Pandas 中的递归定义 [英] Recursive definitions in Pandas

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本文介绍了Pandas 中的递归定义的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

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 ab 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

<小时>

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 <0),你可以写

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IT屋!

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