How to resort every column of a 3d matrix?

I have a 3d matrix "A" with a size of 1440x720x12, which corresponds to Longitude x Latitude x Time. The longitude for this matrix has a range of -180 to 180, while other matrices I'd like to use this matrix with have longitudes with a range of 0 to 360. I'd like to convert the longitudes of matrix A so that I could easily compute each point of the matrix with the other matrices.

So far my code has been:

A.lon=wrapTo360(A.lon);

[A2,I2]=sort(A.lon);

However, I'm having trouble sorting each column of the matrix so that each point corresponds to a longitudanal range of 0 to 360?

[Matt J]

If A is a 3D numeric array, it doesn't make sense that you can dot-index it, as with A.lon.

Additionally, if A is a numeric array, it isn't clear why it matters whether you are measuring lngitude as [0,360] or [-180, 180]. The array entries A(i,j,k) con't contain longitudinal coordinates themselves, but rather, the values of something at those coordinates.

[Shayma Al Ali]

I meant I have a matrix A.sst that's 1440x720x12.

And the reason why it matters is because if I want to multiple A.sst with another matrix B.u10, then it becomes an issue if B(1,1,1)'s coordinates is say (0,90) but A(1,1,1)'s coordinates is (-180,90). Then I'm not really using the same values in the same location.

[Walter Roberson]

sortrows maybe ?

Answer 1

[dpb]

L=-180:60:180

L = 1×7

-180 -120 -60 0 60 120 180

wrapTo360(L)

ans = 1×7

180 240 300 0 60 120 180

[~,ix]=sort(ans)

ix = 1×7

4 5 6 1 7 2 3

M=[1:3].*L.';
M=cat(3,M,2*M)

M = 
M(:,:,1) =

  -180  -360  -540
  -120  -240  -360
   -60  -120  -180
     0     0     0
    60   120   180
   120   240   360
   180   360   540


M(:,:,2) =

        -360        -720       -1080
        -240        -480        -720
        -120        -240        -360
           0           0           0
         120         240         360
         240         480         720
         360         720        1080

M(ix,:,:)

ans = 
ans(:,:,1) =

     0     0     0
    60   120   180
   120   240   360
  -180  -360  -540
   180   360   540
  -120  -240  -360
   -60  -120  -180


ans(:,:,2) =

           0           0           0
         120         240         360
         240         480         720
        -360        -720       -1080
         360         720        1080
        -240        -480        -720
        -120        -240        -360

sortrows doesn't operate on 3D arrays so would have to use it by plane in loop or arrayfun() construct; indexing works by dimension so getting the sorted order along the first dimension lets you rearrange all planes in that order in one step.

The description of the input as going from -180:180 leaves with a possible problem that the wrapTo360 and sort moves the first element to be ahead of the last in the sorted order; they are redundant as are 0 and 360 in the alternate universe so whether this is actually so or only approximately so that one bound ends just before the boundary value we don't know.

If is duplicated, then one could choose to swap the middle indices before rearranging.

Or

circshift(M,-fix(size(M,1)/2),1)

ans = 
ans(:,:,1) =

     0     0     0
    60   120   180
   120   240   360
   180   360   540
  -180  -360  -540
  -120  -240  -360
   -60  -120  -180


ans(:,:,2) =

           0           0           0
         120         240         360
         240         480         720
         360         720        1080
        -360        -720       -1080
        -240        -480        -720
        -120        -240        -360