[SciPy-User] scipy.linalg.solve()'s overwrite option does not work
braingateway
braingateway@gmail....
Sat Nov 6 17:33:30 CDT 2010
Joe Kington :
>
>
> On Sat, Nov 6, 2010 at 4:46 PM, braingateway <braingateway@gmail.com
> <mailto:braingateway@gmail.com>> wrote:
>
> Joe Kington :
> >
> > On Sat, Nov 6, 2010 at 12:13 PM, braingateway
> <braingateway@gmail.com <mailto:braingateway@gmail.com>
> > <mailto: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?
>
>
> Not to take this on too much of a tangent, but since you asked:
>
> "x.base is y.base" usually doesn't work, even when x.base and y.base
> point to the same memory...
>
> Again, consider the transpose of an array combined with a bit of indexing:
> import numpy as np
> x = np.arange(10)
> y = x[:10].T
> x.base is y.base # <-- yields False
> y.base is x # <-- yields False
> np.may_share_memory(x,y) # <-- correctly yields True
>
> The moral of this story is don't use object identity of any sort to
> determine if ndarrays share memory.
>
> I also heard that even may_share_memory is 'True', does not
> necessarily
> mean they share any element.
>
>
> The reason why "may_share_memory" is not a guarantee that the arrays
> actually do is due to situations like this:
> import numpy as np
> x = np.arange(10)
> a = x[::2]
> b = x[1::2]
> np.may_share_memory(a, b) # <-- yields True
>
> However, we could change every element in "a" without affecting "b",
> so they don't _actually_ share memory. The reason why
> "may_share_memory" yields True is essentially due to the fact that "a"
> and "b" are both views into overlapping regions of the same array.
>
> As far as I know, "may_share_memory" is really the only test available
> without dropping down to C to determine if two ndarray objects share
> chunks of memory. It just errs on the side of caution, if you will.
> If it returns True, then the two ndarrays refer to overlapping regions
> of memory.
>
Oh, this is really clarify all the mystery in my mind. I think I will
backup this mail :)
Thanks Sincerely,
LittleBigBrain
> Of course, checking to see if the contents of the array are exactly
> the same is a _completely_ different test, and "(x ==
> y.ravel()).all()" is a fine way to do that. Just keep in mind that
> checking if the elements are the same is very different than checking
> if the arrays share the same memory. For example:
> x = np.arange(10)
> y = x.copy()
> (x == y).all() # <-- correctly yields True
> np.may_share_memory(x, y) # <-- correctly yields False
> They're two different tests, entirely. Of course, for what you're
> doing here, testing to see if the elements are the same is entirely
> reasonable.
>
>
>
>
> Back to the original question: is there anyone actually saw the
> overwrite_a or overwrite_b really showed its effect?
>
>
> I'm afraid I'm not much help on the original question...
>
>
>
> 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.
> >
> > 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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