How do I find 4 or more consecutive zeros and replace these zero's?
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chels19
el 6 de Jul. de 2016
Comentada: Image Analyst
el 3 de Dic. de 2022
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.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/154715/image.png)
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.
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Respuesta aceptada
Guillaume
el 6 de Jul. de 2016
Editada: Guillaume
el 6 de Jul. de 2016
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
3 comentarios
Image Analyst
el 3 de Dic. de 2022
@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.
Más respuestas (3)
Image Analyst
el 6 de Jul. de 2016
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.
0 comentarios
Azzi Abdelmalek
el 6 de Jul. de 2016
Editada: Azzi Abdelmalek
el 6 de Jul. de 2016
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
3 comentarios
Azzi Abdelmalek
el 6 de Jul. de 2016
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)}))
0 comentarios
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