How to apply cubic spline interpolation over long Pandas Series?

Question

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:

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?

Answer 1

pandas calls out to the scipy interpolation routines, I'm not sure why 'cubic' is so memory hungry and slow.

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.

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