# How to change non zero value into 0?

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vikash kumar on 21 Aug 2019
Edited: Jan on 21 Aug 2019
I want to manipulate n*n matrix such the last 10 non-zeros value in every column of matrix and change them into zero.Matrix is in the form of binary[0,1].

#### 1 Comment

madhan ravi on 21 Aug 2019
Can't you mention the version your using?

Andrei Bobrov on 21 Aug 2019
Let A - your binary array (n x n)
m = 10;
B = cumsum(A);
out = A.*(B(end,:) - m >= B);

#### 1 Comment

Jan on 21 Aug 2019
Or in modern Matlab versions:
B = cumsum(A, 1, 'reverse');
out = A .* (B > m);

John D'Errico on 21 Aug 2019
Edited: John D'Errico on 21 Aug 2019
You want to reset the last 10 non-zeros to zero, in each column. So, let me build an example matrix.
A = rand(20,5) > 0.25
A =
20×5 logical array
1 0 0 1 1
1 1 1 1 1
1 1 0 1 1
1 1 1 0 1
0 1 1 1 1
1 1 1 1 1
1 1 0 1 0
1 1 0 1 1
1 1 1 1 1
0 0 1 1 0
0 0 1 0 1
0 0 1 1 1
1 1 0 0 0
0 1 0 1 1
1 1 1 1 1
1 0 0 1 0
1 1 1 0 1
1 0 1 1 1
1 1 1 0 0
1 1 1 1 1
In this kind of boring example, there should be roughly 15 non-zeros in each column. But you want to kill off only the last 10 in each column.
The trivial way to do this is to use a loop. It is always a good idea to think of a SIMPLE way to solve a problem, and only then, if it is a problem, then worry about finding a better solution.
B = A;
for i = 1:size(A,2)
B(find(A(:,i),10,'last'),i) = 0;
end
We can verify that it worked, simply enough.
sum(A,1)
ans =
15 14 13 15 15
sum(B,1)
ans =
5 4 3 5 5
B
B =
20×5 logical array
1 0 0 1 1
1 1 1 1 1
1 1 0 1 1
1 1 1 0 1
0 1 1 1 1
1 0 0 1 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
It is clear the loop worked as expected. It is simple, and sufficient to solve the problem, and not even that inefficient. The code is concise, and easy to understand. It will run in almost any version of MATLAB since it only needs the 'last' option for find, which was introduced into MATLAB, and that must have been well over 20 years since that happened.
If there were issues with time (and ONLY then) I would then consider more sophisticated ideas. Do you have many thousands, or even millions of columns to work on? Will this code run millions of times? If not, then don't bother looking for something more efficient.

KSSV on 21 Aug 2019
If A is your matrix. To get non-zero indices use:
idx = A>0 ;
To replace them to zero use:
A(idx) = 0 ;

John D'Errico on 21 Aug 2019
You are missing the point. The matrix may have some zeros mixed in each column. Your solution will change the last 10 ELEMENTS into zero, even if they were already zero.
KSSV on 21 Aug 2019
This one?
idx = A>0 ;
A(idx) = 0 ;
madhan ravi on 21 Aug 2019
KSSV no, if for instance , last two non zeros in each column to zero:
>> A
A =
4×2 logical array
0 0
1 1
0 1
1 1
>> Expected_result
Expected_result =
4×2 logical array
0 0
0 1
0 0
0 0
>>
Clear now?