问题描述
我有一个包含多个值的时间序列 A.我需要获得一系列以代数方式定义的 B 如下:
I have a time-series A holding several values. I need to obtain a series B that is defined algebraically as follows:
B[t] = a * A[t] + b * B[t-1]
我们可以假设 B[0] = 0，并且 a 和 b 是实数.
where we can assume B[0] = 0, and a and b are real numbers.
有没有办法在 Pandas 中进行这种递归计算?或者我别无选择，只能按照 这个答案?
Is there any way to do this type of recursive computation in Pandas? Or do I have no choice but to loop in Python as suggested in this answer?
以输入为例:
> A = pd.Series(np.random.randn(10,))
0 -0.310354
1 -0.739515
2 -0.065390
3 0.214966
4 -0.605490
5 1.293448
6 -3.068725
7 -0.208818
8 0.930881
9 1.669210
推荐答案
正如我在评论中指出的，您可以使用 scipy.signal.lfilter.在这种情况下(假设 A 是一维 numpy 数组)，您只需要:
As I noted in a comment, you can use scipy.signal.lfilter. In this case (assuming A is a one-dimensional numpy array), all you need is:
B = lfilter([a], [1.0, -b], A)
这是一个完整的脚本:
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))
脚本的输出为:
[ -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 Pandas DataFrame 而不是 numpy 数组:
Another example, in IPython, using a pandas DataFrame instead of a numpy array:
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
并且您想创建一个新列 B，使得 B[k] = A[k] + 2*B[k-1](与B[k] == 0 for k
and you want to create a new column, B, such that B[k] = A[k] + 2*B[k-1] (with B[k] == 0 for k < 0), you can write
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
这篇关于Pandas 中的递归定义的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！