[SciPy-User] scipy.linalg.solve()'s overwrite option does not work
braingateway
braingateway@gmail....
Sat Nov 6 17:18:15 CDT 2010
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'
>>>
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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