How do I find 4 or more consecutive zeros and replace these zero's?
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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.
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採用された回答
Guillaume
2016 年 7 月 6 日
編集済み: Guillaume
2016 年 7 月 6 日
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 件のコメント
Image Analyst
2022 年 12 月 3 日
@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.
その他の回答 (3 件)
Image Analyst
2016 年 7 月 6 日
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 件のコメント
Azzi Abdelmalek
2016 年 7 月 6 日
編集済み: Azzi Abdelmalek
2016 年 7 月 6 日
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 件のコメント
Azzi Abdelmalek
2016 年 7 月 6 日
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)}))
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