In a logical array index the area with the most TRUE values

Below I have a logical vector:

t=[0 1 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 1 0 0 0 1 0];

which has 5 TRUE areas and 6 FALSE areas. The 3rd TRUE area has the most TRUE elements of all TRUE areas.

How can I index it logically? To be more specific I want to get the following result:

t=[0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0];

Thank you!

Answer 1

[Andrei Bobrov]

t=[0 1 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 1 0 0 0 1 0];

ii = 3;
out = cumsum([false;diff(t(:))==1]) == 3 & t(:);

or

a = cumsum([false;diff(t(:))==1]);
aa = a(t>0);
[~,ii]=max(histc(aa,unique(aa))); % or: [~,ii]=max(accumarray(aa,aa,[],@numel));
out = a == 3 & t(:);

or with Image Processing Toolbox

C = bwlabel(t);
Z = regionprops(C,'Area');
[~,ii] = max([Z.Area]);
out = C == ii;

or mix

[~,ii] = min(strfind([0,t(:)'],[0 1])-strfind([t(:)',0],[1 0]));
out = ii == bwlabel(t(:)');

[Giorgos Papakonstantinou]

Thank you! But how you find that ii=3 in this case?

[Andrei Bobrov]

Hi Giorgos! About ii in after or.

[Giorgos Papakonstantinou]

Thank you Andrei!

[Giorgos Papakonstantinou]

Just one remark! In the case that the vector t starts with TRUE and also the area with the most TRUE values is the first area, then:

out = a == 3 & t(:);

fails to give the correct result. This is because

a = cumsum([false;diff(t(:))==1]);

and not

a1 = cumsum(diff([0; t'])==1)

It's better to place a zero and then diff. (as also Azzi does with:

a=find(diff([0 t 0]));

)

[Andrei Bobrov]

I agree. Use a = cumsum(diff([0;t(:)])==1);

[Giorgos Papakonstantinou]

Thank you Andrei for the complete and extensive answer!

Answer 2

[Azzi Abdelmalek]

t=[0 1 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 1 0 0 0 1 0];
a=find(diff([0 t 0]));
idx1=a(1:2:end);
idx2=a(2:2:end)-1;
[~,nmax]=max(idx2-idx1);
t=zeros(1,numel(t));
t(idx1(nmax):idx2(nmax))=1

[Giorgos Papakonstantinou]

Thank you Azzi!