[SciPy-User] scipy.linalg.solve()'s overwrite option does not work
josef.pktd@gmai...
josef.pktd@gmai...
Sat Nov 6 17:24:34 CDT 2010
On Sat, Nov 6, 2010 at 6:18 PM, braingateway <braingateway@gmail.com> wrote:
> josef.pktd@gmail.com :
>> On Sat, Nov 6, 2010 at 5:46 PM, braingateway <braingateway@gmail.com> wrote:
>>
>>> Joe Kington :
>>>
>>>> On Sat, Nov 6, 2010 at 12:13 PM, braingateway <braingateway@gmail.com
>>>> <mailto: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
>>>>
>>>>
>>>> Just because two ndarray objects aren't the same doesn't mean that
>>>> they don't share the same memory...
>>>>
>>>> Consider this:
>>>> import numpy as np
>>>> x = np.arange(10)
>>>> y = x.T
>>>> x is y # --> Yields False
>>>> Nonetheless, x and y share the same data, and storing y doesn't double
>>>> the amount of memory used, as it's effectively just a pointer to the
>>>> same memory as x
>>>>
>>>> Instead of using "is", you should use "numpy.may_share_memory(x, y)"
>>>>
>>> Thanks a lot for pointing this out! I were struggling to figure out
>>> whether the different objects share memory or not. And good to know
>>> a0=numpy.matrix(a) actually did not share the memory.
>>> >>> print 'a0 shares memory with a?', npy.may_share_memory(a,a0)
>>> a0 shares memory with a? False
>>> >>> print 'b0 shares memory with b?', npy.may_share_memory(b,b0)
>>> b0 shares memory with b? False
>>> I also heard that even may_share_memory is 'True', does not necessarily
>>> mean they share any element. Maybe, is 'a0.base is a' usually more
>>> suitable for this purpose?
>>>
>>> Back to the original question: is there anyone actually saw the
>>> overwrite_a or overwrite_b really showed its effect?
>>> If you could show me a repeatable example, not only for
>>> scipy.linalg.solve(), it can also be other functions, who provide this
>>> option, such as eig(). If it does not show any advantage in memory
>>> usage, I might still using numpy.linalg.
>>>
>>
>>
>> import numpy as np
>>
>> a=np.random.randn(20,20)
>> abak = a.copy()
>> x=np.random.randn(20,4)
>> xbak = x.copy()
>> a=np.matrix(a)
>> x=np.matrix(x)
>> b=a*x
>> b = np.array(b)
>> bbak = b.copy()
>>
>> import scipy.linalg as sla
>> a0=np.matrix(a)
>> print a is a0
>> #False
>> b0=np.matrix(b)
>> print b is b0
>> #False
>> X=sla.solve(a,b,overwrite_a=True,debug=True)
>> print X is b
>> #False
>> print (X==b).all()
>> #False
>> print 'a:', (a0==a).all(), (abak==a).all()
>> print 'b:', (b0==b).all(), (bbak==b).all()
>> #
>> Y = sla.solve(a,b,overwrite_a=True,overwrite_b=True,debug=True)
>> print 'a:', (a0==a).all(), (abak==a).all()
>> print 'b:', (b0==b).all(), (bbak==b).all()
>>
>> print (X==Y).all()
>>
>> printout
>> -----------
>> False
>> False
>> solve:overwrite_a= True
>> solve:overwrite_b= False
>> False
>> False
>> a: False False
>> b: True True
>> solve:overwrite_a= True
>> solve:overwrite_b= True
>> a: False False
>> b: True True
>> False
>>
>>
> Thanks a lot! So right now I see it might be some bugs in my Scipy
> version!, After running your code, I got following result:
> >>> sla.__version__
> '0.4.9'
mine:
>>> sla.__version__
'0.4.9'
I have no idea if the overwrite option depends on which Lapack/Blas
implementation is used. I have a generic oldish ATLAS.
Josef
> >>>
> False
> False
> solve:overwrite_a= True
> solve:overwrite_b= False
> False
> False
> a: True True
> b: True True
> solve:overwrite_a= True
> solve:overwrite_b= True
> a: True True
> b: True True
> True
>> The first solve overwrites a, the second solve solves a different
>> problem and the solutions X and Y are not the same.
>> (if the first solve allows overwriting of be instead, then X==Y )
>>
>> I never got a case with overwritten b, and as you said there was no
>> sharing memory with the original array, the copy has always the same
>> result as your original.
>>
>> This was for quick playing with your example, no guarantee on no mistakes.
>>
>> Josef
>>
>>
>>
>>>> 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.
>>>>
>>>> Sincerely,
>>>> LittleBigBrain
>>>> > _______________________________________________
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>>>> > http://mail.scipy.org/mailman/listinfo/scipy-user
>>>> >
>>>>
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