我可以使用 numpy 来加速这个循环吗? [英] Can I use numpy to speed this loop?
问题描述
晚上好,
我正在尝试加速此代码中的循环.我已经通读了 numpy 文档,但无济于事.np.accumulate 看起来几乎是我需要的,但不完全是.
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.
我可以做些什么来加速循环?
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
为了回应 Warren,我更新了我的代码:
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 )
定时代码:
%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
这表明沃伦的方法快了 10 倍以上!非常令人印象深刻!
That shows that Warren's method is over 10 times faster! Very impressive!
推荐答案
您的递推关系是线性的,因此可以将其视为线性过滤器.您可以使用 scipy.signal.lfilter
来计算 s2
.我最近在这里回答了一个类似的问题:python recursive vectorization with timeseries
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
这是一个脚本,展示了如何使用 lfilter
来计算您的系列:
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
输出:
s2 = [ 1. 0.2143694 1.27602566 1.94181186 1.0180607 ]
y = [ 1. 0.2143694 1.27602566 1.94181186 1.0180607 ]
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