How can I divide a linear array into 8 equal parts?

A=[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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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]

This is a long linear array with consecutive 0's and 1's. How can I equally divide this array A into 8 equal parts and count no of 1's from each parts?

[Kevin Chng]

Hi,

 B = reshape(A,5,[])

There are 125 elements in A, hence it cant be equally divide by 8.

[Zara Khan]

How to count no of 1's from each parts ? And what will be the code when the array size is not predefined?

[Kevin Chng]

How to count no of 1's from each parts ?

if let say we divide into 5 rows:

        A=[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  1  1  1  1  1  1  1  1  1  1  1  1  1 1  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];
      B = reshape(A,[],5);
      for i=1:1:5
      partSum(i) = sum(B(:,i))
      end

Does it work for you?

what will be the code when the array size is not predefined?

numEle = numel(A);

for i=1:1:10
        try
               B = reshape(A,i,[]);
        catch
                fprintf('array size is not correct %i \n',i);
        end
end

Then it will tell you what is the array size you can reshape them.

[Zara Khan]

 partSum(i) = sum(B(:,i))
this is always giving same values.

partSum =

     0     5    10     0     0

partSum =

     0     5    10     0     0

partSum =

     0     5    10     0     0

partSum =

     0     5    10     0     0

partSum =

     0     5    10     0     0

I want to count only no of 1's from each of the 5 parts.

[Kevin Chng]

Just noticed the for-loop is not necessary.

B = reshape(A,[],5);
partSum = sum(B)

partSum =

     0     5    10     0     0

0 is the total number of 1's in first part, 5 is the second part, 10........so on.

If want be more specific for better understanding, we could

first part = B(:,1);
second part = B(:,2);
third part = B(:,3);
fourth part = B(:,4);
fifth part = B(:,5);

then either use sum(first part) or nnz(firstpart) to count the number of 1's

Answer 1

[KSSV]

You can get non zeros using nnz.

B = reshape(A,[],5)' ;
N = zeros(size(B,1),1) ;
for i = 1:size(B,1)
    N(i) = nnz(B(i,:)) ;
end

[Zara Khan]

I want to keep each of the part into linear as well.

[KSSV]

I want to keep each of the part into linear as well. ?????example please.

[Zara Khan]

If I am diving a linear array into 8 equal parts then each 8 parts will be also linear. For example [0 0 1 1 0 1], [0 1 1 1 1 ] and so on.

Answer 2

[Kevin Chng]

How to count no of 1's from each parts ?

if let say we divide into 5 rows:

B = reshape(A,[],5);
partSum = sum(B);

you could nnz or sum.

What will be the code when the array size is not predefined?

A=[1 2 3; 4 5 6; 7 8 9]

numEle = numel(A);
for i=1:1:10
        numberofsize = rem(numEle,i);
        if numberofsize ~= 0
            fprintf('array size is not correct for %i \n',i)
        end
end

Then it will tell you what is the array size you can reshape them.

[Zara Khan]

where you are using this numel ?

[Kevin Chng]

To check which size i am allowed to reshape them.

[Zara Khan]

Ok. But in the loop I can't find this numEle.

[Kevin Chng]

Sorry for my mistake. Edited my answer.

[Zara Khan]

NumEle is giving the length of the array. In reshape I think We need to send array name . Here it is A not numEle.

[Kevin Chng]

Ya, you are right, it is my mistake.

NumEle is counting the number of element. Ya, you are right, i should use rem() instead. If the remainder is not 0, the size is not suitable for our reshape size.

[Kevin Chng]

Does it working for you?