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Ian Mallett | 1 Apr 2009 02:20
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Numpy Positional Array

Hello,
I'm trying to make an array of size n*n*2.  It should be of the form:
[[[0,0],[1,0],[2,0],[3,0],[4,0], ... ,[n,0]],
 [[0,1],[1,1],[2,1],[3,1],[4,1], ... ,[n,1]],
 [[0,2],[1,2],[2,2],[3,2],[4,2], ... ,[n,2]],
 [[0,3],[1,3],[2,3],[3,3],[4,3], ... ,[n,3]],
 [[0,4],[1,4],[2,4],[3,4],[4,4], ... ,[n,4]],
   ...   ...   ...   ...   ...   ...   ...
 [[0,n],[1,n],[2,n],[3,n],[4,n], ... ,[n,n]]]
Each vec2 represents the x,y position of the vec in the array itself.  I'm completely stuck on implementing this setup in numpy.  Any pointers?

Thanks,
Ian

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Robert Kern | 1 Apr 2009 02:32
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Re: Numpy Positional Array

2009/3/31 Ian Mallett <geometrian <at> gmail.com>:
> Hello,
> I'm trying to make an array of size n*n*2.  It should be of the form:
> [[[0,0],[1,0],[2,0],[3,0],[4,0], ... ,[n,0]],
>  [[0,1],[1,1],[2,1],[3,1],[4,1], ... ,[n,1]],
>  [[0,2],[1,2],[2,2],[3,2],[4,2], ... ,[n,2]],
>  [[0,3],[1,3],[2,3],[3,3],[4,3], ... ,[n,3]],
>  [[0,4],[1,4],[2,4],[3,4],[4,4], ... ,[n,4]],
>    ...   ...   ...   ...   ...   ...   ...
>  [[0,n],[1,n],[2,n],[3,n],[4,n], ... ,[n,n]]]
> Each vec2 represents the x,y position of the vec in the array itself.  I'm
> completely stuck on implementing this setup in numpy.  Any pointers?

How do you want to fill in the array? If you are typing it in
literally into your code, you would do basically the above, without
the ...'s, and wrap it in numpy.array(...).

Otherwise, you can create empty arrays with numpy.empty((n,n,2)), or
filled in versions using zeros() and ones().

--

-- 
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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Ian Mallett | 1 Apr 2009 02:35
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Re: Numpy Positional Array

On Tue, Mar 31, 2009 at 5:32 PM, Robert Kern <robert.kern <at> gmail.com> wrote:

How do you want to fill in the array? If you are typing it in
literally into your code, you would do basically the above, without
the ...'s, and wrap it in numpy.array(...).
I know that, but in some cases, n will be quite large, perhaps 1000 on a side.  I'm trying to generate an array of that form in numpy entirely for speed and aesthetic reasons.

Ian 
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Robert Kern | 1 Apr 2009 02:39
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Re: Numpy Positional Array

2009/3/31 Ian Mallett <geometrian <at> gmail.com>:
> On Tue, Mar 31, 2009 at 5:32 PM, Robert Kern <robert.kern <at> gmail.com> wrote:
>>
>> How do you want to fill in the array? If you are typing it in
>> literally into your code, you would do basically the above, without
>> the ...'s, and wrap it in numpy.array(...).
>
> I know that, but in some cases, n will be quite large, perhaps 1000 on a
> side.  I'm trying to generate an array of that form in numpy entirely for
> speed and aesthetic reasons.

Again: How do you want to fill in the array? What is the process that
generates the data?

--

-- 
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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Ian Mallett | 1 Apr 2009 02:42
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Re: Numpy Positional Array

The array follows a pattern: each array of length 2 represents the x,y index of that array within the larger array.  

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Robert Kern | 1 Apr 2009 02:48
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Re: Numpy Positional Array

2009/3/31 Ian Mallett <geometrian <at> gmail.com>:
> The array follows a pattern: each array of length 2 represents the x,y index
> of that array within the larger array.

Ah, right. Use dstack(mgrid[0:n,0:n]).

--

-- 
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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Ian Mallett | 1 Apr 2009 02:51
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Re: Numpy Positional Array

Thanks! 

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Partridge, Matthew BGI SYD | 1 Apr 2009 02:45

Re: Numpy Positional Array


> The array follows a pattern: each array of length 2 represents the x,y
index of that array within the larger array.  
	
Is this what you are after?

>>> numpy.array(list(numpy.ndindex(n,n))).reshape(n,n,2)

 
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Ian Mallett | 1 Apr 2009 03:02
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Re: Numpy Positional Array

Same.  Thanks, too.

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Hans-Andreas Engel | 1 Apr 2009 03:40

Re: array of matrices

Robert Kern <robert.kern <at> gmail.com> writes:
> On Sat, Mar 28, 2009 at 23:15, Anne Archibald <peridot.faceted <at> gmail.com>
wrote:
> > 2009/3/28 Geoffrey Irving <irving <at> naml.us>:
> >> On Sat, Mar 28, 2009 at 12:47 AM, Robert Kern <robert.kern <at> gmail.com>
wrote:
> >>> 2009/3/27 Charles R Harris <charlesr.harris <at> gmail.com>:
> >>>>
> >>>> On Fri, Mar 27, 2009 at 4:43 PM, Robert Kern <robert.kern <at> gmail.com>
wrote:
> >>>>>
> >>>>> On Fri, Mar 27, 2009 at 17:38, Bryan Cole <bryan <at> cole.uklinux.net>
wrote:
> >>>>> > I have a number of arrays of shape (N,4,4). I need to perform a
> >>>>> > vectorised matrix-multiplication between pairs of them I.e.
> >>>>> > matrix-multiplication rules for the last two dimensions, usual
> >>>>> > element-wise rule for the 1st dimension (of length N).
> >>>>> >
(...)
> >>
> >> It'd be great if this operation existed as a primitive.  
(...)
> >
> > The infrastructure to support such generalized ufuncs has been added
> > to numpy, but as far as I know no functions yet make use of it.
> 
> I don't think there is a way to do it in general with dot(). Some
> cases are ambiguous. I think you will need separate matrix-matrix,
> matrix-vector, and vector-vector gufuncs, to coin a term.

By the way, matrix multiplication is one of the testcases for the generalized
ufuncs in numpy 1.3 -- this makes playing around with it easy:

  In [1]: N = 10; a = randn(N, 4, 4); b = randn(N, 4, 4)

  In [2]: import numpy.core.umath_tests

  In [3]: (numpy.core.umath_tests.matrix_multiply(a, b) == [dot(ai, bi) for (ai,
bi) in zip(a, b)]).all()
  Out[3]: True

Best, Hansres

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