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A compact way to replace zeros with Inf in a matrix

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Sim
Sim am 16 Okt. 2023
Bearbeitet: Sim am 23 Okt. 2023
Would you be so nice to suggest me a more compact way to replace zeros with Inf in the following matrix? (maybe with just one line of code?)
% Input
A = [0 3 2 5 6;
1 1 4 3 2;
9 0 8 1 1;
5 9 8 2 0;
3 1 7 6 9];
% Replace zeros with Inf
[row,col] = ind2sub(size(A),find(A==0));
for i = 1 : length(row)
A(row(i),col(i))=Inf;
end
% Output
A
A = 5×5
Inf 3 2 5 6 1 1 4 3 2 9 Inf 8 1 1 5 9 8 2 Inf 3 1 7 6 9

Akzeptierte Antwort

J. Alex Lee
J. Alex Lee am 16 Okt. 2023
Bearbeitet: J. Alex Lee am 16 Okt. 2023
You can implicitly index "linearly" for any arrays - it will do all the ind2sub and sub2ind in the background:
% Input
A = [0 3 2 5 6;
1 1 4 3 2;
9 0 8 1 1;
5 9 8 2 0;
3 1 7 6 9];
B = A;
% Replace zeros with Inf
[row,col] = ind2sub(size(A),find(A==0));
for i = 1 : 3
A(row(i),col(i))=Inf;
end
% Output
A
A = 5×5
Inf 3 2 5 6 1 1 4 3 2 9 Inf 8 1 1 5 9 8 2 Inf 3 1 7 6 9
B(B==0) = Inf
B = 5×5
Inf 3 2 5 6 1 1 4 3 2 9 Inf 8 1 1 5 9 8 2 Inf 3 1 7 6 9
isequal(A,B)
ans = logical
1

Weitere Antworten (4)

Les Beckham
Les Beckham am 16 Okt. 2023
Bearbeitet: Les Beckham am 16 Okt. 2023
If you want to retain the non-zero elements of A and replace the zeros with Inf, then this is how I would suggest that you do that.
% Input
A = [0 3 2 5 6;
1 1 4 3 2;
9 0 8 1 1;
5 9 8 2 0;
3 1 7 6 9];
A(A==0) = Inf
A = 5×5
Inf 3 2 5 6 1 1 4 3 2 9 Inf 8 1 1 5 9 8 2 Inf 3 1 7 6 9
Note that your loop doesn't do this, it creates a matrix with Inf in the positions of the zeros in A and zero everywhere else. If that is really what you want then you could do that like this.
A = [0 3 2 5 6;
1 1 4 3 2;
9 0 8 1 1;
5 9 8 2 0;
3 1 7 6 9];
B = zeros(size(A));
B(A==0) = Inf
B = 5×5
Inf 0 0 0 0 0 0 0 0 0 0 Inf 0 0 0 0 0 0 0 Inf 0 0 0 0 0
  3 Kommentare
Les Beckham
Les Beckham am 16 Okt. 2023
Bearbeitet: Les Beckham am 16 Okt. 2023
You are quite welcome.
If you are just getting started with Matlab, I would highly recommend that you take a couple of hours to go through the free online tutorial: Matlab Onramp
Sim
Sim am 17 Okt. 2023
thanks :-)

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Matt J
Matt J am 16 Okt. 2023
Allso just for fun.
A = [0 3 2 5 6;
1 1 4 3 2;
9 0 8 1 1;
5 9 8 2 0;
3 1 7 6 9];
A=A+1./(A~=0)-1
A = 5×5
Inf 3 2 5 6 1 1 4 3 2 9 Inf 8 1 1 5 9 8 2 Inf 3 1 7 6 9

Walter Roberson
Walter Roberson am 23 Okt. 2023
A = [0 3 2 5 6;
1 1 4 3 2;
9 0 8 1 1;
5 9 8 2 0;
3 1 7 6 9];
A(~A) = inf
A = 5×5
Inf 3 2 5 6 1 1 4 3 2 9 Inf 8 1 1 5 9 8 2 Inf 3 1 7 6 9
  2 Kommentare
J. Alex Lee
J. Alex Lee am 23 Okt. 2023
by the way, on huge matrices this is actually faster than testing for zero.
Sim
Sim am 23 Okt. 2023
Bearbeitet: Sim am 23 Okt. 2023
@Walter Roberson Wow!! Thumb up! :-)

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Alexander
Alexander am 16 Okt. 2023
Only for fun. My maybe a bit old-fashoned approach would be:
B=1./A;
B(B==Inf)=0;
C=1./B
  6 Kommentare
Alexander
Alexander am 22 Okt. 2023
Thanks @Stephen23 for the advice and yes, there are precision errors. But I think it depends on the problem you have to solve whether these are significant or not.
Stephen23
Stephen23 am 23 Okt. 2023
"But I think it depends on the problem you have to solve whether these are significant or not."
I can't think of many problems where a more complex, slower, obfuscated approach with precision errors would be preferred over the simpler, clearer, much more robust approach using indexing. Can you give an example?

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