[Numpy-discussion] speeding up the following expression

Sat Nov 12 11:32:58 EST 2011

On Sat, Nov 12, 2011 at 9:59 AM, <josef.pktd at gmail.com> wrote:

> On Sat, Nov 12, 2011 at 10:31 AM, Warren Weckesser
> <warren.weckesser at enthought.com> wrote:
> >
> >
> > On Sat, Nov 12, 2011 at 6:43 AM, <josef.pktd at gmail.com> wrote:
> >>
> >> On Sat, Nov 12, 2011 at 3:36 AM, Geoffrey Zhu <zyzhu2000 at gmail.com>
> wrote:
> >> > Hi,
> >> >
> >> > I am playing with multiple ways to speed up the following expression
> >> > (it is in the inner loop):
> >> >
> >> >
> >> > C[1:(M - 1)]=(a * C[2:] + b * C[1:(M-1)] + c * C[:(M-2)])
> >> >
> >> > where C is an array of about 200-300 elements, M=len(C), a, b, c are
> >> > scalars.
> >> >
> >> > I played with numexpr, but it was way slower than directly using
> >> > numpy. It would be nice if I could create a Mx3 matrix without copying
> >> > memory and so I can use dot() to calculate the whole thing.
> >> >
> >> > Can anyone help with giving some advices to make this faster?
> >>
> >> looks like a np.convolve(C, [a,b,c])  to me except for the boundary
> >> conditions.
> >
> >
> >
> > As Josef pointed out, this is a convolution.  There are (at least)
> > three convolution functions in numpy+scipy that you could use:
> > numpy.convolve, scipy.signal.convolve, and scipy.ndimage.convolve1d.
> > Of these, scipy.ndimage.convolve1d is the fastest.  However, it doesn't
> > beat the simple expression
> >    a*x[2:] + b*x[1:-1] + c*x[:-2]
> > Your idea of forming a matrix without copying memory can be done using
> > "stride tricks", and for arrays of the size you are interested in, it
> > computes the result faster than the simple expression (see below).
> >
> > Another fast alternative is to use one of the inline code generators.
> > This example is a easy to implement with scipy.weave.blitz, and it gives
> > a big speedup.
> >
> > Here's a test script:
> >
> > #----- convolve1dtest.py -----
> >
> >
> > import numpy as np
> > from numpy.lib.stride_tricks import as_strided
> > from scipy.ndimage import convolve1d
> > from scipy.weave import blitz
> >
> > # weighting coefficients
> > a = -0.5
> > b = 1.0
> > c = -0.25
> > w = np.array((a,b,c))
> > # Reversed w:
> > rw = w[::-1]
> >
> > # Length of C
> > n = 250
> >
> > # The original version of the calculation:
> > # Some dummy data
> > C = np.arange(float(n))
> > C[1:-1] = a * C[2:] + b * C[1:-1] + c * C[:-2]
> > # Save for comparison.
> > C0 = C.copy()
> >
> > # Do it again using a matrix multiplication.
> > C = np.arange(float(n))
> > # The "virtual" matrix view of C.
> > V = as_strided(C, shape=(C.size-2, 3), strides=(C.strides[0],
> C.strides[0]))
> > C[1:-1] = np.dot(V, rw)
> > C1 = C.copy()
> >
> > # Again, with convolve1d this time.
> > C = np.arange(float(n))
> > C[1:-1] = convolve1d(C, w)[1:-1]
> > C2 = C.copy()
> >
> > # scipy.weave.blitz
> > C = np.arange(float(n))
> > # Must work with a copy, D, in the formula, because blitz does not use
> > # a temporary variable.
> > D = C.copy()
> > expr = "C[1:-1] = a * D[2:] + b * D[1:-1] + c * D[:-2]"
> > blitz(expr, check_size=0)
> > C3 = C.copy()
> >
> >
> > # Verify that all the methods give the same result.
> > print np.all(C0 == C1)
> > print np.all(C0 == C2)
> > print np.all(C0 == C3)
> >
> > #-----
> >
> > And here's a snippet from an ipython session:
> >
> > In [51]: run convolve1dtest.py
> > True
> > True
> > True
> >
> > In [52]: %timeit C[1:-1] = a * C[2:] + b * C[1:-1] + c * C[:-2]
> > 100000 loops, best of 3: 16.5 us per loop
> >
> > In [53]: %timeit C[1:-1] = np.dot(V, rw)
> > 100000 loops, best of 3: 9.84 us per loop
> >
> > In [54]: %timeit C[1:-1] = convolve1d(C, w)[1:-1]
> > 100000 loops, best of 3: 18.7 us per loop
> >
> > In [55]: %timeit D = C.copy(); blitz(expr, check_size=0)
> > 100000 loops, best of 3: 4.91 us per loop
> >
> >
> >
> > scipy.weave.blitz is fastest (but note that blitz has already been called
> > once, so the time shown does not include the compilation required in
> > the first call).  You could also try scipy.weave.inline, cython.inline,
> > or np_inline (http://pages.cs.wisc.edu/~johnl/np_inline/).
> >
> > Also note that C[-1:1] = np.dot(V, rw) is faster than either the simple
> > expression or convolve1d.  However, if you also have to set up V inside
> > your inner loop, the speed gain will be lost.  The relative speeds also
> > depend on the size of C.  For large C, the simple expression is faster
> > than the matrix multiplication by V (but blitz is still faster).  In
> > the following, I have changed n to 2500 before running convolve1dtest.py:
> >
> > In [56]: run convolve1dtest.py
> > True
> > True
> > True
> >
> > In [57]: %timeit C[1:-1] = a * C[2:] + b * C[1:-1] + c * C[:-2]
> > 10000 loops, best of 3: 29.5 us per loop
> >
> > In [58]: %timeit C[1:-1] = np.dot(V, rw)
> > 10000 loops, best of 3: 56.4 us per loop
> >
> > In [59]: %timeit C[1:-1] = convolve1d(C, w)[1:-1]
> > 10000 loops, best of 3: 37.3 us per loop
> >
> > In [60]: %timeit D = C.copy(); blitz(expr, check_size=0)
> > 100000 loops, best of 3: 10.3 us per loop
> >
> >
> > blitz wins, the simple numpy expression is a distant second, and now
> > the matrix multiplication is slowest.
> >
> > I hope that helps--I know I learned quite a bit. :)
>
> Interesting, two questions
>
> does scipy.signal convolve have a similar overhead as np.convolve1d ?
>

Did you mean np.convolve?  There is no np.convolve1d.   Some of the tests
that I've done with convolution functions are here:
    http://www.scipy.org/Cookbook/ApplyFIRFilter
I should add np.convolve to that page.  For the case considered here,
np.convolve is a bit slower than scipy.ndimage.convolve1d, but for larger
arrays, it looks like np.convolve can be much faster.

> memory:
> the blitz code doesn't include the array copy (D), so the timing might
> be a bit misleading?
>

Look again at my %timeit calls in the ipython snippets.  :)

> I assume the as_strided call doesn't allocate any memory yet, so the
> timing should be correct. (or is this your comment about setting up V
> in the inner loop)
>
>

Yes, that's what I meant; if V has to be created inside the inner loop (so
as_strided is called in the loop), the time it takes to create V eliminates
the benefit of using the matrix approach.

Warren

> Josef
>
> >
> > Warren
> >
> >
> >>
> >> Josef
> >>
> >>
> >> >
> >> > Thanks,
> >> > G
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