[Numpy-discussion] Iterative Matrix Multiplication
Chris Colbert
sccolbert@gmail....
Mon Mar 1 19:15:50 CST 2010
This is how I always do it:
In [1]: import numpy as np
In [3]: tmat = np.array([[0., 1., 0., 5.],[0., 0., 1., 3.],[1., 0., 0.,
2.]])
In [4]: tmat
Out[4]:
array([[ 0., 1., 0., 5.],
[ 0., 0., 1., 3.],
[ 1., 0., 0., 2.]])
In [5]: points = np.random.random((5, 3))
In [7]: hpoints = np.column_stack((points, np.ones(len(points))))
In [9]: hpoints
Out[9]:
array([[ 0.17437059, 0.38693627, 0.201047 , 1. ],
[ 0.99712373, 0.16958721, 0.03050696, 1. ],
[ 0.30653326, 0.62037744, 0.35785282, 1. ],
[ 0.78936771, 0.93692711, 0.58138493, 1. ],
[ 0.29914065, 0.08808239, 0.72032172, 1. ]])
In [10]: np.dot(tmat, hpoints.T).T
Out[10]:
array([[ 5.38693627, 3.201047 , 2.17437059],
[ 5.16958721, 3.03050696, 2.99712373],
[ 5.62037744, 3.35785282, 2.30653326],
[ 5.93692711, 3.58138493, 2.78936771],
[ 5.08808239, 3.72032172, 2.29914065]])
On Mon, Mar 1, 2010 at 6:12 AM, Friedrich Romstedt <
friedrichromstedt@gmail.com> wrote:
> 2010/3/1 Charles R Harris <charlesr.harris@gmail.com>:
> > On Sun, Feb 28, 2010 at 7:58 PM, Ian Mallett <geometrian@gmail.com>
> wrote:
> >> Excellent--and a 3D rotation matrix is 3x3--so the list can remain n*3.
> >> Now the question is how to apply a rotation matrix to the array of vec3?
> >
> > It looks like you want something like
> >
> > res = dot(vec, rot) + tran
> >
> > You can avoid an extra copy being made by separating the parts
> >
> > res = dot(vec, rot)
> > res += tran
> >
> > where I've used arrays, not matrices. Note that the rotation matrix
> > multiplies every vector in the array.
>
> When you want to rotate a ndarray "list" of vectors:
>
> >>> a.shape
> (N, 3)
>
> >>> a
> [[1., 2., 3. ]
> [4., 5., 6. ]]
>
> by some rotation matrix:
>
> >>> rotation_matrix.shape
> (3, 3)
>
> where each row of the rotation_matrix represents one vector of the
> rotation target basis, expressed in the basis of the original system,
>
> you can do this by writing:
>
> >>> numpy.dot(a, rotations_matrix) ,
>
> as Chuck pointed out.
>
> This gives you the rotated vectors in an ndarray "list" again:
>
> >>> numpy.dot(a, rotation_matrix).shape
> (N, 3)
>
> This is just somewhat more in detail what Chuck already stated
> > Note that the rotation matrix
> > multiplies every vector in the array.
>
> my 2 cents,
> Friedrich
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