On 4/23/07, Christopher Barker <Chris.Barker@noaa.gov> wrote:

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:

Sometimes it do, sometimes it don't:

In [8]: x = ones((8,8))

In [9]: y = x[:3,:6]

In [10]: z = y.reshape(2,9)

In [11]: z[...] = 0

In [12]: x
Out[12]:
array([[ 1., 1., 1., 1., 1., 1., 1., 1.],
       [ 1., 1., 1., 1., 1., 1., 1., 1.],
       [ 1., 1., 1., 1., 1., 1., 1., 1.],
       [ 1., 1., 1., 1., 1., 1., 1., 1.],
       [ 1., 1., 1., 1., 1., 1., 1., 1.],
       [ 1., 1., 1., 1., 1., 1., 1., 1.],
       [ 1., 1., 1., 1., 1., 1., 1., 1.],
       [ 1., 1., 1., 1., 1., 1., 1., 1.]])

In [13]: z
Out[13]:
array([[ 0., 0., 0., 0., 0., 0., 0., 0., 0.],
       [ 0., 0., 0., 0., 0., 0., 0., 0., 0.]

As Ann pointed out, sometimes reusing the same array isn't possible. This makes the use of reshaped arrays as l-values subject to subtle errors, so caution is advised.

Chuck