[Numpy-discussion] Matching 0-d arrays and NumPy scalars

Eric Firing efiring at hawaii.edu
Fri Feb 22 00:02:24 EST 2008


Travis E. Oliphant wrote:
>> Travis,
>>
>> You have been getting mostly objections so far;
> I wouldn't characterize it that way, but yes 2 people have pushed back a 
> bit, although one not directly speaking to the proposed behavior.  
> 
> The issue is that [] notation does more than just "select from a 
> container" for NumPy arrays.   In particular, it is used to reshape an 
> array to more dimensions:  [..., newaxis]
> 
> A common pattern is to reduce over a dimension and then re-shape the 
> result so that it can be combined with the un-reduced object.  
> Broadcasting makes this work if the dimension being reduced along is the 
> first dimension.  But, broadcasting is not enough if you want the 
> reduction dimension to be arbitrary:
> 
> Thus,
> 
> y = add.reduce(x, axis=-1)  produces an N-1 array if x is 2-d and a 
> numpy scalar if x is 1-d.

Why does it produce a scalar instead of a 0-d array?  Wouldn't the 
latter take care of your use case, and be consistent with the action of 
reduce in removing one dimension?

I'm not opposed to your suggested change--just trying to understand it. 
I'm certainly sympathetic to your use case, below. I dimly recall 
extensive and confusing (to me) discussions of numpy scalars versus 0-d 
arrays during your heroic push to make numpy gel, and I suspect the 
answer is somewhere back in those discussions.

Eric

> 
> Suppose y needs to be subtracted from x. 
> 
> If x is 2-d, then
> 
>  >>> x - y[...,newaxis]
> 
> is the needed code. But, if x is 1-d,  then
> 
>  >>> x - y[..., newaxis]
> 
> returns an error  and a check must be done to handle the case 
> separately.  If y[..., newaxis]  worked and produced a 1-d array when y 
> was a numpy scalar, this could be avoided.
> 
> 
> -Travis O.
> 
> 
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