# [Numpy-discussion] Difference between shape=() and shape=(1,)

John Reid j.reid@mail.cryst.bbk.ac...
Tue Jul 13 11:54:39 CDT 2010

```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?

Thanks for any tips,
John.

```