正如我在评论指出的那样,你可以使用scipy.signal.lfilter
。在这种情况下(假设A
是一维numpy的阵列),所有你需要的是:
B = lfilter([a], [1.0, -b], A)
下面是一个完整的脚本:
import numpy as np
from scipy.signal import lfilter
np.random.seed(123)
A = np.random.randn(10)
a = 2.0
b = 3.0
# Compute the recursion using lfilter.
# [a] and [1, -b] are the coefficients of the numerator and
# denominator, resp., of the filter's transfer function.
B = lfilter([a], [1, -b], A)
print B
# Compare to a simple loop.
B2 = np.empty(len(A))
for k in range(0, len(B2)):
if k == 0:
B2[k] = a*A[k]
else:
B2[k] = a*A[k] + b*B2[k-1]
print B2
print "max difference:", np.max(np.abs(B2 - B))
脚本的输出是:
[ -2.17126121e+00 -4.51909273e+00 -1.29913212e+01 -4.19865530e+01
-1.27116859e+02 -3.78047705e+02 -1.13899647e+03 -3.41784725e+03
-1.02510099e+04 -3.07547631e+04]
[ -2.17126121e+00 -4.51909273e+00 -1.29913212e+01 -4.19865530e+01
-1.27116859e+02 -3.78047705e+02 -1.13899647e+03 -3.41784725e+03
-1.02510099e+04 -3.07547631e+04]
max difference: 0.0
另一个例子,在IPython中,使用熊猫DataFrame而不是numpy数组:
如果你有
In [12]: df = pd.DataFrame([1, 7, 9, 5], columns=['A'])
In [13]: df
Out[13]:
A
0 1
1 7
2 9
3 5
,你要创建一个新列,B
,使得B[k] = A[k] + 2*B[k-1]
(与B[k] == 0
对于k < 0),你可以写
In [14]: df['B'] = lfilter([1], [1, -2], df['A'].astype(float))
In [15]: df
Out[15]:
A B
0 1 1
1 7 9
2 9 27
3 5 59
这里的开放问题以cythonize它:https://github.com/pydata/pandas/issues/4567,但一些链接也是他们的 – Jeff 2014-10-08 23:26:24
你可以使用'scipy.signal.lfilter'。有关示例,请参阅http://stackoverflow.com/questions/21336794/python-recursive-vectorization-with-timeseries。 – 2014-10-08 23:49:25