Circular indexing of an array

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Wade
Wade 2012-7-31
编辑： amir moin 2021-7-23
I have run into this a number of times in a recent project I'm working on and was wondering what good solutions are out there. What happens is I often have an array representing positions on a line.
X = 1:10;
I want to create a function that indexes the code and circularly wraps around such that when I go through the first iteration of my loop idx-1 = 0 is interpreted as a 10 instead.
for i = 1:10
(X(i-1)+X(i))/2
end
I've tried appending the end element of the array on to the beginning.
X = [10 1:10]
for i = 2:11
...
Seems kind of like an ugly solution though. Any other suggestions?
Thanks!

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采纳的回答

Walter Roberson
Walter Roberson 2012-7-31
编辑：Walter Roberson 2012-7-31
wrapN = @(x, N) (1 + mod(x-1, n));
N = length(X);
for i = 1:10
(X(wrapN(i-1,N))+X(wrapN(i,N)))/2
end
  2 个评论
Abhinav Gupta
Abhinav Gupta 2017-4-30
Just a minor typo in the wrapN function. It should be:
wrapN = @(x, n) (1 + mod(x-1, n));
Walter Roberson
Walter Roberson 2017-4-30
Ah, right. Or to be consistent with my naming,
wrapN = @(x, N) (1 + mod(x-1, N));

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更多回答（6 个）

Oleg Komarov
Oleg Komarov 2012-7-31
I have something different, a vectorized approach:
X = 1:10;
shift = 1;
B1 = (circshift(X,[1 shift]) + X)/2
B2 = (X([end-shift+1 1:end-shift]) + X)/2 %equivalent
isequal(B1,B2) % 1
Teja Muppirala
Teja Muppirala 2012-8-1
X = 1:10;
ifft(fft(X).*fft([1 1]/2,numel(X)))
bym
bym 2012-7-31
编辑：bym 2012-7-31
I don't know if it will help but you can use X as the index variable in the loop
X = [10 1:10]
for X
...
end
alternatively, if inside the loop is what you posted you can do
y = conv(X,[.5 .5]);
y([1 end]) = [];
without a loop
Andrei Bobrov
Andrei Bobrov 2012-8-1
编辑：Andrei Bobrov 2012-8-1
in this case
X = 1:10;
conv(X([end,1:end]),[1 1]/2,'valid');
Wade
Wade 2012-8-2
Thanks all!
amir moin
amir moin 2021-7-23
编辑：amir moin 2021-7-23
Taking you need a circle of length N=10; then you can set the array X(100) to the value of circular index Index(k)
N=10;
Index = @(i) mod(i,N)+(mod(i,N)==0).*N;
X = Index(1:100);
end
  3 个评论
显示 1更早的评论
amir moin
amir moin 2021-7-23
Sorry! I had made a mistake. Now the corrected solution is there. Thank you for your comment.
Walter Roberson
Walter Roberson 2021-7-23
Seems complicated compared to my
wrapN = @(x, N) (1 + mod(x-1, N));
X = wrapN(1:100, 10);

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