Hey Tim,

Here is a short summary:

Reductions and indexing return different types based on the number of
dimensions of the input array:

> >> b = sum(a)
> >> l = len(b) # or whatever

This code works happily if "a" is 2 or more dimensions, but will fail if
it is 1d because the sum(a) will return a scalar in this case.  To write
generic code, you have to put an if/else statement in to check whether b
is a scalar or an array:

> >> b = sum(a)
> >> if type(b) is ArrayType:
...     l = len(b)
... else:
...     l = 1 # or whatever you need

or less verbose but still unpleasant:  

> >> b = asarray(sum(a))
> >> l = len(b)

eric

>  -----Original Message-----
>  From: numpy-discussion-admin@[...].net [mailto:numpy-
>  discussion-admin@[...].net] On Behalf Of Tim Hochberg
>  Sent: Friday, September 13, 2002 3:18 PM
>  To: numpy-discussion@[...].net
>  Subject: Re: [Numpy-discussion] rank-0 arrays
>  
>  
>  
>  
>  [Perry wrote about Travis's proposal]
>  
>  > refresh my memory about how this proposal would work. I've heard
>  > proposals to add new types to Python itself, but that seems out of
>  > the question. Are you talking about adding new scalar types as
>  > module? E.g.
>  [snip]
>  > In any case, doesn't this still have the problem that Eric
complained
>  > about, having to test for whether a result is an array or a scalar
>  > (which was one of the drivers to produce rank-0 results).
>  
>  Hi Perry,
>  
>  I like Travis's proposal the best of those I've seen so far, but I
don't
>  recall the details of Eric's problem. Could you refresh us as to the
>  basics
>  of it?.
>  
>  -tim
>  
>  
>  
>  
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