2007/7/18, Tom Goddard <goddard@cgl.ucsf.edu >:

Does taking a slice of a flatiter always make a copy?  That appears to
be the behaviour in numpy 1.0.3.
For example a.flat[1:3][0] = 5 does not modify the original array a,
even when a is contiguous.  Is this a bug?

>>> import numpy as n
>>> n.version.version
'1.0.3'
>>> a = n.zeros((2,3), n.int32)
>>> a
array([[0, 0, 0],
       [0, 0, 0]])
>>> b = a.flat[1:3]
>>> b[0] = 5
>>> b
array([5, 0])
>>> a
array([[0, 0, 0],
       [0, 0, 0]])
>>> b = None
>>> a
array([[0, 0, 0],
       [0, 0, 0]])

This behavior does not seem to match what is described in the Numpy book
(Dec 7, 2006 version), section 3.1.3 "Other attributes", page 51:

"flat

  Returns an iterator object (numpy.flatiter) that acts like a 1-d
version of the array.
  1-d indexing works on this array and it can be passed in to most
routines as
  an array wherein a 1-d array will be constructed from it.  The new 1-d
array
  will reference this array's data if this array is C-style contiguous,
otherwise,
  new memory will be allocated for the 1-d array, the UPDATEIFCOPY flag
  will be set for the new array, and this array will have its WRITEABLE flag
  set FALSE until the the last reference to the new array disappears.
When the
  last reference to the new 1-d array disappears, the data will be
copied over to
  this non-contiguous array.  This is done so that a.flat effectively
references the
  current array regardless of whether or not it is contiguous or
non-contiguous."

Tom Goddard
UC San Francisco

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