A compact way to replace zeros with Inf in a matrix

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Sim 2023-10-16

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编辑： Sim 2023-10-23

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

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Answer 1

J. Alex Lee 2023-10-16

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编辑：J. Alex Lee 2023-10-16

在 MATLAB Online 中打开

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

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Answer 2

Les Beckham 2023-10-16

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编辑：Les Beckham 2023-10-16

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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

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Les Beckham 2023-10-16

编辑：Les Beckham 2023-10-16

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 2023-10-17

thanks :-)

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Answer 3

Matt J 2023-10-16

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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

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Alexander 2023-10-16

It can't be shorter. Thumbs up.

Sim 2023-10-17

Thumb up! :-)

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Answer 4

Walter Roberson 2023-10-23

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在 MATLAB Online 中打开

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

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J. Alex Lee 2023-10-23

by the way, on huge matrices this is actually faster than testing for zero.

Sim 2023-10-23

编辑：Sim 2023-10-23

@Walter Roberson Wow!! Thumb up! :-)

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Answer 5

Alexander 2023-10-16

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Only for fun. My maybe a bit old-fashoned approach would be:

B=1./A;

B(B==Inf)=0;

C=1./B

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Stephen23 2023-10-17

编辑：Stephen23 2023-10-17

在 MATLAB Online 中打开

As Dyuman Joshi correctly wrote, there definitely are precision issues.

"but "format long" shows the same result"

FORMAT LONG does not have sufficient precision to demonstrate that two values are exactly equal.

"Maybe there is a loss of precision a 100 digits (after the dot) later, but this is more an academic issue than an engineering one (IMHO)"

And yet when we test your method it fails for fairly small integers, showing that it is not just an "academic issue" but a real "engineering" issue caused by not really understanding the behaviors of binary floating point numbers. Anyone who has spent time on this forum (or any other computing forum) will know that floating point precision is much more than just an "academic issue".

Anyway, lets try it right now with a random value I just picked:

format long G
A = 49; % also 93, 98, 99, 103, 105, 107, 117, 123, 186, etc...
B=1./A;
B(B==Inf)=0;
C=1./B % oh no, the alignment is already a warning:
C = 
                        49
D = A;
D(D==0) = Inf % the simple, clear, correct, robust approach.
D = 
    49
isequal(C,D) % oh no, not the same outputs!!!
ans = logical
   0
fprintf('%.999g\n',C,D) % yep, your approach has significant precision issues.
49.00000000000000710542735760100185871124267578125
49

Alexander 2023-10-22

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 2023-10-23

"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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