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

[Martin.Gfeller at swisscom.com]

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? 

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-goettingen.de>
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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.html

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