I have a 5x6 matrix. I want to calculate sum of only those values which are consecutive nonzero along each row. How to do that? The desired result given in description box.

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a = [2 3 1 4 0 0;
0 0 0 0 0 0;
0 0 1 5 7 0;
0 0 0 0 4 0;
4 5 0 0 6 8];
desired_result = [10;0;13;0;23]
desired_result = 4×1
10 0 13 0
  4 Comments
Payel
Payel on 24 Jun 2023
The result of 4th row is not 4 because I want summation of only consecutive non zero values along each row.
The desired result of a should be 17
the cyclist
the cyclist on 24 Jun 2023
I believe a correct re-statement of the rule is, "Sum all non-zeros that are next to at least one other non-zero."

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Accepted Answer

the cyclist
the cyclist on 24 Jun 2023
Edited: the cyclist on 24 Jun 2023
I believe this does what you want:
% Input
a = [2 3 1 4 0 0;
0 0 0 0 0 0;
0 0 1 5 7 0;
0 0 0 0 4 0;
4 5 0 0 6 8];
% Append columns of zeros at both ends, because we are going to check in
% the next line if an element is a "singleton" with zeros on both sides
b = [zeros(height(a),1) a zeros(height(a),1)];
% Identify the non-singleton value (i.e. the ones that do not have zeros on either side)
includeInSum = not((b(:,1:end-2) == 0) & (b(:,3:end)==0));
% Sum the values that should be included
result = sum(a.*includeInSum,2)
result = 5×1
10 0 13 0 23

More Answers (1)

Image Analyst
Image Analyst on 24 Jun 2023
a = [2 3 1 4 0 0;
0 0 0 0 0 0;
0 0 1 5 7 0;
0 0 0 0 4 0;
4 5 0 0 6 8];
desired_result = [10;0;13;0;23]
desired_result = 5×1
10 0 13 0 23
desired_result = sum(a, 2)
desired_result = 5×1
10 0 13 4 23
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  2 Comments
Payel
Payel on 24 Jun 2023
how to get the summation value of 4th row to be zero? as it is asked to find summation of only consecutive nonzero values along each row
Image Analyst
Image Analyst on 24 Jun 2023
Row 4 has one "consecutive' value, so why not sum it in? Do you want to sum in values only if the run of non-zero values is 2 or more? See @the cyclist's answer below. Is it homework? If so you can't use it.
This looks like a homework problem. If you have any questions ask your instructor or read the link below to get started:
Obviously we can't give you the full solution because you're not allowed to turn in our code as your own.

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