[Numpy-discussion] Difference between shape=() and shape=(1,)
Benjamin Root
ben.root@ou....
Tue Jul 13 19:12:19 CDT 2010
On Tue, Jul 13, 2010 at 12:45 PM, Kurt Smith <kwmsmith@gmail.com> wrote:
> On Tue, Jul 13, 2010 at 11:54 AM, John Reid <j.reid@mail.cryst.bbk.ac.uk>
> wrote:
> > Hi,
> >
> > I have some arrays of various shapes in which I need to set any NaNs to
> > 0. I have been doing the following:
> >
> > a[numpy.where(numpy.isnan(a)] = 0.
> >
> >
> >
> > as you can see here:
> >
> > In [20]: a=numpy.ones(2)
> >
> > In [21]: a[1]=numpy.log(-1)
> >
> > In [22]: a
> > Out[22]: array([ 1., NaN])
> >
> > In [23]: a[numpy.where(numpy.isnan(a))]=0.
> >
> > In [24]: a
> > Out[24]: array([ 1., 0.])
> >
> >
> >
> > Unfortunately, I've just discovered that when a.shape == () this doesn't
> > work at all. For example:
> >
> > In [41]: a=numpy.array((1.))
> >
> > In [42]: a.shape
> > Out[42]: ()
> >
> > In [43]: a[numpy.where(numpy.isnan(a))]=0.
> >
> > In [44]: a
> > Out[44]: array(0.0)
> >
> >
> >
> >
> >
> > but if the shape is (1,), everything is ok:
> >
> > In [47]: a=numpy.ones(1)
> >
> > In [48]: a.shape
> > Out[48]: (1,)
> >
> > In [49]: a[numpy.where(numpy.isnan(a))]=0.
> >
> > In [50]: a
> > Out[50]: array([ 1.])
> >
> >
> >
> > What's the difference between the 2 arrays with different shapes?
> >
> > If I pass a scalar into numpy.asarray() why do I get an array of shape
> > () back? In my case this has caused a subtle bug.
> >
> > Is there a better way to set NaNs in an array to 0?
>
> You could make use of np.atleast_1d, and then everything would be
> canonicalized:
>
> In [33]: a = np.array(np.nan)
>
> In [34]: a
> Out[34]: array(nan)
>
> In [35]: a1d = np.atleast_1d(a)
>
> In [36]: a1d
> Out[36]: array([ NaN])
>
> In [37]: a
> Out[37]: array(nan)
>
> In [38]: a1d.base is a
> Out[38]: True
>
> In [39]: a1d[np.isnan(a1d)] = 0.
>
> In [40]: a1d
> Out[40]: array([ 0.])
>
> In [41]: a
> Out[41]: array(0.0)
>
> So Keith's nan_replace would be:
>
> In [42]: def nan_replace(a, fill=0.0):
> ....: a_ = np.atleast_1d(a)
> ....: a_[np.isnan(a_)] = fill
> ....:
>
>
Maybe I am missing something subtle, but what about numpy's nan_to_num()
function?
Ben Root
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