[Numpy-discussion] Difference between shape=() and shape=(1,)
Kurt Smith
kwmsmith@gmail....
Tue Jul 13 12:45:42 CDT 2010
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
....:
>
> Thanks for any tips,
> John.
>
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