running Daniel's test, I got ~0.65 seconds and ~1.05 seconds for the reshape&cumsum and repmat&reshape versions, respectively. However, what you save in compute time is eaten up by doing extra typing.
1 个评论
Daniel Shub
2012-3-28
+1 I knew my answer wouldn't be the fastest. Not only do you have a faster answer, but you also have a faster computer (your answer is faster on my computer also).
Scott
2012-3-28
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