how to formulate logical matrix in a loop?

i am having a matrix and want to separate them in the depending on its value

A=[1,2,3,4,5,6,7,8,9,10]

expected result are

idx1=[1,0,0,0,0,0,0,0,0,0]   % for 1
...
idx10=[0,0,0,0,0,0,0,0,0,1] % for 10

Answer 1

[Andrei Bobrov]

A = 1:10;

idx = A(:) == A(:)'

[Shubham Mohan Tatpalliwar]

what should i change to have a 5 rows with a step of 2

[Andrei Bobrov]

what should i change to have a 5 rows with a step of 2

?

A = 1:2:10;
idx = A(:) == A(:)';

Answer 2

[madhan ravi]

A=[1,2,3,4,5,6,7,8,9,10]
RESULT = zeros(1,numel(A));
RESULT1= RESULT;
for i = 1:numel(A)
idx(i)=A(i)==1;
idx1(i)=A(i)==10;
end
RESULT(idx)  = A(idx)
RESULT1(idx1)  = A(idx1)

[madhan ravi]

No need to use logical indexing , see the above illustration

[Shubham Mohan Tatpalliwar]

how can form a loop for this?

[madhan ravi]

See the edited answer ,problem solved!

[Shubham Mohan Tatpalliwar]

 A=[1,2,3,4,5,6,7,8,9,10]
 idy=zeros(1,length(A));
  for i=1:1:10
  idx22=find(A >=(i) & A<(i+1));
      idy(i,idx22)=1;
end

[madhan ravi]

hows this related to the question??? what did you do in the above comment?

[Shubham Mohan Tatpalliwar]

i have created a array

in which at evry row the numbers are indexed in a step of 1

from this array i can use evry row as a vector

and important is that, its done in a loop so i do not have to repeat the operation 100 times

try the updated code to get a clear idea

[madhan ravi]

I know but you don’t have Use a loop

[madhan ravi]

   A=[1,2,3,4,5,6,7,8,9,10]
   idy=zeros(1,length(A));
    for i=1:1:10
    idx22=find(A >=(i) & A<(i+1));
        idy(i,idx22)=1;
  end

[madhan ravi]

what do you mean by the above code?

[madhan ravi]

What should be the result after the loop?