How to replace negative elements in a Matrix with zeros?

A = [2, 3, -1, 5; -1, 4, -7, -3; -6, 0, 3, 9; 7, 6, -3, 8];
B = [9; 17; 15; -3];
AI = inv(A)
I = A*AI
X = AI*B
A*X

Now I am trying to set up a nested for loop to redefine negative elements in A. I need to replace negative elements in A with a zero. How do I go about doing this?

Answer 1

[Stephen23]

The simplest way is to use max:

 A = max(A,0)

For example:

>> A = [2, 3, -1, 5; -1, 4, -7, -3; -6, 0, 3, 9; 7, 6, -3, 8]
A =
   2   3  -1   5
  -1   4  -7  -3
  -6   0   3   9
   7   6  -3   8
>> A = max(A,0)
A =
   2   3   0   5
   0   4   0   0
   0   0   3   9
   7   6   0   8

[Michael Seitaridis]

Dear Mr. Cobeldick,

Your answer is very usefull and it helped me. So thank you for that!

But I have a question. I did not read in the documentation this syntax, nor I can understand it. Shouldn't

>> A = max(A,0)

produce

A =

2 3 -1 0

-1 0 -7 -3

-6 0 3 0

7 6 -3 0

(replace max number in every row) ?

[DGM]

Doing this:

B = max(A);

returns the maximum values along dim1.

On the other hand, doing this:

B = max(A,0) 

is equivalent to doing

B = max(A,zeros(size(A)))

In these cases, we're comparing each element of A against 0 and picking the largest of the two values.

[Stephen23]

Michael Seitaridis wrote: "I did not read in the documentation this syntax, nor I can understand it"

I just looked in https://www.mathworks.com/help/matlab/ref/max.html and found this:

"Shouldn't A = max(A,0) produce ... "

The max documentation describes it as:

"C = max(A,B) returns an array with the largest elements taken from A or B."

What my answer shows is consistent with that explanation (given scalar expansion). Lets consider element A(1,4), which has value five. Can you explain why you think that the "largest" of zero and five should be zero? As far as I am aware, five is generally considered to be larger than zero.

"(replace max number in every row) ?"

I do not see that written anywhere in max the documentation.

Answer 2

[Jan]

Or:

A(A < 0) = 0

[QuanCCC]

This worked perfectly

[Jerzy Pela]

I compared both methods, since it was one of the bottlenecks in my calculations and max(A,0) was significantly faster. Keep it in mind if you need to do that calculation numerous times in your script. Otherwise both methods are equal

[Josh]

thank you for this extra little insight!

Answer 3

[Johnny Zheng]

A = A*(A>0);

This also works!

Have a summary of possible methods:

A = A*(A>0);
A = max(A,0);
A(A<0) = 0;

[Stephen23]

For non-scalar A (such as that shown in the question) the mtimes operator needs to be replaced with an element-wise times operator otherwise an error or incorrect output is quite likely:

A.*(A>0)

https://www.mathworks.com/help/matlab/matlab_prog/array-vs-matrix-operations.html

Also note that this method changes -Inf values to NaN, which may be an undesired side-effect:

>> A = [-1,0,1,;-Inf,Inf,NaN];
>> A = A.*(A>0)
A =
     0     0     1
   NaN   Inf   NaN

[Adam Danz]

You'd need to multiple element-wise,

A = A.*(A>0);