How can I divide a linear array into 8 equal parts?
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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?
5 comentarios
Kevin Chng
el 8 de Oct. de 2018
Editada: Kevin Chng
el 8 de Oct. de 2018
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
Respuestas (2)
KSSV
el 8 de Oct. de 2018
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
3 comentarios
Kevin Chng
el 8 de Oct. de 2018
Editada: Kevin Chng
el 9 de Oct. de 2018
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.
7 comentarios
Kevin Chng
el 9 de Oct. de 2018
Editada: Kevin Chng
el 9 de Oct. de 2018
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.
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