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 p...@iki.fi 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
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
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 sufficient
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:
I don't
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
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,
I get the same result on 1.4.1
On Thu, Jul 22, 2010 at 1:00 PM, Johann Hibschman
jhibschman+nu...@gmail.com 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
For a
John Salvatier wrote:
I get the same result on 1.4.1
On Thu, Jul 22, 2010 at 1:00 PM, Johann Hibschman
jhibschman+nu...@gmail.com mailto:jhibschman%2bnu...@gmail.com wrote:
I'm trying to understand numpy.subtract.reduce. The documentation
doesn't seem to match the behavior. The
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
___
Pauli Virtanen p...@iki.fi 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
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