[Numpy-discussion] dot function or dot notation, matrices, arrays?
Wayne Watson
sierra_mtnview@sbcglobal....
Sat Dec 19 11:45:02 CST 2009
OK, so what's your recommendation on the code I wrote? Use shape 0xN?
Will that eliminate the need for T?
I'll go back to Tenative Python, and re-read dimension, shape and the like.
Charles R Harris wrote:
>
>
> On Sat, Dec 19, 2009 at 9:45 AM, Wayne Watson
> <sierra_mtnview@sbcglobal.net <mailto:sierra_mtnview@sbcglobal.net>>
> wrote:
>
>
>
> Dag Sverre Seljebotn wrote:
> > Wayne Watson wrote:
> >
> >> I'm trying to compute the angle between two vectors in three
> dimensional
> >> space. For that, I need to use the "scalar (dot) product" ,
> according to
> >> a calculus book (quoting the book) I'm holding in my hands
> right now.
> >> I've used dot() successfully to produce the necessary angle. My
> program
> >> works just fine.
> >>
> >> In the case of the dot(function), one must use np.dev(x.T,x),
> where x is
> >> 1x3.
> >>
> >> I'm not quite sure what your point is about dot()* unless you are
> >> thinking in some non-Euclidean fashion. One can form
> np.dot(a,b) with a
> >> and b arrays of 3x4 and 4x2 shape to arrive at a 3x2 array. That's
> >> definitely not a scalar. Is there a need for this sort of
> calculation in
> >> non-Euclidean geometry, which I have never dealt with?
> >>
> >
> > There's a difference between 1D and 2D arrays that's important
> here. For
> > a 1D array, np.dot(x.T, x) == np.dot(x, x), since there's only one
> > dimension.
> >
> A 4x1, 1x7, and 1x5 would be examples of a 1D array or matrix, right?
>
>
> No, they are all 2D. All matrices are 2D. An array is 1D if it doesn't
> have a second dimension, which might be confusing if you have only
> seen vectors represented as arrays. To see the number of dimensions in
> a numpy array, use shape:
>
> In [1]: array([[1,2],[3,4]])
> Out[1]:
> array([[1, 2],
> [3, 4]])
>
> In [2]: array([[1,2],[3,4]]).shape
> Out[2]: (2, 2)
>
> In [3]: array([1,2, 3,4])
> Out[3]: array([1, 2, 3, 4])
>
> In [4]: array([1,2, 3,4]).shape
> Out[4]: (4,)
>
>
> Are you saying that instead of using a rotational matrix like
> theta = 5.0 # degrees
> m1 = matrix([[2] ,[5]])
> rotCW = matrix([ [cosD(theta), sinD(theta)], [-sinD(theta),
> cosD(theta)] ])
> m2= rotCW*m1
> m1=np.array(m1)
> m2=np.array(m2)
> that I should use a 2-D array for rotCW? So why does numpy have a
> matrix
> class? Is the class only used when working with matplotlib?
>
>
> Numpy has a matrix class because python lacks operators, so where *
> normally means element-wise multiplication the matrix class uses it
> for matrix multiplication, which is different. Having a short form for
> matrix multiplication is sometimes a convenience and also more
> familiar for folks coming to numpy from matlab.
>
>
> To get the scalar value (sum of squares) I had to use a transpose,
> T, on
> one argument.
>
>
> That is if the argument is 2D. It's not strictly speaking a scalar
> product, but we won't go into that here ;)
>
> <snip>
>
> Chuck
>
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--
Wayne Watson (Watson Adventures, Prop., Nevada City, CA)
(121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
"... humans'innate skills with numbers isn't much
better than that of rats and dolphins."
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