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]

採用された回答

Image Analyst
Image Analyst 2022 年 2 月 20 日
編集済み: Image Analyst 2022 年 2 月 20 日
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 件のコメント
Md. Asadujjaman
Md. Asadujjaman 2022 年 2 月 20 日
編集済み: Md. Asadujjaman 2022 年 2 月 20 日
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 2022 年 2 月 20 日
編集済み: Image Analyst 2022 年 2 月 20 日
OK, thanks for Accepting. If you have a newer version of MATLAB you can use the 'omitnan' option like Matt showed below.

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その他の回答 (2 件)

Matt J
Matt J 2022 年 2 月 20 日
sum(1./nonzeros(x),'omitnan')

Jan
Jan 2022 年 2 月 20 日
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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