how to count the number of zeros between 2 one's

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Monika  Kok
Monika Kok am 24 Apr. 2016
Bearbeitet: per isakson am 5 Mai 2017
i am using the code
num = [0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,1,0,0];
num = num(find(num==1, 1, 'first'):find(num==1, 1, 'last'));
% Measure the lengths of the stretches/runs of zeros.
measurements = regionprops(logical(~num), 'Area');
zerospan = [measurements.Area]
a = max(zerospan)
this is giving me answer as 9 which is correct for middle part. but i would also count the end around number of zeros. that is zeros starting from second last position and ending at fourth position. which should give me answer as 6.
thanks monica

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Andrei Bobrov
Andrei Bobrov am 24 Apr. 2016
Bearbeitet: Andrei Bobrov am 24 Apr. 2016
num = [0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,1,0,0]
measurements = regionprops(logical(~num), 'Area')
m = [measurements.Area];
out = [sum(m([1,end])), max(m(2:end-1))]
  1 Kommentar
Image Analyst
Image Analyst am 24 Apr. 2016
"which should give me answer as 6." Just be aware that this answer gives [6,9].

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Azzi Abdelmalek
Azzi Abdelmalek am 24 Apr. 2016
Bearbeitet: Azzi Abdelmalek am 24 Apr. 2016
nu=~num;
s=cumsum(num)+~num(1);
ii=strfind(num,[1 0]);
jj=zeros(size(num));
jj(ii)=1;
s=s+cumsum(jj)
f=accumarray(s',(1:numel(s))',[],@(x) sum(nu(x)));
out=[ nonzeros(f(2:end-1))' sum([f(1) f(end)])]
  2 Kommentare
Monika  Kok
Monika Kok am 24 Apr. 2016
thnq so much i was exactly looking for this only.
can you show me an approach by which i can get the starting location of maximum zero span for example: [1 0 0 0 1 0 0 0 0 0 0 0 1 1 1 1 1 0 1 ] in this case maximum zero span is starting from location 5.
thnx monica
Image Analyst
Image Analyst am 24 Apr. 2016
You almost had it. You just needed to ask for the 'PixelIdxList' when you called regionprops. This gives you the index of every element in each grouping of 0's.
num = [0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,1,0,0];
% Measure the lengths of the stretches/runs of zeros.
measurements = regionprops(logical(~num), 'Area', 'PixelIdxList');
zerospan = [measurements.Area]
% Get index of the largest run in the lengths array:
[longestRun, indexOfLongestRun] = max(zerospan)
% Get the index in the original num array:
indexOfStartOfLongestRun = measurements(indexOfLongestRun).PixelIdxList(1)
You will see that indexOfLongestRun = 9, which is where that run starts at.

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Image Analyst
Image Analyst am 24 Apr. 2016
Just add the first and last area:
result = zerospan(1) + zerospan(end);

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