Interpolate NaN values over a DataFrame as a ring

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I need to interpolate the NaN values over a Dataframe but I want that interpolation to get the first values of the DataFrame in case the NaN value is the last value. Here is an example:

import pandas as pd
import numpy as np
df = pd.DataFrame.from_dict({"a": [1,2,3], "b":[1,2,np.nan]})

So the DataFrame is:

   a    b
0  1  1.0
1  2  2.0
2  3  NaN

But when I interpolate the nan values like:

df.interpolate(method="linear", inplace=True)

I got:

   a    b
0  1  1.0
1  2  2.0
2  3  2.0

The interpolation doesn't use the first value to do it. My desired output wold be to fill in with the value of 1.5 because of that circular interpolation.

pandas numpy python
1 Answers

One possible solution is add first row, interpolate and remove last row:

df = df.append(df.iloc[0]).interpolate(method="linear").iloc[:-1]
print (df)
     a    b
0  1.0  1.0
1  2.0  2.0
2  3.0  1.5

EDIT:

More general solution:

df = pd.DataFrame.from_dict({"a": [1,2,3,4], "b":[np.nan,1,2,np.nan]})

df = pd.concat([df] * 3).interpolate(method="linear").iloc[len(df):-len(df)]

print (df)
   a         b
0  1  1.333333
1  2  1.000000
2  3  2.000000
3  4  1.666667

Or if need working only with last non missing values:

df = pd.DataFrame.from_dict({"a": [1,2,3,4], "b":[np.nan,1,2,np.nan]})

df1 = df.ffill().iloc[[-1]]
df2 = df.bfill().iloc[[0]]
df = pd.concat([df1, df, df2]).interpolate(method="linear").iloc[1:-1]

print (df)
   a    b
0  1  1.5
1  2  1.0
2  3  2.0
3  4  1.5
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