How can I count the sum of inverse value of each non zero elements of a matrix?

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Suppose,
x=[2 1 0 0
0 1 1 1
0 1 1 1
1 0 3 1]
How can I count the sum of inverse value of each non zero elements?
The answer would be = sum (1/2 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/3 + 1/1) = 9.83
How can I count the sum of inverse value of each non zero elements when x also contains NaN?
e.g.,
x=[2 1 0 NaN
0 1 1 1
0 1 1 1
1 0 3 1]

Risposta accettata

Image Analyst
Image Analyst il 20 Feb 2022
Modificato: Image Analyst il 20 Feb 2022
x=[2 1 0 0
0 1 1 1
0 1 1 1
1 0 3 1]
x = 4×4
2 1 0 0 0 1 1 1 0 1 1 1 1 0 3 1
numNonZeros = nnz(x)
numNonZeros = 11
Not sure what you want for the second question since your sum does not include all 16 elements.
x2=x;
x2(x==0) = 1
x2 = 4×4
2 1 1 1 1 1 1 1 1 1 1 1 1 1 3 1
s = sum(1./x2(:))
s = 14.8333
% Sum 11 of the 16 -- only some of them for some reason.
% Not sure which to take and which to exclude.
thesum = (1/2 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/3 + 1/1) % 11 of the 16
thesum = 9.8333
% Now sum all 16
thesum = (1/2 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/3 + 1/1)
thesum = 14.8333
If this is your homework, you're most likely not allowed to turn my code in as your own. And please tag it as homework if it is.
Regarding your edit, to sum up only elements where x is not zero:
x2=x(:); % Make a copy as a column vector.
x2(x==0) = [] % Remove the zeros.
x2 = 11×1
2 1 1 1 1 1 1 3 1 1
s = sum(1./x2(:))
s = 9.8333
  3 Commenti
Md. Asadujjaman
Md. Asadujjaman il 20 Feb 2022
Modificato: Md. Asadujjaman il 20 Feb 2022
I want to do the following sum:
thesum = (1/2 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/3 + 1/1) = 9.8333
% 11 (non zero elements) of the 16
Note: This is not my homework.
I got my answer. Thank you.
Image Analyst
Image Analyst il 20 Feb 2022
Modificato: Image Analyst il 20 Feb 2022
OK, thanks for Accepting. If you have a newer version of MATLAB you can use the 'omitnan' option like Matt showed below.

Accedi per commentare.

Più risposte (2)

Matt J
Matt J il 20 Feb 2022
sum(1./nonzeros(x),'omitnan')

Jan
Jan il 20 Feb 2022
x = [2 1 0 0; ...
0 1 1 1; ...
0 1 1 1; ...
1 0 3 1];
r = sum(1 ./ x(x ~= 0))
r = 9.8333
x = [2 1 0 NaN; ...
0 1 1 1; ...
0 1 1 1; ...
1 0 3 1];
r = sum(1 ./ x(x ~= 0 & ~isnan(x)))
r = 9.8333

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