Subract specific areas from array

I would like to find and define specific areas in an array. For example for the next array;

Array = [0 0 0 0 2 2 1 1 1 2 2 2 2 2 2 2 2 1 1 1 2 2 0 2 2 0 0 0]

i would like to define the area's with connected 2's and divide them in two groups (the area of 2's which which is limited by two 1's, and the 2's which aren't closed by two ones, but for example two zeros or a one and a zero).

In the end i would like to keep the area which is closed in by two ones and dump the rest of my array

What would be the best way to do this?

[Jos (10584)]

What do you mean by "divide in two groups" and "dump the rest"?

It would help if you'd also given an example of the expected outcome, like:

ArrayIn  = [0 0 0 0 2 2 1 1 1 2 2 2 2 2 2 2 2 1 1 1 2 2 0 2 2 0 0 0]
ArrayOut = [0 0 0 0 0 0 1 1 1 2 2 2 2 2 2 2 2 1 1 1 0 0 0 0 0 0 0 0] %?

[Hugo]

Hi Jos,

The outcome i want is something like that, but i would like to create two arrays, so something like this:

 ArrayIn  = [0 0 0 0 2 2 1 1 1 2 2 2 2 2 2 2 2 1 1 1 2 2 0 2 2 0 0 0];
ArrayOut1 = [0 0 0 0 0 0 0 0 0 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 0 0];
ArrayOut2 = [0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 0 2 2 0 0 0];  %

So i want to have one array with the 2's which are locked in between two 1's and i want one array with al 2's which do not meet this condition.

Answer 1

[Azzi Abdelmalek]

A= [1 2 2 1 1 1 2 2 2 2 2 1 1 2 2 0 2 2 0 0 0 1  2 2 2 1 0 1 ];
ArrayOut1=zeros(size(A));
ArrayOut2=ArrayOut1;
idx1=union(strfind(A,[1 2]),strfind(A,[0 2]));
idx2=union(strfind(A,[2 1]),strfind(A,[2 0]))+1;
v=all(A([idx1' idx2']),2);
ArrayOut1(cell2mat(arrayfun(@(x) idx1(x)+1:idx2(x)-1,find(v)','un',0)))=2;
ArrayOut2(cell2mat(arrayfun(@(x) idx1(x)+1:idx2(x)-1,find(~v)','un',0)))=2;

Or in case A start (or/and) ends with 2

A= [2 2 1 1 1 2 2 2 2 2 1 1 2 2 0 2 2 0 0 0 1 2 2 2 1 0 1 2 ];
ArrayOut1=zeros(size(A));
ArrayOut2=ArrayOut1;
A=[0 A 0];
idx1=union(strfind(A,[1 2]),strfind(A,[0 2]));
idx2=union(strfind(A,[2 1]),strfind(A,[2 0]))+1;
v=all(A([idx1' idx2']),2);
ArrayOut1(cell2mat(arrayfun(@(x) idx1(x):idx2(x)-2,find(v)','un',0)))=2
ArrayOut2(cell2mat(arrayfun(@(x) idx1(x):idx2(x)-2,find(~v)','un',0)))=2

[Hugo]

Thanks a lot, this was exactly the thing i searched for!

Answer 2

[Walter Roberson]

nonz = [0 (ArrayIn ~= 0) 0];
begin_groups = strfind(nonz, [0 1]);
end_groups = strfind(nonz, [1 0]);

Now look at the locations indicated by begin_groups and see if [2 2] starts there; likewise look in the corresponding end_groups and see if it ends with [2 2]

Answer 3

[Azzi Abdelmalek]

ArrayIn= [0 0 0 0 2 2 1 1 1 2 2 2 2 2 2 2 2 1 1 1 2 2 0 2 2 0 0 0 ]
B=num2str(ArrayIn);
B=strrep(B,' ','');
[ii1,ii2]=regexp(B,'(?<=1)2+(?=1)','start','end');
[jj1,jj2]=regexp(B,'(?<=0)2+(?=0)|(?<=0)2+(?=1)|(?<=1)2+(?=0) ','start','end');
ArrayOut1=zeros(size(ArrayIn));
ArrayOut2=ArrayOut1;
ArrayOut1(cell2mat(arrayfun(@(x,y) x:y,ii1,ii2,'un',0)))=2
ArrayOut2(cell2mat(arrayfun(@(x,y) x:y,jj1,jj2,'un',0)))=2

Answer 4

[Andrei Bobrov]

[a,b] = regexp(num2str(A(:))','(?<=1)2*(?=1)');
t = false(size(A));
for jj = 1:numel(a)
    t(a(jj):b(jj)) = true; 
end
out = A.*(A==2);

out1=out.*t;
out2 = out.*~t;

or without num2str and regexp

t = [true;diff(A(:))~=0];
n = A(t);
ii = find(t);
m = bsxfun(@plus,strfind(n(:)',[1 2 1]),(1:2)');
tt = false(size(A));
for jj = 1:size(m,2), tt(ii(m(1,jj)):ii(m(2,jj))-1) = true; end

out = A.*(A==2);
out1 = out.*tt;
out2 = out.*~tt;