[Numpy-discussion] Inplace reshape

[Christopher Barker]

Gael Varoquaux wrote:
> Unless I miss something obvious "a.reshape()" doesn't modify a, which is
> somewhat missleading, IMHO.

quite correct. .reshape() creates a new array that shared data with the 
original:

 >>> import numpy
 >>> a = numpy.zeros((2,3))

 >>> help(a.reshape)
Help on built-in function reshape:

reshape(...)
     a.reshape(d1, d2, ..., dn, order='c')

     Return a new array from this one.  The new array must have the same 

     number of elements as self.  Also always returns a view or raises a
     ValueError if that is impossible.;

 >>> a
array([[ 0.,  0.,  0.],
        [ 0.,  0.,  0.]])

 >>> b = a.reshape((6,))
 >>> a
array([[ 0.,  0.,  0.],
        [ 0.,  0.,  0.]])

so a hasn't changed.

 >>> b
array([ 0.,  0.,  0.,  0.,  0.,  0.])

but b is a different shape.

 >>> b[1] = 5
 >>> a
array([[ 0.,  5.,  0.],
        [ 0.,  0.,  0.]])

b and a share data.

if you want to change the shape of a:

 >>> a.shape = (6,)
 >>> a
array([ 0.,  5.,  0.,  0.,  0.,  0.])

-Chris



-- 
Christopher Barker, Ph.D.
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