问题
I am trying to learn pandas but i have been puzzled with the following please. I want to replace NaNs is 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 please, something I'm doing wrong? Is is 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
回答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:
In [11]: 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)
In [12]: df
Out[12]:
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
回答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
回答3:
I'll propose an alternative that involves casting into numpy arrays. Performance wise, I think this is more efficient and probably scales better than the other proposed solutions so far.
The idea being to use an indicator matrix (df.isna().values which is 1 if the element is N/A, 0 otherwise) and broadcast-multiplying that to the row averages.
Thus, we end up with a matrix (exactly the same shape as the original df), which contains the row-average value if the original element was N/A, and 0 otherwise.
We add this matrix to the original df, making sure to fillna with 0 so that, in effect, we have filled the N/A's with the respective row averages.
# setup code
df = pd.DataFrame()
df['c1'] = [1, 2, 3]
df['c2'] = [4, 5, 6]
df['c3'] = [7, np.nan, 9]
# fillna row-wise
row_avgs = df.mean(axis=1).values.reshape(-1,1)
df = df.fillna(0) + df.isna().values * row_avgs
df
giving
c1 c2 c3
0 1.0 4.0 7.0
1 2.0 5.0 3.5
2 3.0 6.0 9.0
来源：https://stackoverflow.com/questions/33058590/pandas-dataframe-replacing-nan-with-row-average