Pandas Dataframe: Replacing NaN with row average

Question

I am trying to learn pandas but I have been puzzled with the following. I want to replace NaNs in a DataFrame with the row average. Hence something like df.fillna(df.mean(axis=1)) should work but for some reason it fails for me. Am I missing anything, is there something wrong with what I'm doing? Is it because its not implemented? see link here

import pandas as pd
import numpy as np
​
pd.__version__
Out[44]:
'0.15.2'

In [45]:
df = pd.DataFrame()
df['c1'] = [1, 2, 3]
df['c2'] = [4, 5, 6]
df['c3'] = [7, np.nan, 9]
df

Out[45]:
    c1  c2  c3
0   1   4   7
1   2   5   NaN
2   3   6   9

In [46]:  
df.fillna(df.mean(axis=1)) 

Out[46]:
    c1  c2  c3
0   1   4   7
1   2   5   NaN
2   3   6   9

However something like this looks to work fine

df.fillna(df.mean(axis=0)) 

Out[47]:
    c1  c2  c3
0   1   4   7
1   2   5   8
2   3   6   9

Answer 1

As commented the axis argument to fillna is NotImplemented.

df.fillna(df.mean(axis=1), axis=1) 

Note: this would be critical here as you don't want to fill in your nth columns with the nth row average.

For now you'll need to iterate through:

m = df.mean(axis=1) for i, col in enumerate(df):     # using i allows for duplicate columns     # inplace *may* not always work here, so IMO the next line is preferred     # df.iloc[:, i].fillna(m, inplace=True)     df.iloc[:, i] = df.iloc[:, i].fillna(m)  print(df)     c1  c2   c3 0   1   4  7.0 1   2   5  3.5 2   3   6  9.0 

An alternative is to fillna the transpose and then transpose, which may be more efficient...

df.T.fillna(df.mean(axis=1)).T 

Answer 2

As an alternative, you could also use an apply with a lambda expression like this:

df.apply(lambda row: row.fillna(row.mean()), axis=1)

yielding also

    c1   c2   c3
0  1.0  4.0  7.0
1  2.0  5.0  3.5
2  3.0  6.0  9.0