[Numpy-discussion] cross product of two 3xn arrays
Gary Ruben
gruben at bigpond.net.au
Thu Feb 16 01:00:04 EST 2006
Hi Travis,
Have you tested this? It appears to give the wrong answer on my system.
I expect to get from this
In [21]: a=array([[[1],[2],[3]],[[4],[5],[6]],[[7],[8],[9]]])
In [22]: b=array([[[1],[2],[4]],[[4],[5],[7]],[[7],[8],[10]]])
the solution
Out[24]:
array([[[ 2],
[-1],
[ 0]],
[[ 5],
[-4],
[ 0]],
[[ 8],
[-7],
[ 0]]])
i.e. the same as my example but with column vectors instead of rows, but
doing cross(a,b,axisa=0,axisb=0,axisc=0) gives
Out[15]:
array([[[ 0],
[ 0],
[-3]],
[[ 0],
[ 0],
[ 6]],
[[ 0],
[ 0],
[-3]]])
Gary R.
Travis Oliphant wrote:
> Ian Harrison wrote:
>
>> Hello,
>>
>> I have two groups of 3x1 arrays that are arranged into two larger 3xn
>> arrays. Each of the 3x1 sub-arrays represents a vector in 3D space. In
>> Matlab, I'd use the function cross() to calculate the cross product of
>> the corresponding 'vectors' from each array. In other words:
>>
>>
>
>
> Help on function cross in module numpy.core.numeric:
>
> cross(a, b, axisa=-1, axisb=-1, axisc=-1)
> Return the cross product of two (arrays of) vectors.
>
> The cross product is performed over the last axis of a and b by default,
> and can handle axes with dimensions 2 and 3. For a dimension of 2,
> the z-component of the equivalent three-dimensional cross product is
> returned.
>
> It's the axisa, axisb, and axisc that you are interested in.
>
> The default is to assume you have Nx3 arrays and return an Nx3 array.
> But you can change the axis used to find vectors.
>
> cross(A,B,axisa=0,axisb=0,axisc=0)
>
> will do what you want. I suppose, a single axis= argument might be
> useful as well for the common situation of having all the other axis
> arguments be the same.
>
> -Travis
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