Re: [Numpy-discussion] bug in round with negative number of decimals

[Charles R H. <cha...@gm...>]

On 9/4/06, Charles R Harris <cha...@gm...> wrote:
>
>
>
> On 9/4/06, Sebastian Haase <ha...@ms...> wrote:
> >
> > Paulo J. S. Silva wrote:
> > > Once again, the information that singed zero is part of IEEE standard
> > is
> > > in the paper I cited in my last message.
> > >
> > > It is very important to be able to compute the sign of an overflowed
> > > quantity in expressions like 1/x when x goes to zero.
> > >
> > > Best,
> > >
> > > Paulo
> > Hi,
> >
> > This is all very interesting ( and confusing (to me, maybe others) at
> > the same time ...) ...
> >
> > Question 0: Are you sure this is not a bug ?
> >   >>> N.array([66]).round(-1)
> > [60]
>
>
> That does look like a bug.
>
> >>> array([66]).round(-1)
> array([60])
> >>> array([66.]).round(-1)
> array([ 70.])
>
> I suspect it is related to the integer data type of the first example. The
> code probably does something like this:
>
> round(66/10)*10
>
> and 66/10 == 6 in integer arithmetic.
>

The problem is somewhere in this bit of code:

// f will be a double
f = PyFloat_FromDouble(power_of_ten(decimals));
if (f==NULL) return NULL;

// op1 will be division
ret = PyObject_CallFunction(op1, "OOO", a, f, out);
if (ret==NULL) goto finish;

// round in place.
tmp = PyObject_CallFunction(n_ops.rint, "OO", ret, ret);
if (tmp == NULL) {Py_DECREF(ret); ret=NULL; goto finish;}

I suspect the problem is the division, which has integer 'a', double 'f',
and integer 'out'. Integer out is probably the killer. Let's see:

# use doubles for output
>>> out = zeros(1)
>>> array([66]).round(decimals=-1, out=out)
array([ 70.])

yep, that's the problem

Chuck.

<snip>