How to replace negative elements in a Matrix with zeros?
显示 更早的评论
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?
采纳的回答
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
3 个评论
Michael Seitaridis
2021-4-29
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
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.
Michael Seitaridis wrote: "I did not read in the documentation this syntax, nor I can understand it"
"Shouldn't A = max(A,0) produce ... "
The max documentation describes it as:
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.
更多回答（2 个）
Jan
2018-1-17
Or:
A(A < 0) = 0
3 个评论
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
Josh
2024-5-4
thank you for this extra little insight!
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;
2 个评论
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)
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
2020-10-14
You'd need to multiple element-wise,
A = A.*(A>0);
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