I have large matrix and i want to find the number of consecutive zeros if there is more than 10 disply(hole) and give the index of the start and end of this vector of zeros,

x=[1 0 1 4 8 4 3 5 9 0 0 0 0 0 0 0 0 0  0 0 0 0  4 4 0 0 0 0 2  4 4 4 4 0 0 0 0 0 0 0 0 0 3 5 6 3 0 0 0 0 4 4 4 0 0 0 0 5 4 1 0 0 0 0 0 4 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 9 ]

how can i find the start and end index of sequence of zeros if the length more than 10?

x=[1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 2 2 0 0 0 1 1 0 0 0 0 4 4   ];
c=0;k=0;z=0;
for i=1:length(x);
    if  x(i)==0;
       c=c+1;
    else
        k=k+x(i);
            if c>10;
        disp(c);
        disp('hole');
    else
       c=0;
    end
    end        
end           

[ali alabodi]

Im so sorry i dont note that i not accept ur answer sorry sir

Answer 1

[Guillaume]

x=[1 0 1 4 8 4 3 5 9 0 0 0 0 0 0 0 0 0  0 0 0 0  4 4 0 0 0 0 2  4 4 4 4 0 0 0 0 0 0 0 0 0 3 5 6 3 0 0 0 0 4 4 4 0 0 0 0 5 4 1 0 0 0 0 0 4 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 9 ]

transitions = find(diff([1, x ~= 0, 1]));
runstarts = transitions(1:2:end);
runends = transitions(2:2:end) - 1;
runlengths = runends - runstarts + 1;
tokeep = runlengths >= 10;
runstarts = runstarts(tokeep)
runends = runends(tokeep)

[ali alabodi]

thanks mr guillaume for answer but can some explain please

more thanks for u

Answer 2

[Rik]

Use this FEX submission if you want to speed up this code.

x=[1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 2 2 0 0 0 1 1 0 0 0 0 4 4];

%use this:
%[b,n]=RunLength(x);
%or this:
ii = [ find(x(1:end-1) ~= x(2:end)) length(x) ];
n = diff([ 0 ii ]);
b = x(ii);

clc
idx= b==0 & n>=10;
for k=find(idx)
    start=1+sum(n(1:(k-1)));stop=start+n(k)-1;
    fprintf('hole from %d to %d\n',start,stop)
end

(the run length encoding is 'borrowed' from here, a FEX submission by Stefan Eireiner)

[ali alabodi]

more more more thanks for u

please can u develope this code if the condition of no; of zeros ( less than 10 ) print (no hole)

[Rik]

Sure, but you did not accept my answer, so I expect you aren't using my code anyway. But here is the edit:

x=[1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 2 2 0 0 0 1 1 0 0 0 0 4 4];
%use this:
%[b,n]=RunLength(x);
%or this:
ii = [ find(x(1:end-1) ~= x(2:end)) length(x) ];
n = diff([ 0 ii ]);
b = x(ii);
clc
idx= b==0 & n>=10;
if any(idx)
    for k=find(idx)
        start=1+sum(n(1:(k-1)));stop=start+n(k)-1;
        fprintf('hole from %d to %d\n',start,stop)
    end
else
    disp('no holes')
end

[ali alabodi]

Im so sorry sir that i m not note that i not accept ur answer , ur answer was very good i attempte to develope it to do that ((if i devide the string of very long number to several equal interval such a 3 intervals or 10 equals intrval each interval equal to (for example 1250 number) and test if the sequence of the zeros exist in select interval ? if exist then count1 = 1, else counte2=1,and repeat the attration more than one then evaluate the no; of exist and absence ?advice me to make this in seperate quastion?or here