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