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Circular indexing of an array

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Wade
Wade am 31 Jul. 2012
Bearbeitet: amir moin am 23 Jul. 2021
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!

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 31 Jul. 2012
Bearbeitet: Walter Roberson am 31 Jul. 2012
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 Kommentare
Abhinav Gupta
Abhinav Gupta am 30 Apr. 2017
Just a minor typo in the wrapN function. It should be:
wrapN = @(x, n) (1 + mod(x-1, n));
Walter Roberson
Walter Roberson am 30 Apr. 2017
Ah, right. Or to be consistent with my naming,
wrapN = @(x, N) (1 + mod(x-1, N));

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Weitere Antworten (6)

Oleg Komarov
Oleg Komarov am 31 Jul. 2012
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 am 1 Aug. 2012
X = 1:10;
ifft(fft(X).*fft([1 1]/2,numel(X)))

bym
bym am 31 Jul. 2012
Bearbeitet: bym am 31 Jul. 2012
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 am 1 Aug. 2012
Bearbeitet: Andrei Bobrov am 1 Aug. 2012
in this case
X = 1:10;
conv(X([end,1:end]),[1 1]/2,'valid');

Wade
Wade am 2 Aug. 2012
Thanks all!

amir moin
amir moin am 23 Jul. 2021
Bearbeitet: amir moin am 23 Jul. 2021
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 Kommentare
amir moin
amir moin am 23 Jul. 2021
Sorry! I had made a mistake. Now the corrected solution is there. Thank you for your comment.
Walter Roberson
Walter Roberson am 23 Jul. 2021
Seems complicated compared to my
wrapN = @(x, N) (1 + mod(x-1, N));
X = wrapN(1:100, 10);

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