Search in array with sequence

I have array like this: y=[00110011100001110011] I need search in array, and if it is sequence [..11..] then i get 5, if it sequence [..111...] then I should get 6. In the answer I should got sequence something like this:[..5665556...] So maybe know how do that?

Answer 1

[Azzi Abdelmalek]

y='00110011100001110011'
ii={'' '5' '6'}
out=char(ii(cellfun(@numel,regexp(y,'1+','match'))))'

%or

y='00110011100001110011';
a=[0 y-'0' 0];
ii5=strfind(a,[0,1,1,0]);
ii6=strfind(a,[0,1,1,1,0]);
v(ii6)=6;
v(ii5)=5;
v(~v)=[];

[Povilas]

Thanks for answer. It works, but if I have that array [00110011100001111111] And I need that result be 5, if it is in interval from 11 to 1111?

[Azzi Abdelmalek]

y='00110011100001111111001111101111'
n=7; %  maximal number of consecutive 1
ii=repmat({'6'},1,n+1)
ii(2:4)={'5'}
out=char(ii(cellfun(@numel,regexp(y,'1+','match'))))'

Answer 2

[Andrei Bobrov]

out = sprintf('%d',strfind([y '0'],'10')-strfind(['0' y],'01')+4)

ADD

out = sprintf('%d',(strfind([y '0'],'10')-strfind(['0' y],'01')>3) + 5)

Answer 3

[Povilas]

but where y='00110011100001111111001111101111' it is not in '' but in [array2] and cant read from that array

Now my code is:

x = [Masyvas];
    x >= 0.14
y1=double(ans)
z=y1'

y=num2str(z)

%y=u(find(~isspace(u)))
r = y(y ~= ' ') %delete space

n=18; %  maximal number of consecutive 1
ii=repmat({'6'},1,n+1);
ii(8:13)={'5'};
out=char(ii(cellfun(@numel,regexp(r,'1+','match'))))

And get error:

??? Index exceeds matrix dimensions.

Error in ==> Untitled at 27
out=char(ii(cellfun(@numel,regexp(r,'1+','match'))))

[Andrei Bobrov]

What is it Masyvas? Please small example for Masyvas.

[Povilas]

masyvas

[Image Analyst]

Do you expect him to type that in? Unlikely. Please make it easy for him to help you by attaching the csv or workbook file.

[Povilas]

add array file.