如何在长 Pandas 系列上应用三次样条插值？
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How to apply cubic spline interpolation over long Pandas Series?
提问by Crolle
I need to replace missing data within pandas Series using cubic spline interpolation. I figured out that I could use the pandas.Series.interpolate(method='cubic')method, which looks like this:
我需要使用三次样条插值替换Pandas系列中的缺失数据。我发现我可以使用这个pandas.Series.interpolate(method='cubic')方法，它看起来像这样：
import numpy as np
import pandas as pd
# create series
size = 50
x = np.linspace(-2, 5, size)
y = pd.Series(np.sin(x))
# deleting data segment
y[10:30] = np.nan
# interpolation
y = y.interpolate(method='cubic')
Although this method works just fine for small series (size = 50), it seems to cause the program to freeze for larger ones (size = 5000). Is there a workaround?
虽然这种方法对小系列（size = 50）效果很好，但对于大系列（）似乎会导致程序冻结size = 5000。有解决方法吗？
回答by chrisb
pandascalls out to the scipyinterpolation routines, I'm not sure why 'cubic'is so memory hungry and slow.
pandas调用scipy插值例程，我不知道为什么'cubic'内存如此饥饿和缓慢。
As a workaround, you could use method='spline'(scipy ref here), which with the right parameters, gives essentially (seems to be some floating point differences?) the same results and is dramatically faster.
作为一种解决方法，您可以使用method='spline'(scipy ref here)，它具有正确的参数，基本上可以提供（似乎是一些浮点差异？）相同的结果并且速度明显更快。
In [104]: # create series
...: size = 2000
...: x = np.linspace(-2, 5, size)
...: y = pd.Series(np.sin(x))
...:
...: # deleting data segment
...: y[10:30] = np.nan
...:
In [105]: %time cubic = y.interpolate(method='cubic')
Wall time: 4.94 s
In [106]: %time spline = y.interpolate(method='spline', order=3, s=0.)
Wall time: 1 ms
In [107]: (cubic == spline).all()
Out[107]: False
In [108]: pd.concat([cubic, spline], axis=1).loc[5:35, :]
Out[108]:
0 1
5 -0.916444 -0.916444
6 -0.917840 -0.917840
7 -0.919224 -0.919224
8 -0.920597 -0.920597
9 -0.921959 -0.921959
10 -0.923309 -0.923309
11 -0.924649 -0.924649
12 -0.925976 -0.925976
13 -0.927293 -0.927293
