How to replace negative elements in a Matrix with zeros?

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Matthew Davern
Matthew Davern 2018-1-17
编辑： Walter Roberson 2024-5-4
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?

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Stephen23
Stephen23 2018-1-17
编辑：Stephen23 2018-12-20
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
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DGM
DGM 2021-4-30
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
Stephen23 2021-4-30
编辑：Stephen23 2021-4-30
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.

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更多回答（2 个）

Jan
Jan 2018-1-17
Or:
A(A < 0) = 0
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Jerzy Pela
Jerzy Pela 2020-2-27
编辑：Jerzy Pela 2020-2-27
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

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Johnny Zheng
Johnny Zheng 2020-10-14
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;
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Stephen23
Stephen23 2020-10-14
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

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