[SciPy-User] scipy.linalg.solve()'s overwrite option does not work

josef.pktd@gmai... josef.pktd@gmai...
Sat Nov 6 12:51:57 CDT 2010


On Sat, Nov 6, 2010 at 1:13 PM, braingateway <braingateway@gmail.com> wrote:
> David Warde-Farley:
>> On 2010-11-05, at 9:21 PM, braingateway wrote:
>>
>>
>>> Hi everyone,
>>> I believe the overwrite option is used for reduce memory usage. But I
>>> did following test, and find out it does not work at all. Maybe I
>>> misunderstood the purpose of overwrite option. If anybody could explain
>>> this, I shall highly appreciate your help.
>>>
>>
>> First of all, this is a SciPy issue, so please don't crosspost to NumPy-discussion.
>>
>>
>>>>>> a=npy.random.randn(20,20)
>>>>>> x=npy.random.randn(20,4)
>>>>>> a=npy.matrix(a)
>>>>>> x=npy.matrix(x)
>>>>>> b=a*x
>>>>>> import scipy.linalg as sla
>>>>>> a0=npy.matrix(a)
>>>>>> a is a0
>>>>>>
>>> False
>>>
>>>>>> b0=npy.matrix(b)
>>>>>> b is b0
>>>>>>
>>> False
>>>
>>
>> You shouldn't use 'is' to compare arrays unless you mean to compare them by object identity. Use all(b == b0) to compare by value.
>>
>> David
>>
>>
> Thanks for reply, but I have to say u did not understand my post at all.
> I did this 'is' comparison on purpose, because I wanna know if the
> overwrite flag is work or not.
> See following example:
>  >>> a=numpy.matrix([0,0,1])
>  >>> a
> matrix([[0, 0, 1]])
>  >>> a0=a
>  >>> a0 is a
> True
> This means a0 and a is actually point to a same object. Then a0 act
> similar to the C pointer of a.
> I compared a0/b0 and a/b by 'is' first to show I did create a new object
> from the original matrix, so the following (a0==a).all() comparison can
> actually prove the values inside the a and b were not overwritten.

even if  "a0 is not a" they can still share the same memory:


>>> aa = np.ones(5)
>>> bb = aa[:,None]
>>> aa is bb
False
>>> bb[0] = 10
>>> aa
array([ 10.,   1.,   1.,   1.,   1.])
>>> (aa == bb.ravel()).all()
True

When I check a variation on your example, I have `a` overwritten but
not `b`.  But the docstring makes only a weak statement, allowing that
it can be overwritten, doesn't necessarily mean it will be overwritten
in every case.


Josef


>
> Sincerely,
> LittleBigBrain
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