On Sat, Nov 12, 2011 at 11:16 AM, <josef.p...@gmail.com> wrote: > On Sat, Nov 12, 2011 at 11:32 AM, Warren Weckesser > <warren.weckes...@enthought.com> wrote: > > > > > > 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. :) > > > > Sorry, I was reading way too fast or selectively (skipping the > imports, namespaces for example). > (I thought convolve1d was numpy not ndimage) > > I tried out your cookbook script a while ago, it's nice, I had only a > rough idea about the timing before. > Compared to the current case the x is much longer, (unless I don't > remember or read it correctly.) >
I just updated the page with np.convolve: http://www.scipy.org/Cookbook/ApplyFIRFilter A big part of the overhead in using np.convolve in that script is that the function only accepts 1-d arrays, so a python loop is required to apply it along an axis of a 2-d array. Warren
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