find sum of the run length of non zero elements between zeros of a matrix

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PRIYAM DEKA
PRIYAM DEKA 2021-7-10
回答： Image Analyst 2021-7-11
suppose my matrix is
a=[ 1 2 3 0 0 0 4 5 6 0 0 0 7 8 0 0 9 0 0 ]
output wanted is
[6 15 15 9]

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采纳的回答

Walter Roberson
Walter Roberson 2021-7-10
Ran in:
a = [ 1 2 3 0 0 0 4 5 6 0 0 0 7 8 0 0 9 0 0 ]
a = 1×19
1 2 3 0 0 0 4 5 6 0 0 0 7 8 0 0 9 0 0
mask = logical(a);
starts = strfind([0 mask], [0 1]);
stops = strfind([mask 0], [1 0]) + 1;
ca = cumsum([0 a]);
ca(stops) - ca(starts)
ans = 1×4
6 15 15 9

更多回答（4 个）

Soniya Jain
Soniya Jain 2021-7-10
a=[ 1 2 3 0 0 0 4 5 6 0 0 0 7 8 0 0 9 0 0 ];
j = 0;
i = 1;
while i ~= (length(a)+1)
if a(i)~=0
sum = 0;
while a(i)~=0
sum = sum + a(i);
i = i + 1;
end
j = j + 1;
res(j) = sum;
end
i = i + 1;
end
Here is the code of it, but if you are not familiar with how to write MATLAB code, then you can start with the MATLAB Onramp tutorial to quickly learn the essentials of MATLAB.
Simon Chan
Simon Chan 2021-7-10
a=[ 1 2 3 0 0 0 4 5 6 0 0 0 7 8 0 0 9 0 0 ];
pos=diff(a)<0; % Position before hitting a zero
c=cumsum(a); % Cumulative sum
d=[0 c(pos)]; % Add a zero in the first position, otherwise first value will be lost
diff(d)
Matt J
Matt J 2021-7-10
编辑：Matt J 2021-7-10
Using this File Exchange submission
https://www.mathworks.com/matlabcentral/fileexchange/78008-tools-for-processing-consecutive-repetitions-in-vectors
it is quite easy.
result=groupFcn(@sum, a, groupTrue(a~=0))
Image Analyst
Image Analyst 2021-7-11
Here's another way:
a = [ 1 2 3 0 0 0 4 5 6 0 0 0 7 8 0 0 9 0 0 ]
% Extract each run:
props = regionprops(logical(a), a, 'PixelValues')
% Sum up each run.
for k = 1 : length(props)
theSums(k) = sum(props(k).PixelValues)
end

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