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