How can I do linear indexing in matlab?

hi, Suppose I have a matrix A=[1 2 3 4;5 6 7 8;9 10 11 12;13 14 15 16]; I want to transfer it into a linear matrix as A=[1 2 5 6 3 4 7 8 9 10 13 14 11 12 15 16]; (i.e. mortan scan order in image processing). Can you help me. All my work will start after this only. I would be very greatful to you

Answer 1

[Image Analyst]

This will get you the desired matrix, whereas the other two don't (yet unless they edit them):

% Reference arrays.
A=[1 2 3 4;5 6 7 8;9 10 11 12;13 14 15 16]
desired = [1 2 5 6 3 4 7 8 9 10 13 14 11 12 15 16]

% Next two solutions don't produce desired.
B1 = A(:)'
B2 = reshape(reshape(A.',2,[]),1,[])

% This will get you the desired.
[rows, columns] = size(A)
B3 = [];
for row = 1 : 2 : rows
  for col = 1 : 2 : columns
    B3 = [B3, A(row, col), A(row, col+1), A(row+1, col), A(row+1, col+1)];
  end
end
B3 % Report values to command window.

[dpb]

"% Next two solutions don't produce desired. B1 = A(:)' B2 = reshape(reshape(A.',2,[]),1,[])"

I'm sure B1 doesn't, but I beg to differ on the B2...

>> A1=[1 2 5 6 3 4 7 8 9 10 13 14 11 12 15 16];
>> all(reshape(reshape(A.',2,[]),1,[]))==all(A1)
ans =
   1
>>

where A1 is the OP's answer pasted into the command window and retitled A1 to not conflict with the original A.

[Image Analyst]

I just copied and pasted his array and your code and I got this:

desired =
     1   2   5   6   3   4   7   8   9  10  13  14  11  12  15  16
B2 =
     1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16

They looked different to me. My B3 gives this:

B3 =
     1   2   5   6   3   4   7   8   9  10  13  14  11  12  15  16

which looks the same as what he desires his final out A to be.

[TAUSIF]

The answer given by Image Analyst really works

Answer 2

[Abdulrazzaq]

try:

B = A(:)';

Best wishes

Abdulrazzaq

Answer 3

[dpb]

One way amongst many --

B=reshape(reshape(A.',2,[]),1,[]);

Consider how Matlab storage is in column-major order and what operation one needs to get to the desired organization from the original...

Answer 4

[Matt J]

Using MAT2TILES (available here),

 B=cellfun(@(c) c(:).',  mat2tiles(A.',[2,2]),'uni',0);

   B=[B{:}]

[TAUSIF]

It produces the desired output.

Answer 5

[Roger Stafford]

If you convert each of the integers 0:15 to binary, permute the digits appropriately, recalculate the numbers from these permuted digits, and finally add 1 to each, you will get the correct linear indexing. The same holds true for larger Z-orders.

[TAUSIF]

thank you all for the answers.....