[Numpy-discussion] Inplace reshape
Christopher Barker
Chris.Barker@noaa....
Mon Apr 23 12:36:26 CDT 2007
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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Chris.Barker@noaa.gov
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