Split a Number Sequence into n equal parts and then replacing the values in the matrix.

How to split a number sequence into n equal parts and then and then replace the value of matrix with it.

For Example:

A 5x5 Matrix(A) is given to us :

Now we will find the maximum element in this matrix using : M = max(A, [], 'all')

Now we want to quantize this max number ie. 29 into n equal parts. Example : Now starting from 0 to 29 ([0:29]) I want to split this sequence to 3 equal parts.

Such that my output now looks like:

0 - 0:9

1 - 10:19

2 - 20:29

Now I want to replace the number that are lying between 0 to 9 with 0, 10 to 19 with 1 and 20 to 29 with 2 from the given matrix.

Such that my final matrix becomes,

Please help with the MATLAB Code?

Answer 1

[Matt J]

result = discretize(A, linspace(0,  max(A(:)), 4)  )-1

[Mark Wood]

Thanks a lot Sir!

[Matt J]

You're welcome, but please Accept-click the answer if it resolved your problem.

[Mark Wood]

This fails,

for example taking the matrix eg.

Mat = [25 6 19 13 7 5; 7 8 10 12 14 17 ; 19 21 16 19 17 16 ; 19 19 17 15 13 11 ; 25 22 15 10 5 3 ; 25 23 20 19 17 1];

Here nrOfIntervals is 5,

0 - 1:5

1 - 6:10

2 - 11:15

3 - 16:20

4 - 21-25

But here the element 5 in the Mat comes in 1 but it should be under 0.

So what I want to say that, here as number of equal parts was 5, so here the interval starts with 1 not 0.

Could you please help in modifying the code!

[Matt J]

I odon't know what your adaptation looks like. Here's what I get,

Mat = [25 6 19 13 7 5; 7 8 10 12 14 17 ; 19 21 16 19 17 16 ; 19 19 17 15 13 11 ; 25 22 15 10 5 3 ; 25 23 20 19 17 1];

result = discretize(Mat, 1:5:26)-1

result = 6×6

4 1 3 2 1 0 1 1 1 2 2 3 3 4 3 3 3 3 3 3 3 2 2 2 4 4 2 1 0 0 4 4 3 3 3 0

[Jonas]

@Matt J i think that was meant to be a comment for another answer below, there is an exact copy of the coment

Answer 2

[Jonas]

i suggest normalizing the matrix by dividing all elements through the biggest value you already have and multiply it afterwards with the number of intervals you want (3 in the example). then you can apply the floor() function. the values equal to the maximum value have to be decreased by one at the end (e.g. 29 will become 3 and then has the be decreased to 2, all other wntries are already correct)

nrOfIntervals=3;
A=A/max(A(:));
A=A*nrOfIntervals;
A=floor(A);
A(A==nrOfIntervals)=nrOfIntervals-1;

edit: a correct version can be found in the comments of this answer

[Mark Wood]

This fails,

for example taking the matrix eg.

Mat = [25 6 19 13 7 5; 7 8 10 12 14 17 ; 19 21 16 19 17 16 ; 19 19 17 15 13 11 ; 25 22 15 10 5 3 ; 25 23 20 19 17 1];

Here nrOfIntervals is 5,

0 - 1:5

1 - 6:10

2 - 11:15

3 - 16:20

4 - 21-25

But here the element 5 in the Mat comes in 1 but it should be under 0.

So what I want to say that, here as number of equal parts was 5, so here the interval starts with 1 not 0.

Could you please help in modifying the code!

[Jonas]

my bad, this should work better:

nrOfIntervals=3;
A=A/max(A(:));
A=A*nrOfIntervals;
A=ceil(A);
A(A>0)=A(A>0)-1;

Answer 3

[Girijashankar Sahoo]

N= 25    % length of sequence
n=sqrt(N)   % break the sequence in n element
X=randi(30,1,25)
A=reshape(X,[n,n])
M=max(A,[],'all')
split=M/3
for i=1:n
    for j=1:n
        Value=A(i,j)
        if Value<(M/split)
            A(i,j)=0;
        elseif (M/split)<Value<(2*M/split)
            A(i,j)=1;
        else Value>(2*M/split)
            A(i,j)=2;
        end
    end
end