Circularly exclude elements.
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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:
, 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.
, 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 
.
. 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 ).
(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)]);
2 个评论
madhan ravi
2019-4-7
explicitly state your desired output
Abhijit Chowdhary
2019-4-7
编辑：Abhijit Chowdhary
2019-4-7
If 
, then I want 
. If 
, then I want 
. Otherwise I want 
, and I'm hoping for a one-line to cover these scenarios.
, then I want . If , then I want . Otherwise I want , and I'm hoping for a one-line to cover these scenarios.采纳的回答
David Goodmanson
2019-4-7
Hi Abhijit,
try
x = 1:n;
x(mod([k-1 k+1]-1,n)+1) = []
3 个评论
Abhijit Chowdhary
2019-4-8
Hello David,
This indeed works for the 
case, however, it seems that it doesn't for other k. Thank you, though!
case, however, it seems that it doesn't for other k. Thank you, though!
David Goodmanson
2019-4-8
编辑：David Goodmanson
2019-4-8
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).
Abhijit Chowdhary
2019-4-8
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