Re: [Numpy-discussion] speeding up the following expression
Hi Warren, On Sat, Nov 12, 2011 at 9:31 AM, Warren Weckesser wrote: > > > On Sat, Nov 12, 2011 at 6:43 AM, wrote: >> >> On Sat, Nov 12, 2011 at 3:36 AM, Geoffrey Zhu 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] > 10 loops, best of 3: 16.5 us per loop > > In [53]: %timeit C[1:-1] = np.dot(V, rw) > 10 loops, best of 3: 9.84 us per loop > > In [54]: %timeit C[1:-1] = convolve1d(C, w)[1:-1] > 10 loops, best of 3: 18.7 us per loop > > In [55]: %timeit D = C.copy(); blitz(expr, check_size=0) > 10 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] > 1 loops, best of 3: 29.5 us per loop > > In [58]: %timeit C[1:-1] = np.dot(V, rw) > 1 loops, best of 3: 56.4 us per loop > > In [59]: %timeit C[1:-1] = convolve1d(C, w)[1:-1] > 1 loops, best of 3: 37.3 us per loop > > In [60]: %timeit D = C.copy(); blitz(expr, check_size=0) > 10 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. :) > > > Warren Thanks for the very helpful response. Fortunately I don't have to set up the matrix every time as I can work on the same C over and over again inside the loop. It's counterintuitive to see that the performance of dot() degrades as n goes up. If I didn'
Re: [Numpy-discussion] speeding up the following expression
On Sat, Nov 12, 2011 at 11:16 AM, wrote: > On Sat, Nov 12, 2011 at 11:32 AM, Warren Weckesser > wrote: > > > > > > On Sat, Nov 12, 2011 at 9:59 AM, wrote: > >> > >> On Sat, Nov 12, 2011 at 10:31 AM, Warren Weckesser > >> wrote: > >> > > >> > > >> > On Sat, Nov 12, 2011 at 6:43 AM, wrote: > >> >> > >> >> On Sat, Nov 12, 2011 at 3:36 AM, Geoffrey Zhu > >> >> 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] > >> > 10 loops, best of 3: 16.5 us per loop > >> > > >> > In [53]: %timeit C[1:-1] = np.dot(V, rw) > >> > 10 loops, best of 3: 9.84 us per loop > >> > > >> > In [54]: %timeit C[1:-1] = convolve1d(C, w)[1:-1] > >> > 10 loops, best of 3: 18.7 us per loop > >> > > >> > In [55]: %timeit D = C.copy(); blitz(expr, check_size=0) > >> > 10 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
Re: [Numpy-discussion] speeding up the following expression
On Sat, Nov 12, 2011 at 11:32 AM, Warren Weckesser wrote: > > > On Sat, Nov 12, 2011 at 9:59 AM, wrote: >> >> On Sat, Nov 12, 2011 at 10:31 AM, Warren Weckesser >> wrote: >> > >> > >> > On Sat, Nov 12, 2011 at 6:43 AM, wrote: >> >> >> >> On Sat, Nov 12, 2011 at 3:36 AM, Geoffrey Zhu >> >> 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] >> > 10 loops, best of 3: 16.5 us per loop >> > >> > In [53]: %timeit C[1:-1] = np.dot(V, rw) >> > 10 loops, best of 3: 9.84 us per loop >> > >> > In [54]: %timeit C[1:-1] = convolve1d(C, w)[1:-1] >> > 10 loops, best of 3: 18.7 us per loop >> > >> > In [55]: %timeit D = C.copy(); blitz(expr, check_size=0) >> > 10 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] >> > 1 loops, best of 3: 29.5 us per loop >> > >> > In [58]: %timeit C[1:-1] = np.dot(V, rw) >> > 1 loops, best of 3: 56.4 us per loop >> > >> > In [59]: %timeit C[1:-1] = convolve1d(C, w)[1:-1] >> > 1 loops, best
Re: [Numpy-discussion] speeding up the following expression
On Sat, Nov 12, 2011 at 9:59 AM, wrote: > On Sat, Nov 12, 2011 at 10:31 AM, Warren Weckesser > wrote: > > > > > > On Sat, Nov 12, 2011 at 6:43 AM, wrote: > >> > >> On Sat, Nov 12, 2011 at 3:36 AM, Geoffrey Zhu > 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] > > 10 loops, best of 3: 16.5 us per loop > > > > In [53]: %timeit C[1:-1] = np.dot(V, rw) > > 10 loops, best of 3: 9.84 us per loop > > > > In [54]: %timeit C[1:-1] = convolve1d(C, w)[1:-1] > > 10 loops, best of 3: 18.7 us per loop > > > > In [55]: %timeit D = C.copy(); blitz(expr, check_size=0) > > 10 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] > > 1 loops, best of 3: 29.5 us per loop > > > > In [58]: %timeit C[1:-1] = np.dot(V, rw) > > 1 loops, best of 3: 56.4 us per loop > > > > In [59]: %timeit C[1:-1] = convolve1d(C, w)[1:-1] > > 1 loops, best of 3: 37.3 us per loop > > > > In [60]: %timeit D = C.copy(); blitz(expr, check_size=0) > > 10 loops, best of 3: 10.3 us per loop > > > > > > blitz wins, the simple numpy expression is a distant second, and now > > the matrix multipli
Re: [Numpy-discussion] speeding up the following expression
On Sat, Nov 12, 2011 at 10:31 AM, Warren Weckesser wrote: > > > On Sat, Nov 12, 2011 at 6:43 AM, wrote: >> >> On Sat, Nov 12, 2011 at 3:36 AM, Geoffrey Zhu 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] > 10 loops, best of 3: 16.5 us per loop > > In [53]: %timeit C[1:-1] = np.dot(V, rw) > 10 loops, best of 3: 9.84 us per loop > > In [54]: %timeit C[1:-1] = convolve1d(C, w)[1:-1] > 10 loops, best of 3: 18.7 us per loop > > In [55]: %timeit D = C.copy(); blitz(expr, check_size=0) > 10 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] > 1 loops, best of 3: 29.5 us per loop > > In [58]: %timeit C[1:-1] = np.dot(V, rw) > 1 loops, best of 3: 56.4 us per loop > > In [59]: %timeit C[1:-1] = convolve1d(C, w)[1:-1] > 1 loops, best of 3: 37.3 us per loop > > In [60]: %timeit D = C.copy(); blitz(expr, check_size=0) > 10 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 s
Re: [Numpy-discussion] speeding up the following expression
On Sat, Nov 12, 2011 at 6:43 AM, wrote: > On Sat, Nov 12, 2011 at 3:36 AM, Geoffrey Zhu 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] 10 loops, best of 3: 16.5 us per loop In [53]: %timeit C[1:-1] = np.dot(V, rw) 10 loops, best of 3: 9.84 us per loop In [54]: %timeit C[1:-1] = convolve1d(C, w)[1:-1] 10 loops, best of 3: 18.7 us per loop In [55]: %timeit D = C.copy(); blitz(expr, check_size=0) 10 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] 1 loops, best of 3: 29.5 us per loop In [58]: %timeit C[1:-1] = np.dot(V, rw) 1 loops, best of 3: 56.4 us per loop In [59]: %timeit C[1:-1] = convolve1d(C, w)[1:-1] 1 loops, best of 3: 37.3 us per loop In [60]: %timeit D = C.copy(); blitz(expr, check_size=0) 10 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. :) 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
Re: [Numpy-discussion] speeding up the following expression
On Sat, Nov 12, 2011 at 3:36 AM, Geoffrey Zhu 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. 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
