During my course I came across that expression:
A(:,end:-1:1)
I have trouble to understand and read the morphemic structure of the 2nd Operand "end;-1;1"
Lets take the example:
A=[1 2 3; 4 5 6; 7 8 9]
I am aware of:
A(:).. Outputs [1 2 3; 4 5 6; 7 8 9] as rows. Operator is :.
A(1,:).. Outputs [1 2 3; 4 5 6; 7 8 9] as columns Operator is , then , .
A(:,1).. Outputs [1 2 3; 4 5 6; 7 8 9] as rows. Operator is , beforehand : .
A(:,end:-1:1)
Output in Matlab show me: 3x3 Matrix.
How am I supposed to read the structure?
end:-1 .. ??1 ..Somehow ":" was for me the operator for show all of the elements.
It makes sense to me that the "Operand1 , Operand2" shows me the 2 Dimensional Matrix.
First Idea:
The end:-1:1 expression seemed to me like a loop. So -1, 0, 1 => **3x Elements** ?
But when I type
A(1,end:3)
it only shows me the 3rd row.
Second Idea:
A(end:-1:1,1)
It shows me the inverted Matrix..
My background:
I am a undergraduate student from the field of language.
I build in my freetime the 8-Bit Sap1 according Ben Eater.
So I am familiar with program memory or instruction memory.
I understand only the result, but not how it is achieved by the MATLAB compiler.
Someone said to me that the "Matrixaddressing is somehow optimized".
Looking forward to a helpful answer in each step. :)
Thanks in advance!
The end keyword in matrix indexing indicates index of last element in the corresponding dimension. So, A(:,end:-1:1) simply means A(:, size(A, 2):-1:1), which in you example (A=[1 2 3; 4 5 6; 7 8 9]), is equivalent to A(:, 3:-1:1).
But to understand what it does, you need to know what 3:-1:1 does. It creates a subrange. You already know 1:3 creates [1, 2, 3]. 1:3 is simplified form of 1:1:3: rangeStrart:increment:rangeEnd. Now, 3:1 or 3:1:1 creates an empty vector, because rangeStart is greater than rangeEnd. To create [3, 2, 1] you need to use a negative step: 3:-1:1.
So, A(:,end:-1:1) means A(:, [3, 2, 1]), which inverts order of rows of A. Also, A(:,end:3) means A(:, 3:3) and eventually A(:, 3), which returns 3rd row of A.
Edit: about your misunderstandings, addressed by @CrisLuengo
I am aware of:
A(:).. Outputs [1 2 3; 4 5 6; 7 8 9] as rows. Operator is :.
A(1,:).. Outputs [1 2 3; 4 5 6; 7 8 9] as columns Operator is , then , .
A(:,1).. Outputs [1 2 3; 4 5 6; 7 8 9] as rows. Operator is ,beforehand : .
A(3, 2) is the element in the 3,2 position (third row, second column) of AA(1, :) is equivalent to A(1, 1:size(A, 2)) and A(1, 1:end) and is the first row of AA(:, 1) is equivalent to A(1:size(A, 1), 1) and A(1:end, 1) and is the first column of AA(:) is equivalent to A(1:numel(A)) and is a single column containing all elements of A