How do I find 4 or more consecutive zeros and replace these zero's?

I have a large matrix and need to loop through it and find where there are instances of 4 or more 0's and replace these 0's with 2's. For example, in the below image I need to replace the 4 or more consecutive 0's (red) with 2's but the other 0 further down is fine. The image on the right is what I'm expecting.

I can count the number of times 0 appears but am stuck on how to alter these 0's. This is what I have so far:

for i = 1:length(x)

    y = x(:,end);

    if y(i) == 0
        count = count + 1;

    else if count >= 4
            lastIndex = i - 1;

            %change 0's in the block of 4 to 2's
            %this is the bit I'm stuck on            

            count = 0; %reset count
        end
        count = 0;
    end
end

Any help would be greatly appreciated.

Answer 1

[Guillaume]

Here is one way to do it, which does not involve looping over the whole vector (only over each run of zero)

transitions = diff([0; x == 0; 0]); %find where the array goes from non-zero to zero and vice versa
runstarts = find(transitions == 1);
runends = find(transitions == -1); %one past the end
runlengths = runends - runstarts;
%keep only those runs of length 4 or more:
runstarts(runlengths < 4) = [];
runends(runlengths < 4) = [];
%expand each run into a list indices:
indices = arrayfun(@(s, e) s:e-1, runstarts, runends, 'UniformOutput', false);
indices = [indices{:}];  %concatenate the list of indices into one vector
x(indices) = 2 %replace the indices with 2

[chels19]

Thank you for your help, it works brilliantly.

[Ancalagon8]

Is it possible to assign the total number of zeros to dates?

[Image Analyst]

@Ancalagon8 I suggest you start a new question with your data attached and say what your desired output is. Include the code where you're trying to do what you want to do.

Answer 2

[Image Analyst]

Here's a way using regionprops to find the areas >= 4 and replace them with 2:

m = [0 0  1 1 0 0 0 0 1 1 1 0 0 1 0 0 0 0 0 0 0 1 1 0 1]'
props = regionprops(bwlabel(m==0), 'Area', 'PixelIdxList');
indexesOf4orMore = find([props.Area] >= 4)
for k = indexesOf4orMore
  theseIndexes = props(k).PixelIdxList
  m(theseIndexes) = 2;
end
m % Echo result to command window.

Answer 3

[Azzi Abdelmalek]

A=[0 0  1 1 0 0 0 0 1 1 1 0 0 1 0 0 0 0 0 0 0 1 1 0 1]'
ii=strfind([1 A'],[1 0])
jj=strfind([A' 1],[0 1])
kk=find(jj-ii+1>=4)
for k=1:numel(kk)
  idx=ii(kk(k)):jj(kk(k))
  A(idx)=2*ones(numel(idx),1)
end

[chels19]

Thank you for your help

[Phani kumar KSV]

Simple and superb code... thankyou

[Amin Azari]

Nice.

Answer 4

[Azzi Abdelmalek]

A=[0 0  1 1 0 0 0 0 1 1 1 0 0 1 0 0 0 0 0 0 0 1 1 0 1]'
a=cumsum(A)+1
v=cell2mat(accumarray(a,(1:numel(a))',[],@(x) {2*(numel(x)>=4 & A(x)==0)+A(x)}))