根据列的值在DataFrame中填充NaN

Question

I have data that resembles the following simplified example:

Col1    Col2    Col3
a       A       10.1
b       A       NaN
d       B       NaN
e       B       12.3    
f       B       NaN
g       C       14.1
h       C       NaN
i       C       NaN

...for many thousands of rows. I need to fillna based upon the value in Col2, using something analogous to the ffill method. The result I'm looking for is this:

Col1    Col2    Col3
a       A       10.1
b       A       10.1
d       B       NaN
e       B       12.3    
f       B       12.3
g       C       14.1
h       C       14.1
i       C       14.1

However, this method ignores the value in Col2. Any ideas?

Answer 1

If I understand correctly then you can groupby on 'Col2' and then call transform on 'Col3' and call ffill :

In [35]:

df['Col3'] = df.groupby('Col2')['Col3'].transform(lambda x: x.ffill())
df
Out[35]:
  Col1 Col2  Col3
0    a    A  10.1
1    b    A  10.1
2    d    B   NaN
3    e    B  12.3
4    f    B  12.3
5    g    C  14.1
6    h    C  14.1
7    i    C  14.1

Answer 2

One answer I found is the following:

df['col3'] = df.groupby('Col2').transform('fillna',method='ffill')['col3']

Any thoughts?

Answer 3

Is this what you're looking for?

import pandas as pd
import numpy as np


df['Col3'] = np.where(df['Col2'] == 'A', df['Col3'].fillna(10.1), df["Col3"])

Of course replace accordingly.

Answer 4

You can take slices of the DataFrame for each element of Col2 , and then concatenate the results.

>>> pd.concat((df.loc[df.Col2 == letter, :].ffill() for letter in df.Col2.unique()))

  Col1 Col2  Col3
0    a    A  10.1
1    b    A  10.1
2    d    B   NaN
3    e    B  12.3
4    f    B  12.3
5    g    C  14.1
6    h    C  14.1
7    i    C  14.1

EDIT: It appears the method presented by @EdChum is the fastest by far.

%timeit pd.concat((df.loc[df.Col2 == letter, :].ffill() for letter in df.Col2.unique()))
100 loops, best of 3: 3.57 ms per loop

%timeit df.groupby('Col2').transform('fillna',method='ffill')['Col3']
100 loops, best of 3: 4.59 ms per loop

%timeit df.groupby('Col2')['Col3'].transform(lambda x: x.ffill())
1000 loops, best of 3: 746 µs per loop