2010/10/5 Chris Withers <ch...@simplistix.co.uk>: > Hi All, > > I can't find any docs on this behavior. > > So, I have a python function. To keep it simple, lets just do addition: > > def add(x,y): > print x,y > retun x+y > > So, I can turn this into a ufunc as follows: > > uadd = np.frompyfunc(add,2,1)
As a side remark, note that this will always return dtype=numpy.object arrays. Maybe numpy.vectorize() is more appropriate for your use-case. > Now, I can apply it to an array: > > >>> uadd.accumulate(np.arange(3,10)) > 3 4 > 7 5 > 12 6 > 18 7 > 25 8 > 33 9 > array([3, 7, 12, 18, 25, 33, 42], dtype=object) > > Okay, but where did the initial 3 come from? > http://docs.scipy.org/doc/numpy/reference/generated/numpy.ufunc.accumulate.html#numpy.ufunc.accumulate > suggests that: > > r = np.empty(len(A)) > t = op.identity # op = the ufunc being applied to A's elements > for i in xrange(len(A)): > t = op(t, A[i]) > r[i] = t > return r I see from the behaviour, that the implementation makes use of the following behaviour of the identity ``op.identity`` with respect to ``op``: >>> op(op.identity, X) == X True So the implementation acts in agnosticism of ``op.identity`` by assuming that the first call of ``t = op(t, A[i])`` in your example will just yield ``A[0]``, while i == 0. This is the work flow: [ t = op.identity ] # in brackets because it cannot be executed, it's pseudo-code i = 0 ===== t = op(t, A[0]) = A[0] r[0] = t = A[0] i = 1 ===== t = op(t, A[1]) = op(A[0], A[1]) r[1] = t i = 2 ===== t = op(t, A[2]) = op(op(A[0], A[1]), A[2]) r[2] = t and so on ... clear now? Friedrich > ...but: > > >>> print uadd.identity > None It's simply not set, and will also not be used. > ...and: > > >>> add(None,3) > None 3 > Traceback (most recent call last): > File "<stdin>", line 1, in <module> > File "<stdin>", line 3, in add > TypeError: unsupported operand type(s) for +: 'NoneType' and 'int' As said, the code in the docs is just pseudo-code, it's not the actual (C?) implementation, it just has the same semantics. > So, where is the reason that the 3 ends up in the output array documented? > > Also, what if I want to specify the identity of my newly created ufunc? > I have a case where I want to specify it as zero: > > >>> uadd.identity = 0 > Traceback (most recent call last): > File "<stdin>", line 1, in <module> > AttributeError: attribute 'identity' of 'numpy.ufunc' objects is not > writable I think you have to write an own class for this. You cannot subclass numpy.ufunc ("it's not an acceptable base class" Python says). _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
