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

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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]

Risposta accettata

Walter Roberson
Walter Roberson il 10 Lug 2021
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

Più risposte (4)

Soniya Jain
Soniya Jain il 10 Lug 2021
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.
  2 Commenti
PRIYAM DEKA
PRIYAM DEKA il 10 Lug 2021
this code works but if the last element is non zero the code doesnot work.
though it can be corrected by defining a=[a 0] before the while loop.
thank you

Accedi per commentare.


Simon Chan
Simon Chan il 10 Lug 2021
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 il 10 Lug 2021
Modificato: Matt J il 10 Lug 2021
Using this File Exchange submission
it is quite easy.
result=groupFcn(@sum, a, groupTrue(a~=0))

Image Analyst
Image Analyst il 11 Lug 2021
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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