[Numpy-discussion] speeding up the following expression
Warren Weckesser
warren.weckesser at enthought.com
Sat Nov 12 10:31:17 EST 2011
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. :)
Warren
> Josef
>
>
> >
> > Thanks,
> > G
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