Can I Use Numpy To Speed This Loop?

Good evening, I am trying to speed up the loop in this code. I have read through the numpy docs but to no avail. np.accumulate looks like it is almost what I need, but not quite.

Solution 1:

Your recurrence relation is linear, so it can be viewed as a linear filter. You can use scipy.signal.lfilter to compute s2. I recently answered a similar question here: python recursive vectorization with timeseries

Here's a script that shows how to use lfilter to compute your series:

import numpy as np
from scipy.signal import lfilter, lfiltic


np.random.seed(123)

N       = 4
AR_part = np.random.randn(N+1)
s2      = np.ndarray(N+1)
s2[0]   = 1.0

beta = 1.3

old_s2  = s2[0]
for t in range( 1, N+1 ):               
    s2_t    = AR_part[ t-1 ] + beta * old_s2
    s2[t]   = s2_t        
    old_s2  = s2_t


# Compute the result using scipy.signal.lfilter.# Transfer function coefficients.# `b` is the numerator, `a` is the denominator.
b = np.array([0, 1])
a = np.array([1, -beta])

# Initial condition for the linear filter.
zi = lfiltic(b, a, s2[:1], AR_part[:1])

# Apply lfilter to AR_part.
y = np.empty_like(AR_part)
y[:1] = s2[:1]y[1:], zo = lfilter(b, a, AR_part[1:], zi=zi)# Compare the results
print "s2 =", s2
print "y  =", y

Output:

s2 = [ 1.          0.21436941.276025661.941811861.0180607 ]
y  = [ 1.          0.21436941.276025661.941811861.0180607 ]

Solution 2:

I am not sure there is much to do to speed up the loop... The only way I see would be to avoid the recursion, ie compute s2[t] "directly" for each t. But this is expensive as well...

You have

s2[t] = AR_part[t-1] + beta * s2[t-1]
= AR_part[t-1] + beta * (AR_part[t-2] + beta * s2[t-2])
= AR_part[t-1] + beta * AR_part[t-2] + beta^2 * s2[t-2]
= np.dot( AR[:t-1], beta_powers[-(t-1):]  )

Where beta_powers contains [beta^1000, beta^999, ... 1.0]. You can create beta_powers this way:

np.power(beta, np,arange(1000))[::-1].

But I can't see a way to compute that stuff faster than what your loop does...

However you can rewrite it:

for t in range(N):
    s2[t+1] = AR_part[t] + beta * s2[t]

Solution 3:

I agree with GHL that you won't get much more performance (though if N was really big, and you were only computing some parts of the vector s2, definitely use his method), but here's a different way to do what you're looking at:

import numpy as np

N       = 1000
AR_part = np.random.randn(N+1)
beta = 1.3defseq_gen(beta, constants, first_element = 1.0):
    next_element = first_element
    yield next_element
    for j in constants:
        next_element = j + beta * next_element
        yield next_element

s2 = np.array([j for j in seq_gen(beta, AR_part, 1.0)])