A compact way to replace zeros with Inf in a matrix

Question

Sim 2023-10-16

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2034259-a-compact-way-to-replace-zeros-with-inf-in-a-matrix

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

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

J. Alex Lee 2023-10-16

2
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2034259-a-compact-way-to-replace-zeros-with-inf-in-a-matrix#answer_1334259

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

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 2

Les Beckham 2023-10-16

3
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2034259-a-compact-way-to-replace-zeros-with-inf-in-a-matrix#answer_1334264

编辑：Les Beckham 2023-10-16

在 MATLAB Online 中打开

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 个评论
显示 1更早的评论隐藏 1更早的评论

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 :-)

请先登录，再进行评论。

Answer 3

Matt J 2023-10-16

3
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2034259-a-compact-way-to-replace-zeros-with-inf-in-a-matrix#answer_1334409

在 MATLAB Online 中打开

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

2 个评论
显示无隐藏无

Alexander 2023-10-16

It can't be shorter. Thumbs up.

Sim 2023-10-17

Thumb up! :-)

请先登录，再进行评论。

Answer 4

Walter Roberson 2023-10-23

3
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2034259-a-compact-way-to-replace-zeros-with-inf-in-a-matrix#answer_1338816

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

2 个评论
显示无隐藏无

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! :-)

请先登录，再进行评论。

Answer 5

Alexander 2023-10-16

2
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2034259-a-compact-way-to-replace-zeros-with-inf-in-a-matrix#answer_1334374

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

B=1./A;

B(B==Inf)=0;

C=1./B

6 个评论
显示 4更早的评论隐藏 4更早的评论

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?

请先登录，再进行评论。

A compact way to replace zeros with Inf in a matrix

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

更多回答（4 个）

3 个评论
显示 1更早的评论隐藏 1更早的评论

2 个评论
显示无隐藏无

2 个评论
显示无隐藏无

6 个评论
显示 4更早的评论隐藏 4更早的评论

另请参阅

类别

标签

Community Treasure Hunt

A compact way to replace zeros with Inf in a matrix

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

更多回答（4 个）

3 个评论 显示 1更早的评论隐藏 1更早的评论

2 个评论 显示 无隐藏 无

2 个评论 显示 无隐藏 无

6 个评论 显示 4更早的评论隐藏 4更早的评论

另请参阅

类别

标签

Community Treasure Hunt

WeChat

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

3 个评论
显示 1更早的评论隐藏 1更早的评论

2 个评论
显示无隐藏无

2 个评论
显示无隐藏无

6 个评论
显示 4更早的评论隐藏 4更早的评论