Circularly exclude elements.
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Abhijit Chowdhary
am 7 Apr. 2019
Kommentiert: Abhijit Chowdhary
am 8 Apr. 2019
Hello all,
I'm trying to compactly write a line of code for a interesting program I'm working on. Fixing , suppose we have the following code:
x = 1:n;
non_adj = zeros(n,n-2);
for k=1:n
non_adj(k,:) = []; %in here comes the confusion
end
In non_adj(k,:) I would like to fill in x without the elements k-1 and k+1. However, these elements are regarded circularily, i.e. if I have , then I want (order of elements irrelevant). I know I can do this with if statements handling the cases, but I was wondering if you all had a nicer way to one liner the construction.
I attempted the following construction
non_adj(k,:) = [k, k+2:mod(k-3,n)+1, mod(k+1,n)+1:k-2];
However this fails, for example, where . The above would construct the array , where in fact I wanted .
Do you have any ideas? If it helps, the context is that I'm trying to construct a vertex set minus the neighbors of vertex k, which happen to be (circularily shifted for ).
EDIT: I've realized I could use set differencing, i.e.: (however, you still might come up with better ideas).
non_adj(k,:) = setdiff([1:n], 1 + [mod(k-2,n) mod(k,n)]);
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Akzeptierte Antwort
David Goodmanson
am 7 Apr. 2019
Hi Abhijit,
try
x = 1:n;
x(mod([k-1 k+1]-1,n)+1) = []
3 Kommentare
David Goodmanson
am 8 Apr. 2019
Bearbeitet: David Goodmanson
am 8 Apr. 2019
Hi Abhijit,
I thought that for a given k, you wanted to keep k, exclude k-1 and k+1 and keep the rest of them, doing so in a circular fashion. Could you give me a quick example of where the code above doesn't work? I am having trouble finding one (n.b. both lines of code have to be run for each new value of k).
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