Interpolating time series in Pandas using Cubic spline

Question

I would like to fill gaps in a column in my DataFrame using a cubic spline. If I were to export to a list then I could use the numpy\'s interp1d function and ap

Answer 1

Most numpy/scipy function require the arguments only to be "array_like", iterp1d is no exception. Fortunately both Series and DataFrame are "array_like" so we don't need to leave pandas:

import pandas as pd
import numpy as np
from scipy.interpolate import interp1d

df = pd.DataFrame([np.arange(1, 6), [1, 8, 27, np.nan, 125]]).T

In [5]: df
Out[5]: 
   0    1
0  1    1
1  2    8
2  3   27
3  4  NaN
4  5  125

df2 = df.dropna() # interpolate on the non nan
f = interp1d(df2[0], df2[1], kind='cubic')
#f(4) == array(63.9999999999992)

df[1] = df[0].apply(f)

In [10]: df
Out[10]: 
   0    1
0  1    1
1  2    8
2  3   27
3  4   64
4  5  125

Note: I couldn't think of an example off the top of my head to pass in a DataFrame into the second argument (y)... but this ought to work too.