[Numpy-discussion] array - dimension size of 1-D and 2-D examples

[Sebastian Berg]

On Tue, 2018-01-09 at 12:27 +0000, Martin.Gfeller at swisscom.com wrote:
> Hi Derek
> 
> I have a related question:
> 
> Given:
> 
> 	a = numpy.array([[0,1,2],[3,4]])
> 	assert a.ndim == 1
> 	b = numpy.array([[0,1,2],[3,4,5]])
> 	assert b.ndim == 2
> 
> Is there an elegant  way to force b to remain a 1-dim object array? 
> 

You will have to create an empty object array and assign the lists to
it.

```
b = np.empty(len(l), dtype=object)
b[...] = l
```

> I have a use case where normally the sublists are of different
> lengths, but I get a completely different structure when they are
> (coincidentally in my case) of the same length.
> 
> Thanks and best regards, Martin
> 
> 
> Martin Gfeller, Swisscom / Enterprise / Banking / Products / Quantax
> 
> Message: 1
> Date: Sun, 31 Dec 2017 00:11:48 +0100
> From: Derek Homeier <derek at astro.physik.uni-goettingen.de>
> To: Discussion of Numerical Python <numpy-discussion at python.org>
> Subject: Re: [Numpy-discussion] array - dimension size of 1-D and 2-D
> 	examples
> Message-ID:
> 	<CC548593-308B-4561-A03C-D3017C707108 at astro.physik.uni-goetting
> en.de>
> Content-Type: text/plain; charset=utf-8
> 
> On 30 Dec 2017, at 5:38 pm, Vinodhini Balusamy <me.vinob at gmail.com>
> wrote:
> > 
> > Just one more question from the details you have provided which
> > from 
> > my understanding strongly seems to be Design [DEREK] You cannot
> > create 
> > a regular 2-dimensional integer array from one row of length 3
> > > and a second one of length 0. Thus np.array chooses the next
> > > most 
> > > basic type of array it can fit your input data in
> 
> Indeed, the general philosophy is to preserve the structure and type
> of your input data as far as possible, i.e. a list is turned into a
> 1d-array, a list of lists (or tuples etc?) into a 2d-array,_ if_ the
> sequences are of equal length (even if length 1).
> As long as there is an unambiguous way to convert the data into an
> array (see below).
> 
> >    Which is the case,  only if an second one of length 0 is given.
> >    What about the case 1 :
> > > > > x12 = np.array([[1,2,3]])
> > > > > x12
> > 
> > array([[1, 2, 3]])
> > > > > print(x12)
> > 
> > [[1 2 3]]
> > > > > x12.ndim
> > 
> > 2
> > > > > 
> > > > > 
> > 
> > This seems to take 2 dimension.
> 
> Yes, structurally this is equivalent to your second example
> 
> > also,
> > > > x12 = np.array([[1,2,3],[0,0,0]])
> > > > print(x12)
> 
> [[1 2 3]
>  [0 0 0]]
> > > > x12.ndim
> 
> 2
> 
> > I presumed the above case and the case where length 0 is provided
> > to be treated same(I mean same behaviour).
> > Correct me if I am wrong.
> > 
> 
> In this case there is no unambiguous way to construct the array - you
> would need a shape (2, 3) array to store the two lists with 3
> elements in the first list. Obviously x12[0] would be
> np.array([1,2,3]), but what should be the value of x12[1], if the
> second list is empty - it could be zeros, or repeating x12[0], or
> simply undefined. np.array([1, 2, 3], [4]]) would be even less
> clearly defined.
> These cases where there is no obvious ?right? way to create the array
> have usually been discussed at some length, but I don?t know if this
> is fully documented in some place. For the essentials, see
> 
> https://docs.scipy.org/doc/numpy/reference/routines.array-creation.ht
> ml
> 
> note also the upcasting rules if you have e.g. a mix of integers and
> reals or complex numbers, and also how to control shape or data type
> explicitly with the respective keywords.
> 
> 					Derek
> 
> 
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