Reputation: 23
Suppose I construct the following 3D array
n = 3;
A = zeros(n,n,n);
A(1:n^3) = 1:n^3;
which gives
>> A
A(:,:,1) =
1 4 7
2 5 8
3 6 9
A(:,:,2) =
10 13 16
11 14 17
12 15 18
A(:,:,3) =
19 22 25
20 23 26
21 24 27
One can see how matlab indexes a 3D array from the above example. Suppose I would like to access (ii = 1, jj = 3, kk = 2) element of this array, which can be done by
>>A(1,3,2)
ans =
16
Alternatively, I can use the following form based on the matlab indexing rule demonstrated above
A(ii + (jj-1)*n + (kk-1)*n^2)
as an example, for ii = 1, jj = 3, kk = 2, I get
>> A(1 + (3-1)*3 + (2-1)*3^2)
ans =
16
To illustrate the problem, I define the following 3D meshgrid (say for the purpose of index manupulations which is not relevant here):
[j1 j2 j3] = meshgrid(1:n);
If I am not wrong, common sense would expect that
A(j1 + (j2-1)*n +(j3-1)*n^2)
to give me the same matrix based on the above discussions, but I get
>> A(j1 + (j2-1)*3 +(j3-1)*3^2)
ans(:,:,1) =
1 2 3
4 5 6
7 8 9
ans(:,:,2) =
10 11 12
13 14 15
16 17 18
ans(:,:,3) =
19 20 21
22 23 24
25 26 27
From this I see that if you want to get the same 3D array you actually need to use
>> A(j2 + (j1-1)*3 +(j3-1)*3^2)
which is very strange to me. I am posting this issue here to see what other people think about this.
Upvotes: 2
Views: 648
Reputation: 36710
There is a unconventional thing in matlab, the order of axis is [Y,X,Z]. Y is the first axis, X the second. As meshgrid returns [X,Y,Z] you must use:
[j2 j1 j3] = meshgrid(1:n);
Then you get the expected result. Alternatively you can switch to ndgrid
which returns the dimensions in order:
[j1 j2 j3] = ndgrid(1:n);
Upvotes: 1