On 7/26/2010 9:41 AM, Johann Hibschman wrote:
> if reduce were defined as a *right* fold, then it would make sense for
> subtract (and divide) to use the right identity
Instead of deviating from the Python definition of reduce,
it would imo make more sense to introduce new functions,
sayfoldl and
Pauli Virtanen writes:
>> Returning a *right* identity for an operation that is otherwise a *left*
>> fold is very odd, no matter how you slice it. That is what looks like
>> special casing...
>
> I think I see your point now.
I know this is unlikely to happen, since it would break things for a
Fri, 23 Jul 2010 11:17:56 -0400, Alan G Isaac wrote:
[clip]
> I also do not understand why there would have to be any special cases.
That's a technical issue: e.g. prod() is implemented via
np.multiply.reduce, and it is not clear to me whether it is possible, in
the ufunc machinery, to leave the
> Fri, 23 Jul 2010 10:29:47 -0400, Alan G Isaac wrote:
>> >>> np.subtract.reduce([])
>> 0.0
>>
>> Getting a right identity for an empty array is surprising. Matching
>> Python's behavior (raising a TypeError) seems desirable. (?)
On 7/23/2010 10:37 AM, Pauli Virtanen wrote
Fri, 23 Jul 2010 10:29:47 -0400, Alan G Isaac wrote:
[clip]
> >>> np.subtract.reduce([])
> 0.0
>
> Getting a right identity for an empty array is surprising. Matching
> Python's behavior (raising a TypeError) seems desirable. (?)
I don't think matching Python's behavior is a suf
On 7/22/2010 4:00 PM, Johann Hibschman wrote:
> I'm trying to understand numpy.subtract.reduce. The documentation
> doesn't seem to match the behavior. The documentation claims
>
>For a one-dimensional array, reduce produces results equivalent to:
>
>r = op.identity
>for i in xrange(l
Pauli Virtanen writes:
> The documentation is incorrect.
Thanks. The observed behavior is more like:
if len(A) == 0:
return op.identity
else:
r = A[0]
for i in xrange(1, len(A):
r = op(r, A[i])
return r
-Johann
___
NumPy-D
Thu, 22 Jul 2010 15:00:50 -0500, Johann Hibschman wrote:
[clip]
> Now, I'm on an older version (1.3.0), which might be the problem, but
> which is "correct" here, the code or the docs?
The documentation is incorrect.
--
Pauli Virtanen
___
NumPy-Discus
John Salvatier wrote:
> I get the same result on 1.4.1
>
> On Thu, Jul 22, 2010 at 1:00 PM, Johann Hibschman
> mailto:jhibschman%2bnu...@gmail.com>> wrote:
>
> I'm trying to understand numpy.subtract.reduce. The documentation
> doesn't seem to match the behavior. The documentation claims
I get the same result on 1.4.1
On Thu, Jul 22, 2010 at 1:00 PM, Johann Hibschman <
jhibschman+nu...@gmail.com > wrote:
> I'm trying to understand numpy.subtract.reduce. The documentation
> doesn't seem to match the behavior. The documentation claims
>
> For a one-dimensional array, reduce prod
I'm trying to understand numpy.subtract.reduce. The documentation
doesn't seem to match the behavior. The documentation claims
For a one-dimensional array, reduce produces results equivalent to:
r = op.identity
for i in xrange(len(A)):
r = op(r,A[i])
return r
However, numpy.subtra
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