How to replace the numbers in matrix?
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Hi,
I'm trying to replace the zeros to the average numbers in between before and after numbers, how to do it?
[1,2,3,0,5,6,0,8,9 ; 1,3,0,0,0,11,13,15,17]
become
[1,2,3,4,5,6,7,8,9 ; 1,3,5,7,9,11,13,15,17]
Thank you
2 Comments
Adam Danz
on 18 Jan 2022
In the updated question, it's not clear how the first two rows below become the last two rows. It looks like you're taking the average in some cases but it's not clear what happens when there are consecutive 0s. Please explain the conversion rule(s).
1 2 3 0 5 6 0 8 9
1 3 0 0 0 10 12 14 16
1 2 3 4 5 6 7 8 9
1 3 5 7 9 10 12 14 16
Accepted Answer
Voss
on 18 Jan 2022
Here is a way to do it using linear interpolation on each row. It will interpolate between the non-zero numbers to fill in the zeros (at least two non-zero numbers are required on a given row in order to interpolate) and leave zeros at the beginning or end of the row unchanged. Demonstrated with a slightly more general case:
A = [1,2,3,0,5,6,0,8,9 ; 1,3,0,0,0,11,13,15,17; 0 0 0 0 0 0 0 0 0; 0 0 3 0 0 7 0 0 0]
c = 1:size(A,2);
for i = 1:size(A,1)
idx = A(i,:) == 0;
if nnz(~idx) < 2
continue
end
A(i,idx) = interp1(c(~idx),A(i,~idx),c(idx),'linear',0);
end
display(A);
4 Comments
Voss
on 18 Jan 2022
@Han-Yun Chiang Sorry, I never use "Import Data", so I don't know if there's a way to apply this code when importing data from a file. You may have to just import the data and then run this code on it afterwards.
More Answers (1)
Adam Danz
on 17 Jan 2022
>I'm trying to replace the zeros to the intermediate numbers
I assume you mean you want to replace 0s with the value immediatly above or below the 0. If the entire columns contains 0s, nothing will change.
A = [1,2,3,0,5,6,0,8,9 ; 1,3,0,0,0,10,12,14,16]
A(A==0) = A(flipud(A)==0)
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