Can I use numpy to speed this loop?

Question

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.

What could I do to speed up the loop?

import numpy as np

N       = 1000
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

In response to Warren, I updated my code:

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

N       = 1000
AR_part = np.random.randn(N+1)

beta = 1.3

def method1( AR_part):
    s2      = np.empty_like(AR_part)
    s2[0]   = 1.0
    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
    return s2

def method2( AR_part):
    y = np.empty_like(AR_part)
    b = np.array([0, 1])
    a = np.array([1, -beta])

    # Initial condition for the linear filter.
    zi = lfiltic(b, a, [1.0], AR_part[:1])

    y[:1] = 1.0
    y[1:], zo = lfilter(b, a, AR_part[1:], zi=zi)

    return y    

s2 = method1( AR_part )
y = method2( AR_part )
np.alltrue( s2==y )

Timing the code:

%timeit method1( AR_part )
100 loops, best of 3: 1.63 ms per loop
%timeit method2( AR_part )
10000 loops, best of 3: 129 us per loop

That shows that Warren's method is over 10 times faster! Very impressive!

Community · Accepted Answer · 2017-05-23 12:29:55Z

5

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.          0.2143694   1.27602566  1.94181186  1.0180607 ]
y  = [ 1.          0.2143694   1.27602566  1.94181186  1.0180607 ]

edited May 23, 2017 at 12:29

CommunityBot

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answered Jan 27, 2014 at 21:53

Warren Weckesser

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1 Comment

Ginger Over a year ago

It's over 10 times faster! Very impressive!

GHL · Accepted Answer · 2014-01-27 21:40:00Z

3

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]

answered Jan 27, 2014 at 21:40

GHL

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Comments

colcarroll · Accepted Answer · 2014-01-27 21:51:56Z

0

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 = 1.3


def seq_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)])

answered Jan 27, 2014 at 21:51

colcarroll

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