[SciPy-User] ndimage.rotate around which anchor point? center?

[Stéfan van der Walt]

Hi Samual

On Sun, Feb 13, 2011 at 9:37 PM, Samuel <scipy at samueljohn.de> wrote:
> My current solution is quite complicated. It rotates an *image* around *center*:
>
>       import scipy as S
>       from scipy import ndimage
>
>       center = S.array(ndimage.center_of_mass( image[:,:,-1] ))
>       orig_shape = S.array(image.shape[0:2],dtype=S.float32)
>       im_center = orig_shape/2.0 # only h,w
>       bigger_shape = orig_shape + S.absolute( center - im_center )
>       bigger_shape = S.array( [ bigger_shape.max(), bigger_shape.max()] ) #square shaped
>       bigger = S.zeros( (bigger_shape[0], bigger_shape[1], image.shape[2]) ,dtype=S.float32)
>       # Now I copy image into top left of bigger
>       bigger[0:orig_shape[0], 0:orig_shape[1],:] = image[:]
>       bigger_center = bigger_shape/2.0
>       image = ndimage.shift( input=bigger,
>                                   shift=S.concatenate( (bigger_center-center, [0]) ),
>                                   order=1 )
>
> The idea is to make enough room, then shift the bigger array such that *center* will be
> in the center of bigger. If you then rotate via ndimage.rotate(image, order=1) the effect
> is as if you would rotate around the point *center*.

The standard solution is pretty much like you describe.  Say we have
transformation matrices T (translate centre to origin) and R (perform
rotation), you want to apply them as follows:

inv(T) R T

In other words, translate centre to origin, rotate, translate back.
The matrices have the forms

R =

[cos(theta) -sin(theta) 0]
[sin(theta) cos(theta) 0]
[0 0 1]

T =

[0 0 tx]
[0 0 ty]
[0 0 1]

After multiplying the matrices, you probably get the solution you
propose earlier on.

You can apply these transformation matrices using the "homography"
function from scikits.image.transform:

https://github.com/stefanv/scikits.image/blob/master/scikits/image/transform/project.py#L19

Regards
Stéfan