On Sat, Nov 12, 2011 at 9:59 AM, <josef.p...@gmail.com> wrote: > On Sat, Nov 12, 2011 at 10:31 AM, Warren Weckesser > <warren.weckes...@enthought.com> wrote: > > > > > > On Sat, Nov 12, 2011 at 6:43 AM, <josef.p...@gmail.com> wrote: > >> > >> On Sat, Nov 12, 2011 at 3:36 AM, Geoffrey Zhu <zyzhu2...@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 > >> > _______________________________________________ > >> > NumPy-Discussion mailing list > >> > NumPy-Discussion@scipy.org > >> > http://mail.scipy.org/mailman/listinfo/numpy-discussion > >> > > >> _______________________________________________ > >> NumPy-Discussion mailing list > >> NumPy-Discussion@scipy.org > >> http://mail.scipy.org/mailman/listinfo/numpy-discussion > > > > > > _______________________________________________ > > NumPy-Discussion mailing list > > NumPy-Discussion@scipy.org > > http://mail.scipy.org/mailman/listinfo/numpy-discussion > > > > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion >
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