I have this problem
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I have a column vector
k =[ 0 0 0 0 0 0 2 5 6 8 7 2 1 0 0 0 0 3 5 4 1 8 6 8 ];
I want to keep the shape of the vector while replacing every 1st and 3rd non zero element with 100 so I end up with
k = [ 0 0 0 0 0 0 100 5 6 8 100 100 1 2 5 100 0 100 5 4 1 100 6 8];
2 个评论
Rik
2018-9-19
How are you counting? I can't reproduce your vector by any way I can think of
% 1 2 3 1 2 3 1 1 2 3 1 2 3 1
% 1 2 3 1 2 3 1 2 3 1 2 3 1 2
% (1 2 3 1 2 3)1 2 3 1 2 3 1(2 3 1 2)3 1 2 3 1 2 3
k=[ 0 0 0 0 0 0 2 5 6 8 7 2 1 0 0 0 0 3 5 4 1 8 6 8 ];
% x x x x x x x x x x
% x x x x x x x x x
% x x x x x x x x x x
None of these patterns result in your second vector. Also, there seem to be some extra values. Please explain what your pattern is, then we can try to help you find a way to implement it.
采纳的回答
The easiest way to do this is with a loop. A second method is less easy to follow, but removes the need for a loop. The final statement shows that the two methods are indeed interchangeable.
% 1 2 3 1 2 3 1 1 2 3 1 2 3 1
k = [ 0 0 0 0 0 0 2 5 6 8 7 2 1 0 0 0 0 3 5 4 1 8 6 8 ];
k2=k;
counter=0;
for n=1:numel(k)
if n>1 && k(n-1)==0
counter=0;%restart for next run
end
counter=counter+1;
if k(n)~=0 && ...
mod(counter,3)~=2 %"not 2" is equivalent to "1 or 3"
k2(n)=100;
end
end
run_start_inds=find([k(1)~=0 diff(k~=0)>0]);
base=ones(size(k));
base(run_start_inds)=base(run_start_inds)-mod(run_start_inds,3)+1;
countlist=cumsum(base);
countlist(k==0)=NaN;
countlist=mod(countlist,3);
k3=k;
k3(countlist==mod(1,3))=100;
k3(countlist==mod(3,3))=100;
isequal(k2,k3)
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