[Numpy-discussion] Understanding mgrid

Robert Kern robert.kern@gmail....
Fri Sep 19 13:22:29 CDT 2008


On Fri, Sep 19, 2008 at 12:59, Brad Malone <brad.malone@gmail.com> wrote:
> Hi, I was wondering if someone could englighten me on what the geometrical
> significance of numpy.mgrid is. I can play around with it and see trends in
> the sizes and number of arrays, but why does it give the output that it
> does? Looking at the example shown below, why does it return a matrix and
> its transpose?

Well, it returns one array. In your example, there is a (2,5,5) array,
which is basically the concatenation of two arrays which *happen* to
be transposes of each other. If you had chosen differently sized axes,
they wouldn't be transposes.

In [14]: mgrid[0:2,0:3]
Out[14]:
array([[[0, 0, 0],
        [1, 1, 1]],

       [[0, 1, 2],
        [0, 1, 2]]])

> Is this a representation of some geometrical grid?

It can be. There are other uses for it.

> Does the
> output imply some sort of connectivity?

It describes an orthogonal grid.

> If so, how do you see it?
>
>  >>> mgrid[0:5,0:5]
>    array([[[0, 0, 0, 0, 0],
>            [1, 1, 1, 1, 1],
>            [2, 2, 2, 2, 2],
>
>            [3, 3, 3, 3, 3],
>
>            [4, 4, 4, 4, 4]],
>    <BLANKLINE>
>           [[0, 1, 2, 3, 4],
>
>            [0, 1, 2, 3, 4],
>            [0, 1, 2, 3, 4],
>            [0, 1, 2, 3, 4],
>
>
>            [0, 1, 2, 3, 4]]])
>
>
>
> I have a cubic grid in 3D space that is spanned by 3 orthogonal vectors. Am
> I able to generate this equivalent grid with mgrid somehow? If so, how is it
> done? I am using mayavi and I need to be able to construct some arrays in
> the same way that mgrid would have constructed them, so this is why I ask.

I would probably use indices() instead of mgrid if you are just given
the x, y, and z vectors. indices([n,m,k]) is equivalent to
mgrid[0:n,0:m,0:k]:


In [19]: x = linspace(0, 1, 3)

In [20]: x
Out[20]: array([ 0. ,  0.5,  1. ])

In [21]: y = linspace(1, 2.5, 4)

In [22]: y
Out[22]: array([ 1. ,  1.5,  2. ,  2.5])

In [23]: z = linspace(3, 5, 5)

In [24]: z
Out[24]: array([ 3. ,  3.5,  4. ,  4.5,  5. ])

In [25]: ix, iy, iz = indices([len(x), len(y), len(z)])

In [26]: x[ix]
Out[26]:
array([[[ 0. ,  0. ,  0. ,  0. ,  0. ],
        [ 0. ,  0. ,  0. ,  0. ,  0. ],
        [ 0. ,  0. ,  0. ,  0. ,  0. ],
        [ 0. ,  0. ,  0. ,  0. ,  0. ]],

       [[ 0.5,  0.5,  0.5,  0.5,  0.5],
        [ 0.5,  0.5,  0.5,  0.5,  0.5],
        [ 0.5,  0.5,  0.5,  0.5,  0.5],
        [ 0.5,  0.5,  0.5,  0.5,  0.5]],

       [[ 1. ,  1. ,  1. ,  1. ,  1. ],
        [ 1. ,  1. ,  1. ,  1. ,  1. ],
        [ 1. ,  1. ,  1. ,  1. ,  1. ],
        [ 1. ,  1. ,  1. ,  1. ,  1. ]]])

In [27]: y[iy]
Out[27]:
array([[[ 1. ,  1. ,  1. ,  1. ,  1. ],
        [ 1.5,  1.5,  1.5,  1.5,  1.5],
        [ 2. ,  2. ,  2. ,  2. ,  2. ],
        [ 2.5,  2.5,  2.5,  2.5,  2.5]],

       [[ 1. ,  1. ,  1. ,  1. ,  1. ],
        [ 1.5,  1.5,  1.5,  1.5,  1.5],
        [ 2. ,  2. ,  2. ,  2. ,  2. ],
        [ 2.5,  2.5,  2.5,  2.5,  2.5]],

       [[ 1. ,  1. ,  1. ,  1. ,  1. ],
        [ 1.5,  1.5,  1.5,  1.5,  1.5],
        [ 2. ,  2. ,  2. ,  2. ,  2. ],
        [ 2.5,  2.5,  2.5,  2.5,  2.5]]])

In [28]: z[iz]
Out[28]:
array([[[ 3. ,  3.5,  4. ,  4.5,  5. ],
        [ 3. ,  3.5,  4. ,  4.5,  5. ],
        [ 3. ,  3.5,  4. ,  4.5,  5. ],
        [ 3. ,  3.5,  4. ,  4.5,  5. ]],

       [[ 3. ,  3.5,  4. ,  4.5,  5. ],
        [ 3. ,  3.5,  4. ,  4.5,  5. ],
        [ 3. ,  3.5,  4. ,  4.5,  5. ],
        [ 3. ,  3.5,  4. ,  4.5,  5. ]],

       [[ 3. ,  3.5,  4. ,  4.5,  5. ],
        [ 3. ,  3.5,  4. ,  4.5,  5. ],
        [ 3. ,  3.5,  4. ,  4.5,  5. ],
        [ 3. ,  3.5,  4. ,  4.5,  5. ]]])

-- 
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless
enigma that is made terrible by our own mad attempt to interpret it as
though it had an underlying truth."
 -- Umberto Eco


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