[Numpy-discussion] problem with vectorized difference equation
francesco82
francesco.barale@gmail....
Fri Apr 6 13:44:41 CDT 2012
Hello everyone,
After reading the very good post
http://technicaldiscovery.blogspot.com/2011/06/speeding-up-python-numpy-cython-and.html
and the book by H. P. Langtangen 'Python scripting for computational
science' I was trying to speed up the execution of a loop on numpy arrays
being used to describe a simple difference equation.
The actual code I am working on involves some more complicated equations,
but I am having the same exact behavior as described below. To test the
improvement in speed I wrote the following in vect_try.py:
#!/usr/bin/python
import numpy as np
import matplotlib.pyplot as plt
dt = 0.02 #time step
time = np.arange(0,2,dt) #time array
u = np.sin(2*np.pi*time) #input signal array
def vect_int(u,y): #vectorized function
n = u.size
y[1:n] = y[0:n-1] + u[1:n]
return y
def sc_int(u,y): #scalar function
y = y + u
return y
def calc_vect(u, func=vect_int):
out = np.zeros(u.size)
for i in xrange(u.size):
out = func(u,out)
return out
def calc_sc(u, func=sc_int):
out = np.zeros(u.size)
for i in xrange(u.size-1):
out[i+1] = sc_int(u[i+1],out[i])
return out
To verify the execution time I've used the timeit function in Ipython:
import vect_try as vt
timeit vt.calc_vect(vt.u) --> 1000 loops, best of 3: 494 us per loop
timeit vt.calc_sc(vt.u) -->10000 loops, best of 3: 92.8 us per loop
As you can see the scalar implementation looping one item at the time
(calc_sc) is 494/92.8~5.3 times faster than the vectorized one (calc_vect).
My problem consists in the fact that I need to iterate the execution of
calc_vect in order for it to operate on all the elements of the input array.
If I execute calc_vect only once, it will only operate on the first slice of
the vectors leaving the remaining untouched. My understanding was that the
vector expression y[1:n] = y[0:n-1] + u[1:n] was supposed to iterate over
all the array, but that's not happening for me. Can anyone tell me what I am
doing wrong?
Thanks!
Francesco
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