[Numpy-discussion] subtract.reduce behavior
Alan G Isaac
aisaac@american....
Fri Jul 23 09:29:47 CDT 2010
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(len(A)):
> r = op(r,A[i])
> return r
>
> However, numpy.subtract.reduce([1,2,3]) gives me 1-2-3==-4, not
> 0-1-2-3==-6.
The behavior does not quite match Python's reduce.
The rule seems to be:
return the *right identity* for empty arrays,
otherwise behave like Python's reduce.
>>> import operator as o
>>> reduce(o.sub, [1,2,3], 0)
-6
>>> reduce(o.sub, [1,2,3])
-4
>>> reduce(o.sub, [])
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: reduce() of empty sequence with no initial value
>>> np.subtract.reduce([])
0.0
Getting a right identity for an empty array is surprising.
Matching Python's behavior (raising a TypeError) seems desirable. (?)
Unfortunately Python's reduce does not make ``initializer`` a
keyword, but maybe NumPy could add this keyword anyway?
(Not sure that's a good idea.)
Alan Isaac
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