[Numpy-discussion] Do ufuncs returned by frompyfunc(), have the out arg?

Ken Basye kbasye1@jhu....
Tue Apr 6 22:48:24 CDT 2010


From: Anne Archibald <peridot.faceted@gmail.com>

On 6 April 2010 15:42, Ken Basye <kbasye1@jhu.edu> wrote:

> > Folks,
> >  I hope this is a simple question.  When I created a ufunc with
> > np.frompyfunc(), I got an error when I called the result with an 'out'
> > argument:
>   

In fact, ordinary ufuncs do not accept names for their arguments. This
is annoying, but fixing it involves rooting around in the bowels of
the ufunc machinery, which are not hacker-friendly.

Anne


> >  >>> def foo(x): return x * x + 1
> >  >>> ufoo = np.frompyfunc(foo, 1, 1)
> >  >>> arr = np.arange(9).reshape(3,3)
> >  >>> ufoo(arr, out=arr)
> > Traceback (most recent call last):
> >  File "<stdin>", line 1, in <module>
> > TypeError: 'out' is an invalid keyword to foo (vectorized)
> >
> > But I notice that if I just put the array there as a second argument, it
> > seems to work:
> >  >>> ufoo(arr, arr)
> > array([[2, 5, 26],
> >       [101, 290, 677],
> >       [1370, 2501, 4226]], dtype=object)
> >
> > # and now arr is the same as the return value
> >
> >
> > Is it reasonable to conclude that there is an out-arg in the resulting
> > ufunc and I just don't know the right name for it?  I also tried putting
> > some other right-shaped array as a second argument and it did indeed get
> > filled in.
> >
> >   Thanks as always,
> >       Ken

Thanks - I hadn't noticed that it's apparently only the array methods 
that can take keyword arguments.  So I assume that if I call a '1-arg' 
ufunc (whether from frompyfunc or an already existing one) with a second 
argument, the second argument will be used as the output location.
   Ken

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