Replace matrix values between two values - matlab

Question

I try to replace values per row that enclosed between two values within accepatable tolerance range/distance between the two values (non-zero). For example:
A=[0 0 0 0 1 *0 *0 *0 *0 *0 -1 *0 *0 *0 *0 *0 *0 *0 *0 *0 *0 *0 *0 *0 *0 *0 -1 *0 *0 *0 1 0 0 0;
0 0 1 *0 *0 *0 *0 *0 *0 *0 -1 *0 *0 *0 *0 *0 *0 *0 *0 *0 *0 *0 *0 *0 *0 *0 -1 *0 *0 *0 1 0 0 0]
* indicates the possibles values to be considered as an enclosure. With tolerance of 6 (for instance), the values inside will be replaced with the first boundary value:
A_replaced=[0 0 0 0 1 *1 *1 *1 *1 *1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 *-1 *-1 *-1 -1 0 0 0;
0 0 1 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 *-1 *-1 *-1 -1 0 0 0]
As you can see, everything that can be considered as an enclosure is replaced by the first boundary value (including the end closure value) since the distance between the two boundaries does not exceeds 6 (smaller than tolerance). Is there vectorized method to do this for large size of matrix? Thanks in advance!

Answer 1

Because I've already answered half of this question, I will do it with two steps. Note that your phrase of the question here is not the fully general case, since you need treating matrices with some continuous non-zeros in the places here that you placed only 1. This is relying on your previous question and comments.
So first step is to find all the zeros sequences that their length is under the tolerance, and replace them with 1. Second step is your previous question- replace all non-zeros with the starting value of the sequence.
Pre-first-step will be to add zeros to the beginning of A, because doing it all with linear indexing and treat A as a long vector is a much comfort, but we don't want to catch zero-sequences that cross the columns. We can remove the zeros at the end.
A=[
0 0 0 0 0 0.5 0.9 -1 0.1 0.3 1 1 0 0 0 0.9 -1 0.1 0 0 0 0 0.3 -1 -1 0 0 0 0 0;
0 0 0 0 0 0.1 0.3 1 -1 -1 -1 -1 0 0 0 0 0 -1 -1 -1 0 0 1 1 1 0 0 0 0 0 ];
tol=6;
A=[zeros(size(A,1),tol), A]; % add zeros at beginning of each row
A=A.'; % working on transposed A let us index continuously the sequences
zero_starts=strfind([A(1)==0; A(:)~=0].',[1 0]).'; % indices where `0` sequences starts
zero_starts(end)=[]; % last sequance is not enclosed
zero_length=diff([0; find(diff(find(~A))>1)]); % length of each of these zero-sequences
zero_tol=find(zero_length<=tol); % zero-sequences that shorter than tolerance
zero_tol(strfind(diff([0; zero_tol].')-1>0,[1 0])+1)=[]; % remove each second of continuous zero-seqeunces
% replace all short zero-seqeunces with `1`
sl=arrayfun(#(x)0:x-1, zero_length(zero_tol),'uni',0);
A(repelem(zero_starts(zero_tol),zero_length(zero_tol))+cat(2,sl{:}).')=1;
Now, the second step is to replace each non-zero sequences with the first element of each sequence:
B=find(A);
A(A~=0)=repelem(A(B([0 ;find(diff(B)>1)]+1)),diff([0;find(diff(B)>1) ;length(B)]));
Lastly, we can reverse back A to its shape and size-
A=A(tol+1:end,:).'
A=[
0 0 0 0 0 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0 0 0 0 0.3 -1 -1 0 0 0 0 0;
0 0 0 0 0 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0 0 1 1 1 0 0 0 0 0 ];